Wikibuku idwikibooks https://id.wikibooks.org/wiki/Halaman_Utama MediaWiki 1.47.0-wmf.10 first-letter Media Istimewa Pembicaraan Pengguna Pembicaraan Pengguna Wikibuku Pembicaraan Wikibuku Berkas Pembicaraan Berkas MediaWiki Pembicaraan MediaWiki Templat Pembicaraan Templat Bantuan Pembicaraan Bantuan Kategori Pembicaraan Kategori Resep Pembicaraan Resep Wisata Pembicaraan Wisata TimedText TimedText talk Modul Pembicaraan Modul Acara Pembicaraan Acara 0 1761 117488 113031 2026-07-13T02:06:01Z Exploraz 43531 Membatalkan revisi [[Special:Diff/45766|45766]] oleh [[Special:Contributions/114.79.37.122|114.79.37.122]] ([[User talk:114.79.37.122|bicara]]) 117488 wikitext text/x-wiki {{DISPLAYTITLE:<span style="display:block;text-align:center;font-size:200%;font-style:bold;background: #E5FFFF;line-height:1em;-moz-border-radius: 15px; -webkit-border-radius: 15px; border-radius: 15px; {{gradient|#07867F|Turquoise|vertical}}">Rahasia Otak Super</span>}} <div style="border:0; -moz-box-shadow: 0 1px 3px rgba(0, 0, 0, 0.35); -webkit-box-shadow: 0 1px 3px rgba(0, 0, 0, 0.35); box-shadow: 0 1px 3px rgba(0, 0, 0, 0.35); -moz-border-radius: 7px; -webkit-border-radius: 7px; border-radius: 7px; background: #fff; background: -moz-linear-gradient(top, #fff 75%, #F5F5F5 100%); background: -webkit-gradient(linear, left top, left bottom, color-stop(75%,#fff), color-stop(100%,#F5F5F5)); background: -webkit-linear-gradient(top, #fff 75%,#F5F5F5 100%); background: -o-linear-gradient(top, #fff 75%,#F5F5F5 100%); background: -ms-linear-gradient(top, #fff 75%,#F5F5F5 100%); background: linear-gradient(top, #fff 75%,#fff 100%); height:auto; padding-left:10px; padding-right:10px; padding-bottom:5px; padding-top:5px; margin:5px 5px 5px 5px; {{{style|}}}"> {{cquote|bgcolor=#F0FFF0|Copyright (c) 2006, 2007 Abd Shomad, Nanang MH, Faiza Yahya and others. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled "GNU Free Documentation License".}} :''Tidak ada istilah lemah otak, yang ada hanyalah otak yang belum terlatih.'' Buku ini dipersembahkan untuk mencerdaskan rakyat Indonesia dengan cara-cara yang sederhana dan menyenangkan, silakan disebarkan. Buku ini dikarang dengan lisensi ''GNU Free Document License''. * Jika Anda ingin mencetak isi buku ini dengan jumlah lebih dari 100 eksemplar, silakan mempelajari isi dokumen [http://www.gnu.org/licenses/fdl.txt GNU Free Document License]. * Jika Anda ingin menyebarkan isi buku ini dalam bentuk media elektronik, jangan lupa mencantumkan salinan verbatim dari lisensi [http://www.gnu.org/licenses/fdl.txt GNU Free Document License] dan mencantumkan alamat universal (''URL'') dari buku ini, yaitu : [http://id.wikibooks.org/wiki/Rahasia_Otak_Super http://id.wikibooks.org/wiki/Rahasia_Otak_Super]. Anda juga bisa langsung menyunting (''edit'') isi buku ini kapan saja Anda mau. Langsung saja klik '''sunting''' atau '''''edit''''' di seluruh bagian dari buku ini. Selamat berkarya. == Daftar Isi == *[[Rahasia Otak Super/Pendahuluan|Pendahuluan]] *[[Rahasia Otak Super/Bukti-Bukti Teknik Mengingat Sederhana|Bukti-Bukti Teknik Mengingat Sederhana]] *[[Rahasia Otak Super/Uji Ingatan Anda|Uji Ingatan Anda]] *[[Rahasia Otak Super/Teknik Menghafal|Teknik Menghafal]] *[[Rahasia Otak Super/Uji Ulang Ingatan Anda|Uji Ulang Ingatan Anda]] *[[Rahasia Otak Super/Cara Cepat Menghafal|Cara Cepat Menghafal]] *[[Rahasia Otak Super/Cara Cepat Menghitung|Cara Cepat Menghitung]] **[[Rahasia Otak Super/Cara Cepat/Menghitung/Rata Rata|Menghitung Rata Rata]] **[[Rahasia Otak Super/Cara Cepat Menghitung/Teknik Vedic Math|Teknik Vedic Math]] *[[Rahasia Otak Super/Mempertajam Ingatan|Mempertajam Ingatan]] *[[Rahasia Otak Super/Sejarah Mnemonic|Sejarah Mnemonic]] *[[Rahasia Otak Super/Testimonial|Testimonial]] *[[Rahasia Otak Super/Senarai|Senarai]] *[[Rahasia Otak Super/Credits|Kredit]] == Alamat Group Diskusi == Untuk pengguna GMail (lebih diutamakan) Group name : Rahasia-Otak-Super Group home page : http://groups.google.com/group/Rahasia-Otak-Super Group email : Rahasia-Otak-Super@googlegroups.com Description : Diskusi tentang bagaimana meningkatkan kemampuan otak Untuk pengguna Yahoo Group name : Otak-Super Group home page : http://groups.yahoo.com/group/Otak-Super Group email : Otak-Super@yahoogroups.com Description : Diskusi tentang bagaimana meningkatkan kemampuan otak Post message : Otak-Super@yahoogroups.com Subscribe : Otak-Super-subscribe@yahoogroups.com Unsubscribe : Otak-Super-unsubscribe@yahoogroups.com List owner : Otak-Super-owner@yahoogroups.com [[Kategori:Rahasia Otak Super]] jny7u9s2rvi9pgwy5zfcwz827o251es 0 8458 117489 117077 2026-07-13T04:01:44Z ~2026-39479-46 43532 117489 wikitext text/x-wiki ===Power Berry 1=== Setelah mendapatkan 1001 medali dalam pacuan kuda, kamu dapat menukarkannya dengan power berry pada walikota. ===Power Berry 2=== Jika memancing di laut di musim winter (jauh dari dermaga), kamu akan mendapatkan power berry saat memancing tapi butuh waktu yang lama. ===Power Berry 3=== Pergi ke tambang dekat hot spring dan mulailah menggali dengan cangkulmu untuk mendapatkan power berry. biasanya > 3 lantai ke bawah. ===Power Berry 4=== Pergi ke Winter Mine dan galilah Power Berry di sana. ===Power Berry 5=== Pergi ke Cedar Tree yang tumbuh terpisah dari pohon lainnya di sekitar gunung yang terdapat hamparan bunga. Cobalah tebang pohonnya lalu si pohon akan berkata"aku sudah berakar 100 tahun di sini, apakah kau akan menebangku?", jika kamu menjawab ya, maka gotz akan muncul dan memarahimu dan kau tidak akan dapat power berry. Jika kamu tidak menebangnya ia akan memberimu Power Berry. Jika kamu telah menerima Berry tapi masih tetap juga menebangnya, kamu akan pingsan. ===Power Berry 6=== Setelah mendapatkan semua peralatan yang ditawarkan dari TV Shopping Network, mereka akan menawarkan Power Berry seharga 5000G, kamu harus pergi ke restoran dan menggunakan telepon untuk memesan berry tersebut. Tiga hari kemudian pesananmu akan tiba. ===Power Berry 7=== Pergi ke belakang Winter Mine lalu teruslah tekan tombol X dekat sudut kiri atas tambang. ===Power Berry 8=== Jika kamu menang di Swimming Festival dan menjadi juara pertama, kamu akan mendapatkan power berry karena telah berenang dengan sangat baik. Agar bisa menang, cobalah untuk mencari buku tentang festival renang di perpustakaan lantai 1, dan usahakan sebelum lomba renang, stamina mu masih utuh. ===Power Berry 9=== Bebas Di Musim Apapapun kecuali Winter, jika kamu menanam sedikitnya 100 bunga (12 kantung), Anna akan datang pada siang hari dan ingin memetik beberapa tangkai. Jika kamu mengijinkannya, Anna akan memberikan power berry padamu. ===Power Berry 10=== Power Berry ini didapatkan dengan cara memberikan hasil peternakan/tani kepada dewi panen di air terjun dekat hotspring. berikan 1 hasil tani di pagi hari mulai pukul 8 sampai 12 siang. Lemparkan ke air terjun dengan posisi membelakangi pintu tambang. berikan 5 hasil panen berturut-turut di saat hari cerah. Hari hujan/festival dewi panen tidak akan muncul. ===Secret Power Berry 11=== Pastikan kamu membawa 3 mentimun ke gunung pada saat musim semi, panas, atau gugur. Pergilah ke Mother’s Lake, berdiri di sisi kiri danau di depan 2 pohon. (Kamu harus berdiri di tempat yang tepat agar berhasil). Jika kamu melempar ketiga mentimun tersebut ke kolam, Kappa akan muncul. Syarat jam 11.00 am sampai jam 05.00 pm (namun ada juga player yang mencoba di luar jam tersebut dan berhasil). Jika tidak muncul berarti kamu berdiri di tempat yang salah, geser sedikit dan lempar lagi 3 mentimun. Kappa akan memberimu special power berry. Berry ini akan mengurangi peluang kamu terserang penyakit. Dengan berry ini kamu dapat bekerja keras 2 kali lipat walaupun cuaca sedang buruk. [[Kategori:Harvest Moon, Back to Nature]] 1hg5ejqcwy6e8qidzshoigb9dizfkrc 0 11844 117490 99508 2026-07-13T04:02:53Z ~2026-39479-46 43532 117490 wikitext text/x-wiki '''ikan legendaris dapat anda temukan di gua misterius didekat gunung, namun sebelum itu anda harus memiliki kail pancing level 2 (setelah menerima pancing pertama dari greg/kakek di dermaga, anda harus mempunyai 50 ikan di kolam anda. Dan otomatis si kakek akan memberikan pancing level 2) 1. Squid – Mineral Beach (Summer) Lempar small fish(ikan kecil) ke laut sebelum memancing. (Ingat satu small fish hanya untuk satu hari memancing, jika gagal maka silakan mencoba lagi keesokan harinya dengan melemparkan satu small fish lagi). 2. Char – Waterfall(air terjun) atau sungai (semua musim ) Harus membuat resep Grilled Fish, Sashimi dan Sushi terlebih dahulu, baru bisa memancing ikan jenis ini. 3. Angler – Mineral beach(Winter) Memancinglah di malam hari (di atas 08.00 pm sampai 06.00 am). 4. Catfish – Winter Mine Pond (Winter) Memancinglah di kolam di dasar Winter Mine. Jangan frustasi jika sulit, biasanya membutuhkan waktu lebih dari 30 menit(bukan jam game) 5. Sea Bream - bisa didapatkan di semua musim kecuali summer. Jual 200 ikan atau lebih (jualnya harus lempar pake tangan ga bisa pake keranjang, ukuran ikan bebas, bisa kecil sedang atau besar) , lalu pancing di laut pada musim spring, fall, winter (tidak bisa saat summer) 6. Carp – Mother Hill Lake (semua musim kecuali Winter). Jika pada musim spring/summer/fall lokasi bebas selama masih di area danau, tapi jika anda ingin mencoba di musim winter, mancingnya dari jembatan mau ke arah gunung. Syaratnya harus menangkap kelima ikan legendaris lain terlebih dahulu. [[Kategori:Harvest Moon, Back to Nature]] 9j8hf41d4163gzub3fl3insdr27g2dn 0 23568 117483 117479 2026-07-13T00:48:43Z Akuindo 8654 117483 wikitext text/x-wiki contoh soal <ol start=1> <li>Berapa hasil dari <math>\sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Misalkan 2020 = p} \\ \sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64} &= \sqrt{(2020-5) \cdot (2020-3) \cdot (2020+3) \cdot (2020+5) + 64} \\ &= \sqrt{(p-5) \cdot (p-3) \cdot (p+3) \cdot (p+5) + 64} \\ &= \sqrt{(p-5) \cdot (p+5) \cdot (p-3) \cdot (p+3) + 64} \\ &= \sqrt{(p^2-25) \cdot (p^2-9) + 64} \\ &= \sqrt{p^4-34p^2+ 225 + 64} \\ &= \sqrt{p^4-34p^2+ 289} \\ &= \sqrt{(p^2-17)^2} \\ &= p^2-17 \\ &= 2020^2-17 \\ &= (2000+20)^2-17 \\ &= 4.000.000+80.000+400-17 \\ &= 4.080.383 \\ \end{align} </math> </div></div> <ol start=2> <li>Berapa nilai x dari <math>\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} = \frac{9}{10}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} &= \frac{9}{10} \\ \text{misalkan untuk } \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} = p \\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} &= p \\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}} &= p^2 \\ x^2-x-p &= p^2 \\ x^2-2x+1+x-1 &= p^2+p \\ (x-1)^2+(x-1) &= p^2+p \\ x-1 &= p \\ \text{misalkan untuk } \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}} &= q^3 \\ x^2 q &= q^3 \\ x^2 &= q^2 \\ x &= q \\ \frac{x-1}{x} &= \frac{9}{10} \\ x &= 10 \\ \end{align} </math> </div></div> <ol start=3> <li>Berapa nilai x dari <math>(\frac{x}{x+10})^{x+10}=\frac{1}{1024}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{x+10}{x})^{-(x+10)} &= (1024)^{-1} \\ (\frac{x+10}{x})^{x+10} &= 1024 \\ (\frac{x+10}{x})^{x+10} &= 2^{10} \\ (\frac{x+10}{x})^{\frac{x+10}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (\frac{1}{2})^{-1} \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (1+(-\frac{1}{2}))^{(1+(-\frac{2}{1}))} \\ \frac{10}{x} &= -\frac{1}{2} \\ x &= -20 \\ \end{align} </math> </div></div> <ol start=4> <li>Berapa nilai x dari <math>x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{2}+\sqrt{x+\frac{1}{4}} &= (\sqrt{x+\frac{1}{4}})^2+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+(\frac{1}{2})^2 \\ &= (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 \\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}} &= 4 \\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2} &= 4 \\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 4 \\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 &= 4 \\ \sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 2 \\ \sqrt{x+\frac{1}{4}} &= \frac{3}{2} \\ x+\frac{1}{4} &= \frac{9}{4} \\ x &= 2 \\ \end{align} </math> </div></div> <ol start=5> <li>Berapa nilai x dari <math>\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^3}{\sqrt{8-x^2}}+x^2-8 &= 0 \\ \frac{x^3}{\sqrt{8-x^2}} &= 8-x^2 \\ x^3 &= (8-x^2)^{\frac{3}{2}} \\ x &= (8-x^2)^{\frac{1}{2}} \\ x^2 &= 8-x^2 \\ 2x^2-8 &= 0 \\ x^2-4 &= 0 \\ (x-2)(x+2) &= 0 \\ \text{membuktikan } \\ x=2 \text{ maka hasilnya 0 } \\ x=-2 \text{ maka hasilnya -8 } \\ \text{jadi } x=2 \\ \end{align} </math> </div></div> <ol start=6> <li>Berapa nilai x dari <math>\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} = 5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} &= 5 \\ \frac{x^{50}+x^{60}+x^{70}}{31}} &= 5^5 \\ x^{50}+x^{60}+x^{70} &= 5^5 \cdot 31 \\ x^{50}(1+x^{10}+x^{20}) &= 5^5 \cdot 31 \\ (x^{10}^5)(1+x^{10}+(x^{10}^2) &= 5^5 \cdot 31 \\ \text{ misalkan } x^{10} = a \\ a^5(1+a+a^2) &= 5^5 \cdot 31 \\ a &= 5 \\ x^{10} &= 5 \\ x &= ^5 log 10 \\ \end{align} </math> </div></div> <ol start=7> <li>Berapa nilai x dari <math>\sqrt{3x+5+\sqrt{4x+5}} = x</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{3x+5+\sqrt{4x+5}} &= x \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \text{misalkan } \sqrt{4x+5}=y \text{ dan } 4x+5=y^2 \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \sqrt{y^2+y-x} &= x \\ y^2+y &= x^2+x \\ y=x \\ 4x+5 &= y^2 \\ 4x+5 &= x^2 \\ x^2-4x-5 &= 0 \\ (x-5)(x+1) &= 0 \\ x=5 &\text{ atau } x=-1 \text{ (TM) } \\ \end{align} </math> </div></div> <ol start=8> <li>Berapa nilai x dari <math>\sqrt{1+\sqrt{1+x}} = \sqrt[3]{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{1+\sqrt{1+x}} &= \sqrt[3]{x} \\ \sqrt[3]{x} &= n \\ x &= n^3 \\ \sqrt{1+\sqrt{1+n^3}} &= n \\ 1+\sqrt{1+n^3} &= n^2 \\ \sqrt{1+n^3} &= n^2-1 \\ 1+n^3 &= n^4-2n^2+1 \\ n^4-n^3-2n^2 &= 0 \\ n^2(n^2-n-2) &= 0 \\ n^2(n-2)(n+1) &= 0 \\ n=0, n=2 \text{ atau } n=-1 \\ n &= 0 \\ x &= 0^3 \\ &= 0 \\ n &= 2 \\ x &= 2^3 \\ &= 8 \\ n &= -1 \\ x &= (-1)^3 \\ &= -1 \\ \text{yang paling mungkin untuk nilai x adalah } 8 \\ \end{align} </math> </div></div> <ol start=9> <li>Berapa nilai x dari <math>\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} &= \frac{\sqrt{1+x}}{\sqrt{x}} \\ \sqrt{x}(\sqrt{1+x}+\sqrt{x}) &= (\sqrt{1+x}-\sqrt{x})\sqrt{1+x} \\ \sqrt{x(1+x)}+x &= 1+x-\sqrt{x(1+x)} \\ 2\sqrt{x(1+x)} &= 1 \\ \sqrt{x(1+x)} &= \frac{1}{2} \\ x(1+x) &= \frac{1}{4} \\ x^2+x &= \frac{1}{4} \\ 4x^2+4x &= 1 \\ 4x^2+4x-1 &= 0 \\ x &= \frac{-4 \pm \sqrt{4^2-4(4)(-1)}}{2(4)} \\ &= \frac{-4 \pm \sqrt{32}}{8} \\ &= \frac{-4 \pm 4\sqrt{2}}{8} \\ &= \frac{-1 \pm \sqrt{2}}{2} \\ \text{karena akar x harus minimal nol jadi } x = \frac{-1+\sqrt{2}}{2} \\ \end{align} </math> </div></div> <ol start=10> <li>Berapa nilai x dari <math>\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x-\sqrt{x+1}}{x+\sqrt{x+1}} &= \frac{11}{19} \\ \text{misalkan } \sqrt{x+1}=y \text{ dan } x=y^2-1 \\ \frac{y^2-1-y}{y^2-1+y} &= \frac{11}{19} \\ 19(y^2-y-1) &= 11(y^2+y-1) \\ 19y^2-19y-19 &= 11y^2+11y-11 \\ 8y^2-30y-8 &= 0 \\ 4y^2-15y-4 &= 0 \\ (4y+1)(y-4) &= 0 \\ y=-\frac{1}{4} \text{ (TM) atau } & y=4 \\ x &= 4^2-1 \\ &= 15 \\ \end{align} </math> </div></div> <ol start=11> <li>Berapa nilai x dari <math>\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} &= 98 \\ \text{misalkan } \sqrt{x^2-1}=y \\ \frac{x+y}{x-y}+\frac{x-y}{x+y} &= 98 \\ \frac{(x+y)^2+(x-y)^2}{(x-y)(x+y)} &= 98 \\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} &= 98 \\ \frac{2(x^2+y^2)}{x^2-y^2} &= 98 \\ \frac{x^2+y^2}{x^2-y^2} &= 49 \\ x^2+y^2 &= 49(x^2-y^2) \\ x^2+y^2 &= 49x^2-49y^2 \\ 48x^2 &= 50y^2 \\ 24x^2 &= 25y^2 \\ 24x^2 &= 25(\sqrt{x^2-1})^2 \\ 24x^2 &= 25(x^2-1) \\ 24x^2 &= 25x^2-25 \\ x^2 &= 25 \\ x &= \pm 5 \\ \end{align} </math> </div></div> <ol start=12> <li>Berapa nilai x dari <math>\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \sqrt{x}=y \text{ dan } x=y^2 \\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} &= \frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\sqrt{\frac{y^2}{y^2+y}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\frac{y}{\sqrt{y^2+y}} \\ y^2+y-\sqrt{(y^2+y)(y^2-y)} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^4-y^2} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^2(y^2-1)} &= \frac{5}{4}y \\ y(y+1)-y\sqrt{y^2-1} &= \frac{5}{4}y \\ y+1-\sqrt{y^2-1} &= \frac{5}{4} \\ -\sqrt{y^2-1} &= \frac{1}{4}-y \\ y^2-1 &= (\frac{1}{4}-y)^2 \\ y^2-1 &= \frac{1}{16}-\frac{1}{2}y+y^2 \\ -1 &= \frac{1}{16}-\frac{1}{2}y \\ \frac{1}{2}y &= \frac{1}{16}+1 \\ \frac{1}{2}y &= \frac{17}{16} \\ y &= \frac{17}{8} \\ x &= (\frac{17}{8})^2 \\ &= \frac{289}{64} \\ \end{align} </math> </div></div> <ol start=13> <li>Berapa nilai x dari <math>\sqrt[4]{62+x}+\sqrt[4]{275-x}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ misalkan } \sqrt[4]{62+x}=a, 62+x=a^4, \sqrt[4]{275-x}=b \text{ dan } 275-x=b^4 \\ a+b &= 7 \\ (a+b)^2 &= 49 \\ a^2+b^2+2ab &= 49 \\ a^2+b^2 &= 49-2ab \\ a^4+b^4 &= 62+x+275-x \\ (a^2+b^2)^2-2(ab)^2 &= 337 \\ (49-2ab)^2-2(ab)^2 &= 337 \\ 2401-196ab+4(ab)^2-2(ab)^2 &= 337 \\ 2(ab)^2-196ab+2064 &= 0 \\ (ab)^2-98ab+1032 &= 0 \\ (ab-12)(ab-86) &= 0 \\ ab = 12 \text{ atau } & ab = 86 \text{ (TM) karena hasil kali maksimum yaitu 12 } \\ ab =12 \text{ dan } a+b=7 \\ a+b &= 7 \\ b &= 7-a \\ ab &= 12 \\ a(7-a) &= 12 \\ -a^2+7a &= 12 \\ a^2-7a+12 &= 0 \\ (a-3)(a-4) &= 0 \\ a=3 \text{ atau } & a=4 \\ a=3, b=4 \\ 62+x &= a^4 \\ 62+x &= (3)^4 \\ 62+x &= 81 \\ x &= 19 \\ a=4, b=3 \\ 62+x &= a^4 \\ 62+x &= (4)^4 \\ 62+x &= 256 \\ x &= 194 \\ \end{align} </math> </div></div> <ol start=14> <li>Berapa nilai x dari <math>\sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2} &= 7 \\ (\sqrt[3]{8+x})^2-\sqrt[3]{8+x} \sqrt[3]{27-x}+(\sqrt[3]{27-x})^2 &= 7 \\ \text{misalkan } \sqrt[3]{8+x}=a, 8+x=a^3, \sqrt[3]{27-x}=b \text{ dan } 27-x=b^3 \\ a^2-ab+b^2 &= 7 \\ a^3+b^3 &= 8+x+27-x \\ &= 35 \\ a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ 35 &= (a+b)(7) \\ a+b &= 5 \\ b &= 5-a \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ 5^3 &= 35+3ab(5) \\ 125 &= 35+15ab \\ 80 &= 15ab \\ ab &= 6 \\ a(5-a) &= 6 \\ 5a-a^2 &= 6 \\ a^2-5a+6 &= 6 \\ (a-2)(a-3) &= 6 \\ a=2 &\text{ atau } a=3 \\ a=2, b=3 \text{ dan } a=3,b=2 \\ 8+x &= a^3 \\ &= 2^3 \\ &= 8 \\ x &= 0 \\ 8+x &= a^3 \\ &= 3^3 \\ &= 27 \\ x &= 19 \\ \end{align} </math> </div></div> <ol start=15> <li>Berapa nilai x dari <math>3^x+5^x-9^x+15^x-25^x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3^x+5^x-9^x+15^x-25^x &= 1 \\ 3^x+5^x-(3^2)^x+(3 \cdot 5)^x-(5^2)^x &= 1 \\ 3^x+5^x-(3^x)^2+(3^x \cdot 5^x)-(5^x)^2 &= 1 \\ \text{misalkan } 3^x=a \text{ dan } 5^x=b \\ a+b-a^2+ab-b^2 &= 1 \\ a^2-ab+b^2-a-b+1 &= 0 \\ 2a^2-2ab+2b^2-2a-2b+2 &= 0 \\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1 &= 0 \\ (a-b)^2+(a-1)^2+(b-1)^2 &= 0 \\ a-b=0; a-1=0; b-1 &= 0 \\ a=b &= 1 \\ 3^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> <ol start=16> <li>Berapa nilai x dari <math>^6log x^2+^{6x}log \frac{6}{x}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ \text{misalkan } 6x=a \text{ maka } x=\frac{a}{6} \\ ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ ^6log (\frac{a}{6})^2+^{6 \frac{a}{6}}log \frac{6}{\frac{a}{6}} &= 1 \\ ^6log \frac{a^2}{6^2}+^alog \frac{6^2}{a} &= 1 \\ ^6log a^2-^6log 6^2+^alog 6^2-^alog a &= 1 \\ 2 ^6log a-2 ^6log 6+2 ^alog 6-^alog a &= 1 \\ 2 ^6log a-2+2 \frac{1}{^6log a}-1 &= 1 \\ 2 ^6log a+2 \frac{1}{^6log a}-4 &= 0 \\ 2 ^6log^2 a-4 ^6log a+2 &= 0 \\ ^6log^2 a-2 ^6log a+1 &= 0 \\ (^6log a-1)^2 &= 0 \\ ^6log a &= 1 \\ a &= 6 \\ x &= \frac{a}{6} \\ &= \frac{6}{6} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=17> <li>Berapa nilai x dari (x+500)³+x=20?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+500)^3+x &= 20 \\ \text{misalkan } a=x+500 \text{ maka } x=a-500 \\ a^3+a-500 &= 20 \\ a^3+a &= 520 \\ a(a^2+1) &= 8 \cdot 65 \\ a(a^2+1) &= 8(64+1) \\ a(a^2+1) &= 8(8^2+1) \\ a &= 8 \\ x &= 8-500 \\ &= -492 \\ \end{align} </math> </div></div> <ol start=18> <li>Berapa nilai x dari <math>\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2} &= 0 \\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}} &= \frac{1}{2} \\ \frac{x^n+4^n}{x^n+16^n} &= (\frac{1}{2})^n \\ \frac{x^n+4^n}{x^n+16^n} &= \frac{1}{2^n} \\ 2^n(x^n+4^n) &= x^n+16^n \\ 2^n(x^n+2^{2n}) &= x^n+2^{4n} \\ 2^n \cdot x^n+2^{3n} &= x^n+2^{4n} \\ 2^n \cdot x^n-x^n &= 2^{4n}-2^{3n} \\ x^n(2^n-1) &= 2^{3n}(2^n-1) \\ x^n &= 2^{3n} \\ x^n &= (2^3)^n \\ x^n &= 8^n \\ x &= 8 \\ \end{align} </math> </div></div> <ol start=19> <li>Berapa hasil dari <math>\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ x &= \frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5 \cdot 6}+\sqrt{5 \cdot 5}+\sqrt{6 \cdot 4}+\sqrt{5 \cdot 4}}{\sqrt{5 \cdot 4}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5} \cdot \sqrt{6}+\sqrt{5} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{4}}{2 \cdot \sqrt{5}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{6} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{5}+\sqrt{5} \cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}} \\ &= \frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ &= \frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ \frac{1}{x} &= \frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}} \\ &= \frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ &= \frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1} \\ &= \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5} \\ &= \sqrt{6}-\sqrt{4} \\ &= \sqrt{6}-2 \\ x &= \frac{1}{\sqrt{6}-2} \\ &= \frac{\sqrt{6}+2}{6-4} \\ &= \frac{\sqrt{6}+2}{2} \\ &= 1+\frac{\sqrt{6}}{2} \\ \end{align} </math> </div></div> <ol start=20> <li>Berapa hasil dari <math>(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5 \\ \text{misalkan } x=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ x &= \frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ &= \frac{\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}} \\ &= \frac{\sqrt{3}+1}{4} \\ 4x &= \sqrt{3}+1 \\ 4x-1 &= \sqrt{3} \\ (4x-1)^2 &= 3 \\ 16x^2-8x+1 &= 3 \\ 16x^2 &= 8x+2 \\ 8x^2 &= 4x+1 \\ x^2 &= \frac{4x+1}{8} \\ *cara 1 \\ x^3 &= x \cdot x^2 \\ &= x(\frac{4x+1}{8}) \\ &= \frac{4x^2+x}{8} \\ &= \frac{4x^2}{8}+\frac{x}{8} \\ &= \frac{4(\frac{4x+1}{8})}{8}+\frac{x}{8} \\ &= \frac{16x+4}{64}+\frac{x}{8} \\ &= \frac{4x+1}{16}+\frac{x}{8} \\ &= \frac{4x+1+2x}{16} \\ &= \frac{6x+1}{16} \\ x^5 &= x^2 \cdot x^3 \\ &= (\frac{4x+1}{8})(\frac{6x+1}{16}) \\ &= \frac{24x^2+10x+1}{128} \\ &= \frac{24x^2}{128}+\frac{10x+1}{128} \\ &= \frac{24(\frac{4x+1}{8})}{128}+\frac{10x+1}{128} \\ &= \frac{96x+24}{1024}+\frac{10x+1}{128} \\ &= \frac{96x+24+80x+8}{1024} \\ &= \frac{176x+32}{1024} \\ &= \frac{176x}{1024}+\frac{32}{1024} \\ &= \frac{176}{1024}(\frac{\sqrt{3}+1}{4})+\frac{32}{1024} \\ &= \frac{44(\sqrt{3}+1)}{1024}+\frac{32}{1024} \\ &= \frac{44\sqrt{3}+44}{1024}+\frac{32}{1024} \\ &= \frac{76+44\sqrt{3}}{1024} \\ &= \frac{19+11\sqrt{3}}{256} \\ *cara 2 \\ x^4 &= (x^2)^2 \\ &= (\frac{4x+1}{8})^2 \\ &= \frac{16x^2+8x+1}{64} \\ &= \frac{16x^2}{64}+\frac{8x}{64}+\frac{1}{64} \\ &= \frac{x^2}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{\frac{4x+1}{8}}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{4x}{32}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{8}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{4}+\frac{3}{64} \\ x^5 &= x \cdot x^4 \\ &= (\frac{\sqrt{3}+1}{4})(\frac{x}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\frac{\sqrt{3}+1}{4}}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\sqrt{3}+1}{16}+\frac{3}{64}) \\ &= \frac{(\sqrt{3}+1)^2}{64}+(\frac{\sqrt{3}+1}{4})\frac{3}{64} \\ &= \frac{3+2\sqrt{3}+1}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{4+2\sqrt{3}}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{16+8\sqrt{3}}{256}+\frac{3\sqrt{3}+3}{256} \\ &= \frac{19+11\sqrt{3}}{256} \\ \end{align} </math> </div></div> <ol start=21> <li>Berapa hasil dari <math>\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4} &= \frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots) \\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots &= \frac{1}{1-\frac{1}{4}} \\ &= \frac{4}{3} \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{4}{3}) \\ &= \frac{1}{4}+\frac{16}{3} \\ &= \frac{67}{12} \\ x &= \frac{67}{9} \\ \end{align} </math> </div></div> <ol start=22> <li>Berapa nilai y-x jika <math>\frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} = \frac{x}{y}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} &= \frac{x}{y} \\ \frac{\frac{106 \times 107}{2}}{\frac{106}{2}(4+109)} &= \frac{x}{y} \\ \frac{53 \times 107}{53 \times 113} &= \frac{x}{y} \\ y-x &= 113-107 = 6 \\ \end{align} </math> </div></div> <ol start=23> <li>Berapa angka satuan dari hasil 17²⁰²⁴?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan angka satuannya} \\ 17^1 &= 7 \\ 17^2 &= 9 \\ 17^3 &= 3 \\ 17^4 &= 1 \\ 17^5 &= 7 \\ 17^6 &= 9 \\ 17^7 &= 3 \\ 17^8 &= 1 \\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1} \end{align} </math> </div></div> <ol start=24> <li>Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ 1! + 2! + 3! + 4! + \dots + 2024! &= 1 + (1x2) + (1x2x3) + (1x2x3x4) + \dots + 2024! \\ &= 1 + 2 + 6 + 24 + 120 + 720 + \dots + 2024! \\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka } 1+2+6+24 = 33 \text{ jadi angka satuannya adalah } 3 \end{align} </math> </div></div> <ol start=25> <li>Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ \frac{1! + 2! + 3! + 4! + \dots + 2024!}{12} &= \frac{1 + 1x2 + 1x2x3 + 1x2x3x4 + \dots + 2024!}{12} \\ &= \frac{1 + 2 + 6 + 24 + \dots + 2024!}{12} \\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi } 1+2+6 = 9 \end{align} </math> </div></div> <ol start=26> <li>Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{penjumlahan} \\ 1+1+1+1+ \dots \text{ (sebanyak 88 buah) }+x+y &= A \\ 88+x+y &= A \\ \text{perkalian} \\ 1 \times 1 \times 1 \times \dots \text{ (sebanyak 88 buah) }\times x \times y &= A \\ x \times y &= A \\ 88+x+y &= xy \\ xy-y &= 88+x \\ y(x-1) &= 88+x \\ y &= \frac{88+x}{x-1} \\ \text{uji selidiki untuk x=2} \\ y &= \frac{88+2}{2-1} \\ &= 90 \\ \text{buktikan} \\ 88+x+y &= xy \\ 88+2+90 &= 2(90) \\ 180 &= 180 \\ \text{terbukti} \\ \text{nilai A adalah } 180 \\ \end{align} </math> </div></div> <ol start=27> <li>Berapakah nilai x, y dan z dari <math>x+y-z=1, x^2+y^2-z^2=-5 \text{ dan } x^3+y^3-z^3=-53</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+y-z &= 1 \\ x+y &= z+1 \\ x^2+2xy+y^2 &= z^2+2z+1 \\ x^2+y^2-z^2 &= 2z+1-2xy \\ -5 &= 2z+1-2xy \\ 2xy &= 2z+6 \\ xy &= z+3 \\ x^2+y^2-z^2 &= -5 \\ x^2+y^2 &= z^2-5 \\ x^3+y^3-z^3 &= -53 \\ (x+y)(x^2-xy+y^2)-z^3+53 &= 0 \\ (x+y)(x^2+y^2-xy)-z^3+53 &= 0 \\ (z+1)(z^2-5-(z+3))-z^3+53 &= 0 \\ (z+1)(z^2-z-8)-z^3+53 &= 0 \\ z^3-z^2-8z+z^2-z-8-z^3+53 &= 0 \\ -9z+45 &= 0 \\ -9z &= -45 \\ z &= 5 \\ x+y &= 5+1 \\ x+y &= 6 \\ x &= 6-y \\ xy &= 5+3 \\ xy &= 8 \\ (6-y)y &= 8 \\ 6y-y^2 &= 8 \\ y^2-6y+8 &= 0 \\ (y-4)(y-2) &= 0 \\ y=4 \text{ atau } y=2 \\ \text{jika } y=4 \\ x+y &= z+1 \\ x+4 &= 5+1 \\ x &= 2 \\ \text{jika } y=2 \\ x+y &= z+1 \\ x+2 &= 5+1 \\ x &= 4 \\ \end{align} </math> </div></div> <ol start=28> <li>Berapakah nilai titik koordinat (x,y) dari <math>\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}</math> dan <math>\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x+y}+\sqrt{x-y} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+y}-\sqrt{x-y} &= \sqrt{\frac{52y}{3x}} \\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y}) &= \sqrt{\frac{432x}{13y}} \cdot \sqrt{\frac{52y}{3x}} \\ x+y-x+y &= \sqrt{\frac{432x \cdot 52y}{13y \cdot 3x}} \\ 2y &= \sqrt{144 \cdot 4} \\ 2y &= \sqrt{576} \\ 2y &= 24 \\ y &= 12 \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13(12)}} \\ x+12+x-12+2 \cdot \sqrt{x+12} \cdot \sqrt{x-12} &= \frac{36x}{13} \\ 2x+2 \sqrt{x^2-144} &= \frac{36x}{13} \\ 2(x+\sqrt{x^2-144}) &= \frac{36x}{13} \\ x+\sqrt{x^2-144} &= \frac{18x}{13} \\ \sqrt{x^2-144} &= \frac{5x}{13} \\ x^2-144 &= \frac{25x^2}{169} \\ \frac{144x^2}{169}-144 &= 0 \\ \frac{x^2}{169}-1 &= 0 \\ x^2-169 &= 0 \\ (x-13)(x+13) &= 0 \\ x_1=13 &\text{ atau } x_2=-13 \text{ (TM) karena } x>y \\ \end{align} </math> jadi titik koordinat (13,12) </div></div> <ol start=29> <li>Berapakah nilai dari <math>x^2-7x</math> jika <math>(x-2)^2+\frac{1}{(x-2)^2} = 11</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x-2)^2+\frac{1}{(x-2)^2} &= 11 \\ (x-2)^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2)^2} &= 11-2 \\ (x-2-\frac{1}{x-2})^2 &= 9 \\ x-2-\frac{1}{x-2} &= 3 \\ (x-2)^2-1 &= 3(x-2) \\ x^2-4x+4-1 &= 3x-6 \\ x^2-7x &= -9 \\ \end{align} </math> </div></div> <ol start=30> <li>Berapakah nilai dari <math>\frac{(x+y)^2(x+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)}</math> jika xy+yz+xz=1?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy+yz+xz &= 1 \\ x^2+xy+yz+xz &= x^2+1 \\ x(x+y)+z(x+y) &= x^2+1 \\ (x+y)(x+z) &= x^2+1 \\ \text{dengan pola yang sama } \\ (y+x)(y+z) &= y^2+1 \\ (x+z)(y+z) &= z^2+1 \\ \frac{(x+y)^2(y+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)} &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)} \\ &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)^2(y+z)^2(x+z)^2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=31> <li>Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} w+5 &= w+x+y+z+5 \\ x+4 &= w+x+y+z+5 \\ y+3 &= w+x+y+z+5 \\ z+2 &= w+x+y+z+5 \\ \text{jumlahkan keempat persamaan } \\ w+x+y+z+14 &= 4(w+x+y+z+5) \\ w+x+y+z+14 &= 4(w+x+y+z)+20 \\ 3(w+x+y+z) &= -6 \\ w+x+y+z &= -2 \\ \end{align} </math> </div></div> <ol start=32> <li>Berapakah nilai dari <math>\frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2}</math> jika <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3</math> dan x+y+z=xyz?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2} &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2} \\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) \\ 3^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz}) \\ 9 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz}) \\ &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2 \\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} &= 7 \\ \end{align} </math> </div></div> <ol start=33> <li>Berapakah nilai dari <math>\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}</math> jika <math>x^2+y^2+z^2 = -2(ab+bc+ac)</math>?>/li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+z^2 &= -2(xy+yz+xz) \\ x^2+y^2+z^2+2(xy+yz+xz) &= 0 \\ (x+y+z)^2 &= 0 \\ x+y+z &= 0 \\ x+y &= -z \\ x+z &= -y \\ y+z &= -x \\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z} &= \frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x} \\ &= -2-(-5)-(-7) \\ &= 10 \\ \end{align} </math> </div></div> <ol start=34> <li>Berapakah nilai dari <math>\frac{20xyz}{xy+yz+xz}</math> jika <math>16^x = 256^y = 625^z = 40</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 16^x = 256^y = 625^z &= 40 \\ 2^{4x} = 4^{4y} = 5^{4z} &= 40 \\ 2^{4x} &= 40 \\ 2 &= 40^{\frac{1}{4x}} \\ 4^{4y} &= 40 \\ 4 &= 40^{\frac{1}{4y}} \\ 5^{4z} &= 40 \\ 5 &= 40^{\frac{1}{4z}} \\ 2 \cdot 4 \cdot 5 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z}} \\ 1 &= \frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z} \\ 4 &= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{20xyz}{xy+yz+xz} &= 20 \cdot \frac{xyz}{xy+yz+xz} \\ &= 20 \cdot (\frac{xy+yz+xz}{xyz})^{-1} \\ &= 20 \cdot (\frac{1}{z} + \frac{1}{x} + \frac{1}{y})^{-1} \\ &= 20 \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{-1} \\ &= 20 \cdot (4)^{-1} \\ &= 20 \cdot \frac{1}{4} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=35> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+3x^2+1}</math> jika <math>6x^2+25x+6=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 6x^2+25x+6 &= 0 \\ 6x+25+\frac{6}{x} &= 0 \\ 6(x+\frac{1}{x}) &= -25 \\ x+\frac{1}{x} &= \frac{-25}{6} \\ (c+\frac{1}{x})^2 &= (\frac{-25}{6})^2 \\ x^2+2+\frac{1}{x^2} &= \frac{625}{36} \\ x^2+\frac{1}{x^2} &= \frac{625}{36}-2 \\ x^2+\frac{1}{x^2} &= \frac{553}{36} \\ \frac{x^2}{x^4+3x^2+1} &= \frac{1}{x^2+3+\frac{1}{x^2}} \\ &= \frac{1}{a^2+\frac{1}{x^2}+3} \\ &= \frac{1}{\frac{553}{36}+3} \\ &= \frac{1}{\frac{661}{36}} \\ &= \frac{36}{661} \\ \end{align} </math> </div></div> <ol start=36> <li>Berapakah nilai dari <math>\frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5} &= \frac{(9+2\sqrt{20})^{1013}}{((2)^3+3(2)^2(\sqrt{5})+3(2)(\sqrt{5})^2+(\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{((2+\sqrt{5})^2)^{1013}}{((2+\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{(2+\sqrt{5})^{2026}}{(2+\sqrt{5})^{2025}}+6-\sqrt{5} \\ &= 2+\sqrt{5}+6-\sqrt{5} \\ &= 8 \\ \end{align} </math> </div></div> <ol start=37> <li>Berapakah nilai dari <math>27x^3+\frac{8}{x^3}</math> jika <math>3x+\frac{2}{x}=6</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3x+\frac{2}{x} &= 6 \\ (3x+\frac{2}{x})^3 &= 6^3 \\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3} &= 216 \\ 27x^3+18(6)+\frac{8}{x^3} &= 216 \\ 27x^3+108+\frac{8}{x^3} &= 216 \\ 27x^3+\frac{8}{x^3} &= 108 \\ \end{align} </math> </div></div> <ol start=38> <li>Berapakah nilai dari <math>x^6+\frac{8}{x^3}</math> jika <math>x^3+\frac{1}{x^3}=8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3+\frac{1}{x^3} &= 8 \\ x^3 &= 8-\frac{1}{x^3} \\ x^6 &= 8x^3-1 \\ x^6+\frac{8}{x^3} &= 8x^3-1+\frac{8}{x^3} \\ &= 8x^3+\frac{8}{x^3}-1 \\ &= 8(x^3+\frac{1}{x^3})-1 \\ &= 8(8)-1 \\ &= 63 \\ \end{align} </math> </div></div> <ol start=39> <li>Berapakah nilai dari <math>4x+\frac{25}{x}</math> jika <math>2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2\sqrt{x}+\frac{5}{\sqrt{x}} &= 4x-\frac{25}{x} \\ 2\sqrt{x}+\frac{5}{\sqrt{x}} &= (2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}}) \\ 1 &= 2\sqrt{x}-\frac{5}{\sqrt{x}} \\ 1^2 &= (2\sqrt{x}-\frac{5}{\sqrt{x}})^2 \\ 1 &= 4x-20+\frac{25}{x} \\ 4x+\frac{25}{x} &= 21 \\ \end{align} </math> </div></div> <ol start=40> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2-x+1}{x^2+x+1} &= \frac{5}{6} \\ \frac{x^2+1-x}{x^2+1+x} &= \frac{5}{6} \\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} &= \frac{5}{6} \\ \text{ misalkan } x+\frac{1}{x} &= y \\ \frac{y-1}{y+1} &= \frac{5}{6} \\ 6(y-1) &= 5(y+1) \\ 6y-6 &= 5y+5 \\ y &= 11 \\ x+\frac{1}{x} &= 11 \\ \end{align} </math> </div></div> <ol start=41> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt{x}+x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x}+x &= 1 \\ x-1 &= -\sqrt{x} \\ (x-1)^2 &= (-\sqrt{x})^2 \\ x^2-2x+1 &= x \\ x^2-3x+1 &= 0 \\ x-3+\frac{1}{x} &= 0 \\ x+\frac{1}{x} &= 3 \\ \end{align} </math> </div></div> <ol start=42> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt[3]{x}-\sqrt[3]{x-36}=3</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x}-\sqrt[3]{x-36} &= 3 \\ (\sqrt[3]{x}-\sqrt[3]{x-36})^3 &= 3^3 \\ x-(x-36)-3 \sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36}) &= 27 \\ 36-3 \sqrt[3]{x(x-36)}3 &= 27 \\ -9 \sqrt[3]{x(x-36)} &= -9 \\ \sqrt[3]{x(x-36)} &= 1 \\ x(x-36) &= 1 \\ x^2-36x-1 &= 0 \\ x-36-\frac{1}{x} &= 0 \\ x-\frac{1}{x} &= 36 \\ \end{align} </math> </div></div> <ol start=43> <li>Berapakah nilai dari <math>x+\frac{16}{x}</math> jika <math>x-3\sqrt{x}=4</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x-3\sqrt{x} &= 4 \\ x-4 &= 3\sqrt{x} \\ x^2-8x+16 &= 9x \\ x^2-17x+16 &= 0 \\ x-17+\frac{16}{x} &= 0 \\ x+\frac{16}{x} &= 17 \\ \end{align} </math> </div></div> <ol start=44> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+4}</math> jika <math>x^2-7x+2=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2-7x+2 &= 0 \\ x^2+2 &= 7x \\ x+\frac{2}{x} &= 7 \\ x^2+4+\frac{4}{x^2} &= 49 \\ x^2+\frac{4}{x^2} &= 45 \\ \frac{x^4+4}{x^2} &= 45 \\ \frac{x^2}{x^4+4} &= \frac{1}{45} \\ \end{align} </math> </div></div> <ol start=45> <li>Berapakah nilai dari <math>x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}</math> jika <math>x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}} &= 25 \\ x^{\frac{1}{2}}+x^{-\frac{1}{2}} &= 23 \\ x+2+x^{-1} &= 529 \\ x+x^{-1} &= 527 \\ x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+x^{-\frac{3}{4}} &= 110 \\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1} &= x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}} \\ &= 527+110 \\ &= 637 \\ \end{align} </math> </div></div> <ol start=46> <li>Berapakah nilai dari <math>\sqrt{8x^6+x^5+x^4+5x^3+1}</math> jika <math>\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5} &= 0 \\ \frac{x^2+x+1}{x^5} &= 0 \\ x^2+x+1 &= 0 \\ x^2+x+1 &= 0 \\ (x-1)(x^2+x+1) &= 0(x-1) \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ \sqrt{8x^6+x^5+x^4+5x^3+1} &= \sqrt{(2x^3)^2+x^3x^2+x^3x+5x^3+1} \\ &= \sqrt{(2(1))^2+(1)x^2+(1)x+5(1)+1} \\ &= \sqrt{(2)^2+x^2+x+5+1} \\ &= \sqrt{4+x^2+x+1+5} \\ &= \sqrt{4+0+5} \\ &= \sqrt{9} \\ &= 3 \\ \end{align} </math> </div></div> <ol start=47> <li>Berapakah nilai dari <math>f(1)+f(2)+f(3)+ \dots + f(99)</math> jika <math>f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{1}{\sqrt{x+1}+\sqrt{x}} \\ &= \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} \\ &= \sqrt{x+1}-\sqrt{x} \\ f(1)+f(2)+f(3)+ \dots + f(98)+f(99) &= \sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+ \cdot + \sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99} \\ &= \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+ \cdot + \sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ &= \sqrt{100}-\sqrt{1} \\ &= 10-1 \\ &= 9 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots + \frac{2024}{2025})</math> jika <math>h(x)=\frac{3}{3+9^x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} h(x) &= \frac{3}{3+9^x} \\ h(1-x) &= \frac{3}{3+9^{1-x}} \\ &= \frac{3}{3+\frac{9}{9^x}} \\ &= \frac{9^x}{3+9^x} \\ h(x)+h(1-x) &= \frac{3}{3+9^x}+\frac{9^x}{3+9^x} \\ &= \frac{3+9^x}{3+9^x} \\ &= 1 \\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots +(1-\frac{2}{2025})+(1-\frac{1}{2025})) \\ & 5(1+1+1+ \dots +1+1) \text{ sebanyak 1012 kali } \\ & 5(1012) \\ & 5060 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>\frac{7^{2025} - 7^{2023} + 432}{7^{2024} + 7^{2023} + 72}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72} &= \frac{7^{2023}7^{2}-7^{2023} + 48 \times 9}{7^{2023}7^1+7^{2023}+8 \times 9} \\ &= \frac{7^{2023}(7^{2}-1)+48 \times 9}{7^{2023}(7^1+1)+8 \times 9} \\ &= \frac{7^{2023}(49-1)+48 \times 9}{7^{2023}(7+1) + 8 \times 9} \\ &= \frac{7^{2023} \times 48+48 \times 9}{7^{2023} \times 8+8 \times 9} \\ &= \frac{48(7^{2023}+9)}{8(7^{2023}+9)} \\ &= \frac{48}{8} \\ &= 6 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>tan (x+\frac{\pi}{4})</math> jika <math>\frac{1}{cos x}-tan x = \frac{4}{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{cos x}-tan x &= \frac{4}{5} \\ sec x-tan x &= \frac{4}{5} \\ sec^2 x-tan^2 x &= 1 \\ (sec x+tan x)(sec x-tan x) &= 1 \\ (sec x+tan x)\frac{4}{5} &= 1 \\ sec x+tan x &= \frac{5}{4} \\ \text{kedua persamaan dengan cara metode eliminasi } \\ 2 tan x &= \frac{5}{4}-\frac{4}{5} \\ 2 tan x &= \frac{9}{20} \\ tan x &= \frac{9}{40} \\ tan (x+\frac{\pi}{4}) &= \frac{tan x+tan \frac{\pi}{4}}{1-tan x \cdot tan \frac{\pi}{4}} \\ &= \frac{\frac{9}{40}+1}{1-\frac{9}{40} \cdot 1} \\ &= \frac{\frac{49}{40}}{\frac{31}{40}} \\ &= \frac{49}{31} \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>sin^3 x+csc^3 x</math> jika <math>sin x-csc x = 8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ Dengan menggunakan rumus: } (a-b)^3 &= a^3-b^3-3ab(a-b) \\ (sin x-csc x)^3 &= sin^3 x-csc^3 x-3sin x csc x(sin x-csc x) \\ 8^3 &= sin^3 x-csc^3 x-3sin x (\frac{1}{sin x})(8) \\ 512 &= sin^3 x-csc^3 x-24 \\ sin^3 x-csc^3 x &= 512+24 \\ sin^3 x-csc^3 x &= 536 \\ \end{align} </math> </div></div> <ol start=52> <li>Berapakah nilai dari <math>(sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2</math> jika <math>\frac{1}{sin x}+\frac{1}{cos x} = 10</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{sin x}+\frac{1}{cos x} &= 10 \\ \frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} &= 100 \\ (sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2 &= sin^2 x+\frac{2sin x}{cos x}+\frac{1}{cos^2 x}+cos^2 x+\frac{2cos x}{sin x}+\frac{1}{sin^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2(sin^2 x+cos^2 x)}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+100 \\ &= 101 \\ \end{align} </math> </div></div> <ol start=53> <li>Berapakah nilai dari (x-1)⁶ jika <math>x=\frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin 80^\circ &= cos 10^\circ \\ sin 80^\circ-cos 10^\circ &= 0 \\ x &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+2 sin 20^\circ cos 20^\circ}{cos 10^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+4 sin 10^\circ cos 10^\circ cos 20^\circ}{cos 10^\circ} \\ &= 4 cos 55^\circ cos 25^\circ+4 sin 10^\circ cos 20^\circ \\ &= 2(2 cos 55^\circ cos 25^\circ+2 sin 10^\circ cos 20^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ+sin (-10)^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ-sin 10^\circ) \\ &= 2(cos 80^\circ-sin 10^\circ+cos 30^\circ+sin 30^\circ) \\ &= 2(cos 80^\circ-sin (90^\circ-80^\circ)+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(cos 80^\circ-cos 80^\circ+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= \sqrt{3}+1 \\ x-1 &= \sqrt{3} \\ (x-1)^6 &= (\sqrt{3})^6 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=54> <li>Berapakah nilai dari x jika <math>x=\frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ} \\ 2x sin 20^\circ-x sin 40 ^\circ &= x sin 20^\circ-x^2 sin 10^\circ \\ x^2 sin 10^\circ+x sin 20^\circ-x sin 40 ^\circ &= 0 \\ x(x sin 10^\circ+sin 20^\circ-sin 40 ^\circ) &= 0 \\ x = 0 &\text{ atau } x sin 10^\circ+sin 20^\circ-sin 40 ^\circ = 0 \\ x sin 10^\circ+sin 20^\circ-sin 40 ^\circ &= 0 \\ x sin 10^\circ &= sin 40 ^\circ-sin 20^\circ \\ x &= \frac{sin 40 ^\circ-sin 20^\circ}{sin 10^\circ} \\ &= \frac{2 cos 30 ^\circ sin 10^\circ}{sin 10^\circ} \\ &= 2 cos 30 ^\circ \\ &= \frac{2 \sqrt{3}}{2} \\ &= \sqrt{3} \\ \end{align} </math> </div></div> <ol start=55> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x^2}{x^2-16y^2} = \frac{625}{49}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2}{x^2-16y^2} &= \frac{625}{49} \\ \frac{x^2-16y^2}{x^2} &= \frac{49}{625} \text{ (terbalik posisinya)} \\ 1-\frac{16y^2}{x^2} &= \frac{49}{625} \\ \frac{16y^2}{x^2} &= 1 - \frac{49}{625} \\ (\frac{4y}{x})^2 &= \frac{576}{625} \\ (\frac{4y}{x})^2 &= (\frac{24}{25})^2 \\ \frac{4y}{x} &= \frac{24}{25} \\ \frac{y}{x} &= \frac{6}{25} \\ \frac{x}{y} &= \frac{25}{6} \\ \end{align} </math> </div></div> <ol start=56> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x}{y}+\frac{x+10y}{y+10x} = 2</math> serta bilangan real untuk x dan y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x}{y}+\frac{x+10y}{y+10x} &= 2 \\ \frac{x}{y}+\frac{\frac{x}{y}+10}{1+10\frac{x}{y}} &= 2 \\ \text{misalkan } \frac{x}{y} = a \\ a+\frac{a+10}{1+10a} &= 2 \\ a(1+10a)+a+10 &= 2(1+10a) \\ 10a^2+a+a+10 &= 2+20a \\ 10a^2-18a+8 &= 0 \\ 5a^2-9a+4 &= 0 \\ (5a-4)(a-1) &= 0 \\ a = \frac{4}{5} &\text{ atau } a = 1 \\ \text{jadi } \frac{x}{y} = {\frac{4}{5}, 1} \\ \end{align} </math> </div></div> <ol start=57> <li>Berapakah nilai dari xy jika <math>x^4+y^4+x^2y^2=15 \text{ dan } x^2+y^2+xy=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+xy &= 5 \\ x^2+y^2 &= 5-xy \\ x^4+y^4+x^2y^2 &= 15 \\ (x^2)^2+(y^2)^2+2x^2y^2-x^2y^2 &= 15 \\ (x^2+y^2)^2-x^2y^2 &= 15 \\ (5-xy)^2-x^2y^2 &= 15 \\ 25-10xy+x^2y^2-x^2y^2 &= 15 \\ 25-10xy &= 15 \\ 10xy &= 10 \\ xy &= 1 \\ \end{align} </math> </div></div> <ol start=58> <li>Berapakah nilai dari x jika <math>4^x = 63(4^3+1)(4^6+1)(4^{12}+1)+1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 4^x &= 63(4^3+1)(4^6+1)(4^{12}+1)+1 \\ 4^x-1 &= 63(4^3+1)(4^6+1)(4^{12}+1) \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{4^3-1} \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{63} \\ &= (4^3+1)(4^6+1)(4^{12}+1)(4^3-1) \\ &= (4^3-1)(4^3+1)(4^6+1)(4^{12}+1) \\ &= (4^6-1)(4^6+1)(4^{12}+1) \\ &= (4^{12}-1)(4^{12}+1) \\ &= 4^{24}-1 \\ 4^x &= 4^{24} \\ x &= 24 \\ \end{align} </math> </div></div> <ol start=59> <li>Berapakah nilai dari <math>\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}</math> jika <math>x=\sqrt{9+4\sqrt{5}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \sqrt{9+4\sqrt{5}} \\ x &= 2+\sqrt{5} \\ x^2 &= 9+4\sqrt{5} \\ x^2-4x &= 9+4\sqrt{5}-4(2+\sqrt{5}) \\ x^2-4x &= 1 \\ x^2 &= 4x+1 \\ x^3 &= x \cdot x^2 \\ &= x(4x+1) \\ &= 4x^2+x \\ &= 4(4x+1)+x \\ &= 16x+4+x \\ &= 17x+4 \\ x^4 &= x \cdot x^3 \\ &= x(17x+4) \\ &= 17x^2+4x \\ &= 17(4x+1)+4x \\ &= 68x+17+4x \\ &= 72x+17 \\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1} &= \frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1} \\ &= \frac{72x+17-85x-20+8x+2+5x+3}{2} \\ &= \frac{2}{2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=60> <li>Berapakah nilai dari <math>\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}</math> jika 2x-1=<math>\sqrt{61}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \frac{x^3+1}{x^5-x^4-x^3+x^2} = p \\ p &= \frac{x^3+1}{x^5-x^4-x^3+x^2} \\ &= \frac{x^3+1}{x^5-x^4-(x^3-x^2)} \\ &= \frac{x^3+1}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)} \\ &= \frac{x^2-x+1}{x^2(x-1)^2} \\ &= \frac{x^2-x+1}{(x(x-1))^2} \\ &= \frac{x(x-1)+1}{(x(x-1))^2} \\ 2x-1 &= \sqrt{61} \\ x &= \frac{\sqrt{61}+1}{2} \\ x-1 &= \frac{\sqrt{61}-1}{2} \\ x(x-1) &= (\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2}) \\ &= \frac{61-1}{4} \\ &= \frac{60}{4} \\ &= 15 \\ p &= \frac{x(x-1)+1}{(x(x-1))^2} \\ &= \frac{15+1}{15^2} \\ &= \frac{16}{15^2} \\ \sqrt{p} &= \sqrt{\frac{16}{15^2}} \\ &= \frac{4}{15} \\ \end{align} </math> </div></div> <ol start=61> <li>Berapakah nilai dari <math>(\frac{x-3}{x})^{25}</math> jika <math>x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\sqrt[5]{8}+\sqrt[5]{2} &= 1+\sqrt[5]{16}+\sqrt[5]{4} \\ x+(\sqrt[5]{2})^3+\sqrt[5]{2} &= 1+(\sqrt[5]{2})^4+(\sqrt[5]{2})^2 \\ x &= (\sqrt[5]{2})^4-(\sqrt[5]{2})^3+(\sqrt[5]{2})^2-\sqrt[5]{2}+1 \\ \text{misalkan } \sqrt[5]{2} = p \\ x &= p^4-p^3+p^2-p+1 \\ x &= \frac{p^5+1}{p+1} \\ (\frac{x-3}{x})^{25} &= (1-\frac{3}{x})^{25} \\ &= (1-\frac{3}{\frac{p^5+1}{p+1}})^{25} \\ &= (1-\frac{3(p+1)}{p^5+1})^{25} \\ &= (1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2})^5+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{2+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-3\sqrt[5]{2}-3)}{3})^{25} \\ &= (-\sqrt[5]{2})^{25} \\ &= (-2)^5 \\ &= -32 \\ \end{align} </math> </div></div> <ol start=62> <li>Berapakah nilai dari <math>x^{50}+x^{49}+x^{48}+x^{47}+x^{46}</math> jika <math>x^2+x+1=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+x+1 &= 0 \\ x^2+x &= -1 \\ \frac{x^3-1}{x-1} &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46} &= x^{48}(x^2+x+1)+x^{45}(x^2+x) \\ &= x^{48}(0)+(x^3)^{15}(-1) \\ &= 0+(1)^{15}(-1) \\ &= -1 \\ \end{align} </math> </div></div> <ol start=63> <li>Berapakah 2²⁴ dari <math>8^7+8^6+8^5+8^4+8^3+8^2+8+1=A</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= A \\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1) &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8 &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= 8A+1 \\ 8^8+A &= 8A+1 \\ 8^8 &= 7A+1 \\ (2^3)^8 &= 7A+1 \\ 2^{24} &= 7A+1 \\ \end{align} </math> </div></div> <ol start=64> <li>Berapakah nilai dari <math>x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1</math> jika <math>x+\frac{1}{x}=\sqrt{3}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{x} &= \sqrt{3} \\ x^2+2+\frac{1}{x^2} &= 3 \\ x^2-1+\frac{1}{x^2} &= 0 \\ x^2(x^2-1+\frac{1}{x^2}) &= x^2(0) \\ x^4-x^2+1 &= 0 \\ (x^2+1)(x^4-x^2+1) &= (x^2+1)0 \\ x^6-x^4+x^2+x^4-x^2+1 &= 0 \\ x^6+1 &= 0 \\ x^6 &= -1 \\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1 &= {x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1 \\ &= (-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2-1+1 \\ &= -1+1-1+1-1+1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=65> <li>Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari <math>\frac{c-b}{a-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= ax^2+bx+c \\ f(2) &= a(2)^2+2b+c = 4 \\ &= 4a+2b+c = 4 \\ f(7) &= a(7)^2+7b+c = 49 \\ &= 49a+7b+c = 49 \\ 49a+7b+c &= 49 \\ 4a+2b+c &= 4 \\ 45a+5b &= 45 \text{ (f(7) dikurangi f(2)) } \\ 9a+b &= 9 \\ b &= -9a+9 \\ 4a+2b+c &= 4 \\ 4a+2(-9a+9)+c &= 4 \\ 4a-18a+18+c &= 4 \\ -14a+18+c &= 4 \\ c &= 14a-14 \\ \frac{c-b}{a-1} &= \frac{14a-14-(-9a+9)}{a-1} \\ &= \frac{14(a-1)+9(a-1)}{a-1} \\ &= \frac{(14+9)(a-1)}{a-1} \\ &= 23 \\ \end{align} </math> </div></div> <ol start=66> <li>Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+y)^3 &= x^3+y^3+3xy(x+y) \\ 11^3 &= 242+3xy(11) \text{ (dibagi 11)} \\ 11^2 &= 22+3xy \\ 121 &= 22+3xy \\ 99 &= 3xy \\ xy &= 33 \\ (x-y)^2 &= x^2+y^2-2xy \\ &= ((x+y)^2-2xy)-2xy \\ &= (x+y)^2-4xy \\ &= 11^2-4(33) \\ &= 121-132 \\ &= -11 \\ \end{align} </math> </div></div> <ol start=67> <li>Berapa f(1)+f(-1) jika <math>f(\frac{ax-b}{bx-a})</math>=x²-5x+6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ jika} f(1) = f(\frac{ax-b}{bx-a}) \\ 1 &= \frac{ax-b}{bx-a} \\ bx-a &= ax-b \\ (b-a)x &= -b+a \\ &= -(b-a) \\ &= -1 \\ f(1) &= x^2-5x+6 \\ &= (-1)^2-5(-1)+6 \\ &= 12 \\ \text{ jika} f(-1) = f(\frac{ax-b}{bx-a}) \\ -1 &= \frac{ax-b}{bx-a} \\ -(bx-a) &= ax-b \\ -bx+a &= ax-b \\ (-b-a)x &= -b-a \\ &= 1 \\ f(-1) &= x^2-5x+6 \\ &= (1)^2-5(1)+6 \\ &= 2 \\ f(1)+f(-1) &= 12+2 \\ &= 14 \\ \end{align} </math> </div></div> <ol start=68> <li>berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)-x &= f(x-1) \\ f(x)-f(x-1) &= x \\ x=1 ; f(1)-f(0) &= 1 \\ x=2 ; f(2)-f(1) &= 2 \\ x=3 ; f(3)-f(2) &= 3 \\ x=4 ; f(4)-f(3) &= 4 \\ \dots \\ x=200 ; f(200)-f(199) &= 200 \\ \text{ jumlahkan tersebut menjadi } \\ f(200)-f(0) &= 1+2+3+4+\dots+200 \\ &= \frac{200 \cdot 201}{2} \\ &= 20.100 \\ f(200)-1 &= 20.100 \\ &= 20.101 \\ \end{align} </math> </div></div> <ol start=69> <li>Misalkan f(x) adalah fungsi rekursif yang berlaku ∀x ∈ R sebagai berikut: : f(x)+f(15-x) = 2024 : f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)+f(15-x) &= 2024 \\ f(15+x) &= f(x)+2020 \\ *cara 1 \\ \text{ganti x dengan 15+x } \\ f(15+x)+f(-x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ *cara 2 \\ \text{ganti x dengan -x } \\ f(-x)+f(15+x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ \end{align} </math> </div></div> <ol start=70> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>2f(\frac{2002}{x}) + f(x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2f(\frac{2002}{x}) + f(x) &= 3x \\ \text{ganti x dengan 2 } \\ 2f(\frac{2002}{2}) + f(2) &= 3(2) \\ 2f(1001) + f(2) &= 6 \\ \text{ganti x dengan 1001 } \\ 2f(\frac{2002}{1001}) + f(1001) &= 3(1001) \\ 2f(2) + f(1001) &= 3003 \\ 2f(2) + f(1001) &= 3003 \\ f(1001) &= 3003 - 2f(2) \\ 2f(1001) + f(2) &= 6 \\ 2(3003 - 2f(2)) + f(2) &= 6 \\ 6006 - 4f(2) + f(2) &= 6 \\ 3f(2) &= 6000 \\ f(2) &= 2000 \\ \end{align} </math> </div></div> <ol start=71> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>f(\frac{1}{x}) + \frac{1}{x}f(-x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{1}{x})+\frac{1}{x}f(-x) &= 3x \\ \text{ganti x dengan 1/3 } \\ f(3)+3f(-\frac{1}{3}) &= 1 \\ \text{ganti x dengan -3 } \\ f(-\frac{1}{3}) - \frac{1}{3}f(3) &= -9 \\ \text{dikalikan 3 } \\ 3f(-\frac{1}{3})-f(3) &= -27 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ 2f(3) &= 28 \\ f(3) &= 14 \\ \end{align} </math> </div></div> <ol start=72> <li>Diketahui polinom <math>f(7^b-1)=7^{3b}-10</math>. tentukan nilai f(5)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(5) &= f(7^b-1) \\ 5 &= 7^b-1 \\ 7^b &= 6 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(6-1) &= 6^3-10 \\ f(5) &= 216-10 \\ &= 206 \\ *cara 2 \\ \text{misalkan } 7^b-1=a \text{ maka } 7^b=a+1 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(a) &= (a+1)^3-10 \\ f(5) &= (5+1)^3-10 \\ &= 6^3-10 \\ &= 216-10 \\ &= 206 \\ \end{align} </math> </div></div> <ol start=73> <li>Diketahui polinom <math>f(6^b-7)=6^{3b}-2 \cdot 6^{2b}-4</math>. tentukan nilai f(-2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(-2) &= f(6^b-7) \\ -2 &= 6^b-7 \\ 6^b &= 5 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(5-7) &= 5^3-2 \cdot 5^2-4 \\ f(-2) &= 125-50-4 \\ &= 71 \\ *cara 2 \\ \text{misalkan } 6^b-7=a \text{ maka } 6^b=a+7 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(a) &= (a+7)^3-2(a+7)^2-4 \\ f(-2) &= (-2+7)^3-2(-2+7)^2-4 \\ &= 5^3-2(5)^2-4 \\ &= 125-50-4 \\ &= 71 \\ \end{align} </math> </div></div> <ol start=74> <li>Jika <math>f(xy)=\frac{f(x)}{y}</math> dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(10) &= 7 \\ f(2 \cdot 5) &= 7 \\ f(xy) &= \frac{f(x)}{y} \\ f(2 \cdot 5) &= \frac{f(2)}{5} \\ 7 &= \frac{f(2)}{5} \\ f(2) &= 35 \\ \end{align} </math> </div></div> <ol start=75> <li>Jika <math>f(xy)=\frac{f(x+y)}{xy}</math> dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(15) &= 16 \\ f(3 \cdot 5) &= 16 \\ f(xy) &= \frac{f(x+y)}{xy} \\ f(3 \cdot 5) &= \frac{f(3+5)}{3 \cdot 5} \\ f(15) &= \frac{f(8)}{15} \\ 16 &= \frac{f(8)}{15} \\ f(8) &= 240 \\ \end{align} </math> </div></div> <ol start=76> <li>Jika <math>f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15</math> maka tentukan nilai f(16)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x+\frac{1}{x}+6) &= x^2+\frac{1}{x^2}+15 \\ &= (x+\frac{1}{x})^2-2+15 \\ &= (x+\frac{1}{x})^2+13 \\ \text{misalkan } x+\frac{1}{x} &= p \\ f(x+\frac{1}{x}+6) &= (x+\frac{1}{x})^2+13 \\ f(p+6) &= p^2+13 \\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 } \\ f(p+6) &= p^2+13 \\ f(10+6) &= 10^2+13 \\ f(16) &= 100+13 \\ &= 113 \\ \end{align} </math> </div></div> <ol start=77> <li>Tentukan nilai x jika <math>f(x)=\frac{4}{4-x}</math> dan <math>f(x \cdot f(x))^{\frac{f(4x)}{f(x)}}=256</math>!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{4}{4-x} \\ f(4x) &= \frac{4}{4-4x} \\ \frac{f(4x)}{f(x)} &= \frac{\frac{4}{4-4x}}{\frac{4}{4-x}} \\ &= \frac{4-x}{4-4x} \\ f(x \cdot f(x)) &= f(x(\frac{4}{4-x})) \\ &= f(\frac{4x}{4-x}) \\ &= \frac{4}{4-(\frac{4x}{4-x})} \\ &= \frac{4}{\frac{16-4x-4x}{4-x}} \\ &= \frac{4}{\frac{16-8x}{4-x}} \\ &= \frac{4(4-x)}{4(4-4x)} \\ &= \frac{4-x}{4-4x} \\ \text{misalkan } \frac{4-x}{4-4x} &= a \\ f(x \cdot f(x))^{\frac{f(4x)}{f(x)}} &= 256 \\ a^a &= 256 \\ a^a &= 4^4 \\ a &= 4 \\ \frac{4-x}{4-4x} &= 4 \\ 4-x &= 16-16x \\ 15x &= 12 \\ x &= \frac{4}{5} \\ \end{align} </math> </div></div> <ol start=78> <li>Fungsi <math>f(x) = \frac{kx}{2x+1} \text{dengan } x \neq -\frac{1}{2}</math>. Dengan f(f(x)) = x maka tentukan nilai k!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{kx}{2x+1} \\ f(f(x)) &= x \\ f(\frac{kx}{2x+1}) &= x \\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1} &= x \\ \frac{\frac{k^2x}{2x+1}}{\frac{2kx+2x+1}{2x+1}} &= x \\ \frac{k^2x}{2kx+2x+1} &= x \\ \frac{k^2}{2kx+2x+1} &= 1 \\ k^2 &= 2kx+2x+1 \\ k^2-2kx &= 2x+1 \\ k^2-2kx+x^2 &= x^2+2x+1 \\ (k-x)^2 &= (x+1)^2 \\ (k-x)^2-(x+1)^2 &= 0 \\ (k-x+x+1)(k-x-(x+1)) &= 0 \\ k=-1 &\text{ atau } k=2x+1 &\text{ (TM) } \\ \end{align} </math> </div></div> <ol start=79> <li>Jika n = 2023²+2024² maka berapa hasil dari <math>\sqrt{2n-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n &= 2023^2+2024^2 \\ &= 2023^2+(2023+1)^2 \\ \text{misalkan 2023 = p} \\ n &= p^2+(p+1)^2 \\ &= p^2+p^2+2p+1 \\ &= 2p^2+2p+1 \\ \sqrt{2n-1} &= \sqrt{2(2p^2+2p+1)-1} \\ &= \sqrt{4p^2+4p+2-1} \\ &= \sqrt{4p^2+4p+1} \\ &= \sqrt{(2p+1)^2} \\ &= 2p+1 \\ &= 2(2023)+1 \\ &= 4046+1 \\ &= 4047 \\ \end{align} </math> </div></div> <ol start=80> <li>Tentukan nilai dari a+b+c merupakan bilangan bulat positif jika ab = 2, bc = 3 dan ac = 6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab \cdot bc \cdot ac &= 2 \cdot 3 \cdot 6 \\ (abc)^2 &= 36 \\ abc &= \pm 6 \\ abc &= 6 \\ \frac{abc}{ab} &= c = \frac{6}{2} = 3 \\ \frac{abc}{bc} &= a = \frac{6}{3} = 2 \\ \frac{abc}{ac} &= b = \frac{6}{6} = 1 \\ a+b+c &= 6 \\ \end{align} </math> </div></div> <ol start=81> <li>Tentukan nilai dari (a-c)^b jika <math>\frac{ab}{a+b} = \frac{1}{3}</math>, <math>\frac{bc}{b+c} = \frac{1}{4}</math> dan <math>\frac{ac}{a+c} = \frac{1}{9}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{ab}{a+b} &= \frac{1}{3} \\ \frac{a+b}{ab} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{b} + \frac{1}{a} &= 3 \\ \frac{bc}{b+c} &= \frac{1}{4} \\ \frac{b+c}{bc} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{b} &= 4 \\ \frac{ac}{a+c} &= \frac{1}{9} \\ \frac{a+c}{ac} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{a} &= 9 \\ \text{Misalkan 1/a = x, 1/b = y dan 1/c = z} \\ x+y &= 3 \\ y+z &= 4 \\ x+z &= 9 \\ x+y &= 3 \\ y+z &= 4 \\ x-z &= -1 \\ x-z &= -1 \\ x+z &= 9 \\ 2x &= 8 \\ x &= 4 \\ x-z &= -1 \\ 4-z &= -1 \\ z &= 5 \\ x+y &= 3 \\ 4+y &= 3 \\ y &= -1 \\ \frac{1}{a} &= 4 \\ a &= \frac{1}{4} \\ \frac{1}{b} &= -1 \\ b &= -1 \\ \frac{1}{c} &= 5 \\ c &= \frac{1}{5} \\ (a-c)^b &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=82> <li>Tentukan nilai dari a, b dan c jika <math>\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}</math> dan a+2b+3c=28?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan k untuk semua ketiga persamaan tersebut } \\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5} &= k \\ a+b &= 2k \\ a+c &= 4k \\ b+c &= 5k \\ 2a+b+c &= 6k \\ 2a+5k &= 6k \\ k &= 2a \\ a &= \frac{k}{2} \\ b &= \frac{3k}{2} \\ c &= \frac{7k}{2} \\ a+2b+3c &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ a &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ b &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ c &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=83> <li>Tentukan nilai dari (b+c)^a jika <math>\frac{a+b+c}{2} = \sqrt{a-2}+\sqrt{b-1}+\sqrt{c}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{a+b+c}{2} &= \sqrt{a-2}+\sqrt{b-1}+\sqrt{c} \\ a+b+c &= 2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}) \\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c} &= 0 \\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1 &= 0 \\ (\sqrt{a-2}-1)^2+(\sqrt{b-1}-1)^2+(\sqrt{c}-1)^2 &= 0 \\ (\sqrt{a-2}-1)^2 &= 0 \\ \sqrt{a-2}-1 &= 0 \\ \sqrt{a-2} &= 1 \\ a-2 &= 1 \\ a &= 3 \\ (\sqrt{b-1}-1)^2 &= 0 \\ \sqrt{b-1}-1 &= 0 \\ \sqrt{b-1} &= 1 \\ b-1 &= 1 \\ b &= 1 \\ (\sqrt{c}-1)^2 &= 0 \\ \sqrt{c}-1 &= 0 \\ \sqrt{c} &= 1 \\ c &= 1 \\ (b+c)^a &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=84> <li>x dan y merupakan bilangan tak nol. Jika xy = <math>\frac{x}{y}</math> = x-y maka berapa nilai x+y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy &= \frac{x}{y} \\ y^2 &= 1 \\ y^2 - 1 &= 0 \\ (y-1)(y+1) &= 0 \\ y = 1 &\text{ atau } y = -1 \\ \frac{x}{y} &= x-y \\ x &= xy-y^2 \\ x-xy &= -y^2 \\ x(1-y) &= -y^2 \\ x &= \frac{-y^2}{1-y} \\ \text{cek y=1 } \\ x &= \frac{-1^2}{1-1} \\ \text{tidak memenuhi syarat } \\ \text{cek y=-1 } \\ x &= \frac{-(-1)^2}{1-(-1)} \\ &= \frac{-1}{2} \\ x+y &= -1-\frac{1}{2} \\ &= -\frac{3}{2} \\ \end{align} </math> </div></div> <ol start=86> <li>Berapa nilai x dari <math>(\frac{a}{b})^3+(\frac{b}{a})^3 = 2\sqrt{x}</math> jika <math>\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b} \\ \frac{a+b}{ab} &= \frac{1}{a+b} \\ (a+b)^2 &= ab \\ a^2+2ab+b^2 &= ab \\ a^2+b^2 &= -ab \\ \text{misalkan } \frac{a}{b}+\frac{b}{a} = n \\ \frac{a}{b}+\frac{b}{a} &= n \\ \frac{a^2+b^2}{ab} &= n \\ a^2+b^2 &= nab \\ n &= -1 \\ \frac{a}{b}+\frac{b}{a} &= n \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3(\frac{a}{b}+\frac{b}{a}) &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3n &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3 &= n^3-3n \\ &= (-1)^3-3(-1) \\ &= 2 \\ 2\sqrt{x} &= 2 \\ \sqrt{x} &= 1 \\ x &= 1 \\ \end{align} </math> </div></div> <ol start=87> <li>Berapa nilai m dari <math>x^2-mx-1=0</math> jika <math>\sqrt[3]{x_1}+\sqrt[3]{x_2}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x_1} &= a \\ x_1 &= a^3 \\ \sqrt[3]{x_2} &= b \\ x_2 &= b^3 \\ \sqrt[3]{x_1}+\sqrt[3]{x_2} &= 1 \\ a+b &= 1 \\ x^2-mx-1 &= 0 \\ x_1+x_2 &= m \\ x_1 \cdot x_2 &= -1 \\ x_1+x_2 &= m \\ a^3+b^3 &= m \\ x_1 \cdot x_2 &= -1 \\ a^3 \cdot b^3 &= -1 \\ (ab)^2 &= (-1)^3 \\ ab &= -1 \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ (1)^3 &= m+3(-1)(1) \\ 1 &= m-3 \\ m &= 4 \\ \end{align} </math> </div></div> <ol start=88> <li>Berapa nilai <math>\frac{x_1}{x_2}</math> dari <math>ax^2-18x-b=0</math> jika <math>ab=45</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab &= 45 \\ b &= \frac{45}{a} \\ ax^2-18x-b &= 0 \\ ax^2-18x-\frac{45}{a} &= 0 \\ a^2x^2-18ax-45 &= 0 \\ (ax-3)(ax-15) &= 0 \\ ax-3 &= 0 \\ x &= \frac{3}{a} \\ ax-15 &= 0 \\ x &= \frac{15}{a} \\ \frac{x_1}{x_2} &= \frac{\frac{3}{a}}{\frac{15}{a}} \\ &= \frac{3}{15} \\ &= \frac{1}{5} \\ \frac{x_1}{x_2} &= \frac{\frac{15}{a}}{\frac{3}{a}} \\ &= \frac{15}{3} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=89> <li>Jika <math>\frac{u_3}{u_1+u_2} = \frac{7}{8}</math> merupakan barisan aritmetika maka berapa dari <math>\frac{u_2+u_3}{u_1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{u_3}{u_1+u_2} &= \frac{7}{8} \\ \frac{a+2b}{a+a+b} &= \frac{7}{8} \\ \frac{a+2b}{2a+b} &= \frac{7}{8} \\ 8(a+2b) &= 7(2a+b) \\ 8a+16b &= 14a+7b \\ 9b &= 6a \\ b &= \frac{2a}{3} \\ \frac{u_2+u_3}{u_1} &= \frac{a+b+a+2b}{a} \\ &= \frac{2a+3b}{a} \\ &= \frac{2a+3(\frac{2a}{3})}{a} \\ &= \frac{2a+2a}{a} \\ &= \frac{4a}{a} \\ &= 4 \\ \end{align} </math> </div></div> <ol start=90> <li>Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7p+q}{2p+q} &= \frac{17p+q}{7p+q} \\ (7p+q)^2 &= (17p+q)(2p+q) \\ 49p^2+14pq+q^2 &= 34p^2+19pq+q^2 \\ 15p^2 &= 5pq \\ 3p &= q \\ \frac{7p+q}{2p+q} &= \frac{7p+3p}{2p+3p} \\ &= \frac{10p}{5p} \\ &= 2 \\ \end{align} </math> </div></div> <ol start=91> <li>Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{rataan geometris } \\ \sqrt{a \cdot b} &= b-24 \\ a \cdot b &= (b-24)^2 \\ \text{rataan aritmatik } \\ \frac{a+b}{2} &= a+15 \\ a+b &= 2(a+15) \\ a+b &= 2a+30 \\ a &= b-30 \\ a \cdot b &= (b-24)^2 \\ (b-30)b &= (b-24)^2 \\ b^2-30b &= b^2-48b+576 \\ 18b &= 576 \\ b &= 32 \\ a &= b-30 \\ &= 32-30 \\ &= 2 \\ a+b &= 32+2 \\ &= 34 \\ \end{align} </math> </div></div> <ol start=91> <li>Segitiga lancip ABC dengan <math>\frac{a^4+b^4+c^4+a^2b^2}{c^2(a^2+b^2)}=2</math>. tentukan nilai sudut C?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{syarat segitiga lancip semua sudut masing-masing kurang dari } 90^\circ \\ c^2 &= a^2+b^2-2ab cos C \\ cos C &= \frac{a^2+b^2-c^2}{2ab} \\ a^4+b^4+c^4+a^2b^2 &= 2c^2(a^2+b^2) \\ a^4+b^4+a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-2c^2(a^2+b^2)+(c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= (ab)^2 \\ a^2+b^2-c^2 &= \pm ab \\ cos C &= \pm \frac{ab}{2ab} \\ &= \pm \frac{1}{2} \\ &= \frac{1}{2} \text{ (karena sudut harus kurang dari } 90^\circ) \\ C &= 60^\circ \\ \end{align} </math> </div></div> <ol start=92> <li>Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan panjang DA dan EA } = x \text{ dan panjang AB } = y \\ \text{luas segitiga CAB } &= \frac{CA \cdot AB}{2} \\ 45 &= \frac{(x+9)(x+5)}{2} \\ 90 &= x^2+14x+45 \\ x^2+14x &= 45 \\ y^2 &= (x+9)^2+(x+5)^2 \\ &= x^2+18x+81+x^2+10x+25 \\ &= 2x^2+28x+106 \\ &= 2(x^2+14x)+106 \\ &= 2(45)+106 \\ &= 196 \\ y &= 14 \\ \end{align} </math> jadi panjang BC adalah 14 cm </div></div> <ol start=93> <li>Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kita cari ukuran GC yaitu } \\ \frac{GC}{AD} &= \frac{CE}{DE} \\ \frac{GC}{15} &= \frac{6}{18} \\ GC &= 5 \\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC } \\ &= \frac{1}{2} \cdot BC \cdot CF - \frac{1}{2} \cdot GC \cdot CE \\ &= \frac{1}{2} \cdot 15 \cdot 18 - \frac{1}{2} \cdot 5 \cdot 6 \\ &= 135 - 15 \\ &= 120 \\ \end{align} </math> jadi luas daerah BFEG adalah 120 cm² </div></div> <ol start=94> <li>Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ AI &= \frac{1}{5} AF \\ &= \frac{1}{5} 10 \\ &= 2 \\ IF &= AF-AI \\ &= 10-2 \\ &= 8 \\ \text{luas trapesium AFGD } &= \frac{(AD+EF) \cdot AF}{2} \\ &= \frac{(4+6)10}{2} \\ &= 50 \\ \text{luas segitiga AID } &= \frac{AI \cdot AF}{2} \\ &= \frac{(2)4}{2} \\ &= 4 \\ \text{luas segitiga IFG } &= \frac{IF \cdot FG}{2} \\ &= \frac{(8)6}{2} \\ &= 24 \\ \text{luas daerah segitiga IGD } &= \text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG } \\ &= 50-4-24 \\ &= 22 \\ \end{align} </math> jadi luas daerah segitiga IGD adalah 22 cm² </div></div> <ol start=96> <li>Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-<math>\pi</math>) cm³ maka berapa luas permukaan kerucut tersebut?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ \text{volume balok} \\ V_b &= x^2(12) \\ \text{volume kerucut} \\ V_b &= \frac{1}{3}\pi x^2(12) \\ &= 4\pi x^2 \\ V_{b-k} &= Vb-Vk \\ 100(3-\pi) &= 12x^2-4\pi x^2 \\ 100(3-\pi) &= 4x^2(3-\pi) \\ x^2 &= 25 \\ x &= 5 \\ s &= \sqrt{12^2+5^2} \\ &= \sqrt{144+25} \\ &= \sqrt{169} \\ &= 13 \\ \text{luas permukaan kerucut } &= \pi r(r+s) \\ &= \pi(5)(5+13) \\ &= 90\pi \\ \end{align} </math> jadi luas daerah permukaan kerucut adalah 90<math>\pi</math> cm² </div></div> <ol start=97> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 3a+1 \\ B &= 3b+2 \\ A(A+1)+3B \\ (3a+1)(3a+1+1)+3(3b+2) \\ (3a+1)(3a+2)+9b+6 \\ 9a^2+9a+2+9b+6 \\ 9a^2+9a+9b+8 \\ 9(a^2+a+b)+8 \\ \text{sisa pembagiannya adalah } 8 \\ \end{align} </math> </div></div> <ol start=98> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 9a+7 \\ B &= 9b+8 \\ A(A-5)+9B \\ (9a+7)(9a+7-5)+9(9b+8) \\ (9a+7)(9a+2)+81b+72 \\ 81a^2+81a+14+81b+72 \\ 81a^2+81a+81b+86 \\ 81a^2+81a+81b+81+5 \\ 81(a^2+a+b+1)+5 \\ \text{sisa pembagiannya adalah } 5 \\ \end{align} </math> </div></div> <ol start=99> <li>Jika <math>\begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix}</math> maka berapa hasil dari A²¹+A²⁵+A⁴⁶?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A^2 &= A \cdot A \\ &= \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \\ A^3 &= A^2 \cdot A \\ &= \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ &= - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= -I \\ A^{21}+A^{25}+A^{46} &= A^{21} \cdot (I+A^4+A^{25}) \\ &= A^{21} \cdot (I+A^3 \cdot A +A^{24} \cdot A) \\ &= (A^3)^7 \cdot (I+A^3 \cdot A +(A^3)^8 \cdot A) \\ &= (-I)^7 \cdot (I-I \cdot A +(-I)^8 \cdot A) \\ &= -I \cdot (I-A+A) \\ &= -I \cdot I \\ &= -I \\ &= -\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ \end{align} </math> </div></div> <ol start=100> <li>Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida?</li></ol> : bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15 : ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9). : ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8) : maka Ida menulis bilangan 8. <ol start=101> <li>Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9?</li></ol> : Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu. : untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja. : untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja. : Jadi banyaknya bilangan mungkin 2. <ol start=102> <li>Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli!</li></ol> ;cara 1 # 8¹ = 1 # 8² = 1 (8²=8¹x8¹ sama dengan 1x1) # 8³ = 1 (8³=8¹x8² sama dengan 1x1) # 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2) # 8⁵ = 1 # 8ⁿ = 1 (semua n untuk bilangan asli) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli ;cara 2 # 8ⁿ = b mod 7 # 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil) # (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya) # 8ⁿ = 1ⁿ mod 7 # 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli <ol start=103> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 5?</li></ol> ;cara 1 # 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 # 7¹ = 7 (sisa 1) # 7² = 49 (sisa 2) # 7³ = 343 (sisa 3) # 7⁴ = 2,401 (sisa 0) # 7⁵ = 16,807 # 7⁶ = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3 ;cara 2 :17¹ = 2 :17² = 4 :17³ = 3 :17⁴ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3 :17⁹⁹ = (17⁴)²⁴ x 17³ Untuk 17⁴ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 3. Jadi 17⁹⁹ dibagi 7 bersisa 3 ;cara 3 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 17⁴ ::17⁴ = 1 mod 5 ::(17⁴)²⁴ = 1²⁴ mod 5 ::17⁹⁶ = 1²⁴ mod 5 ::17⁹⁶ = 1 mod 5 ::17⁹⁶ x 17³ = 1 x 17³ mod 5 ::17⁹⁹ = 17³ mod 5 ::17⁹⁹ = 17 x 17 x 17 mod 5 ::17⁹⁹ = 2 x 2 x 2 mod 5 ::17⁹⁹ = 8 mod 5 ::17⁹⁹ = 3 mod 5 Jadi 17⁹⁹ dibagi 5 bersisa 3 <ol start=104> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 7?</li></ol> ;cara 1 :17¹ = 3 :17² = 2 :17³ = 6 :17⁴ = 4 :17⁵ = 5 :17⁶ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3 :17⁹⁹ = (17⁶)¹⁶ x 17³ Untuk 17⁶ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 6. Jadi 17⁹⁹ dibagi 7 bersisa 6 ;cara 2 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 17⁶ ::17⁶ = 1 mod 7 ::(17⁶)¹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1 mod 7 ::17⁹⁶ x 17³ = 1 x 17³ mod 7 ::17⁹⁹ = 17³ mod 7 ::17⁹⁹ = 17 x 17 x 17 mod 7 ::17⁹⁹ = 3 x 3 x 3 mod 7 ::17⁹⁹ = 27 mod 7 ::17⁹⁹ = 6 mod 7 Jadi 17⁹⁹ dibagi 7 bersisa 6 <ol start=105> <li>Berapa hasil sisa dari 41²⁰²⁴ dibagi 33?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 41^{2024} &= 41^{2024} \text{ mod } 33 \\ &= (33 \times 3 + 2)^{2024} \text{ mod } 33 \\ &= 2^{2024} \text{ mod } 33 \\ &= 2^{2020} 2^4 \text{ mod } 33 \\ &= (2^5)^{404} 2^4 \text{ mod } 33 \\ &= (33 - 1)^{404} 2^4 \text{ mod } 33 \\ &= (-1)^{404} 2^4 \text{ mod } 33 \\ &= 2^4 \text{ mod } 33 \\ &= 16 \text{ mod } 33 \\ \text{Jadi hasil sisa adalah } 16 \\ \end{align} </math> </div></div> <ol start=106> <li>Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 99^{99} &= (3^2 \times 11)^{99} \\ &= 3^{198} \times 11^{99} \\ 243^n &= (3^5)^n \\ &= 3^{5n} \\ \text{agar bisa membagi, maka} \\ 5n &= 198 \\ n &= 39.6 \\ \text{jadi bilangan n terbesar adalah } 39 \\ \end{align} </math> </div></div> <ol start=107> <li>Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸?</lu></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 88^{88} &= (8 \times 11)^{88} \\ &= 8^{88} \times 11^{88} \\ &= 8^{87} \times 8 \times 11^{88} \\ &= (8^3)^{29} \times 8 \times 11^{88} \\ &= 512^{29} \times 8 \times 11^{88} \\ 512^n &= 512^{29} \\ \text{jadi bilangan n terbesar adalah } 29 \\ \end{align} </math> </div></div> <ol start=108> <li>Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6!</li></ol> ; cara 1 : KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. : N dibagi 3 sisa 1 : N dibagi 5 sisa 2 : N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a : KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70 : KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21 : KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97 ; cara 2 : Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35 : kemudian ditambahkan sisa masing-masing sesuai dengan KPK. : KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, <b>97</b> : KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, <b>97</b> : KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, <b>97</b> Jadi bilangan bulat positif adalah 97 :: NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 <ol start=109> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |- | 4 || 0 || Semua isi ember B dibuang |- | 1 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. <ol start=110> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |} nah ada ember A berisi 4 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |- | 1 || 0 || Semua isi ember B dibuang |- | 0 || 1 || Tuangkan 1 l dari ember A ke B sehingga ember A kosong |- | 3 || 1 || Isikan 3 l ke ember A |- | 0 || 4 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |} nah ada ember B berisi 4 liter. # Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kemungkinan 1} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot +3 \cdot (\frac{4}{3}^x)^2 &= 7 \cdot \frac{4}{3}^x \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-6}></math>? [[Kategori:Soal-Soal Matematika]] k10nd8frnfjdi905zyf9p5r2frrmr0s 117484 117483 2026-07-13T00:56:01Z Akuindo 8654 117484 wikitext text/x-wiki contoh soal <ol start=1> <li>Berapa hasil dari <math>\sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Misalkan 2020 = p} \\ \sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64} &= \sqrt{(2020-5) \cdot (2020-3) \cdot (2020+3) \cdot (2020+5) + 64} \\ &= \sqrt{(p-5) \cdot (p-3) \cdot (p+3) \cdot (p+5) + 64} \\ &= \sqrt{(p-5) \cdot (p+5) \cdot (p-3) \cdot (p+3) + 64} \\ &= \sqrt{(p^2-25) \cdot (p^2-9) + 64} \\ &= \sqrt{p^4-34p^2+ 225 + 64} \\ &= \sqrt{p^4-34p^2+ 289} \\ &= \sqrt{(p^2-17)^2} \\ &= p^2-17 \\ &= 2020^2-17 \\ &= (2000+20)^2-17 \\ &= 4.000.000+80.000+400-17 \\ &= 4.080.383 \\ \end{align} </math> </div></div> <ol start=2> <li>Berapa nilai x dari <math>\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} = \frac{9}{10}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} &= \frac{9}{10} \\ \text{misalkan untuk } \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} = p \\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} &= p \\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}} &= p^2 \\ x^2-x-p &= p^2 \\ x^2-2x+1+x-1 &= p^2+p \\ (x-1)^2+(x-1) &= p^2+p \\ x-1 &= p \\ \text{misalkan untuk } \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}} &= q^3 \\ x^2 q &= q^3 \\ x^2 &= q^2 \\ x &= q \\ \frac{x-1}{x} &= \frac{9}{10} \\ x &= 10 \\ \end{align} </math> </div></div> <ol start=3> <li>Berapa nilai x dari <math>(\frac{x}{x+10})^{x+10}=\frac{1}{1024}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{x+10}{x})^{-(x+10)} &= (1024)^{-1} \\ (\frac{x+10}{x})^{x+10} &= 1024 \\ (\frac{x+10}{x})^{x+10} &= 2^{10} \\ (\frac{x+10}{x})^{\frac{x+10}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (\frac{1}{2})^{-1} \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (1+(-\frac{1}{2}))^{(1+(-\frac{2}{1}))} \\ \frac{10}{x} &= -\frac{1}{2} \\ x &= -20 \\ \end{align} </math> </div></div> <ol start=4> <li>Berapa nilai x dari <math>x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{2}+\sqrt{x+\frac{1}{4}} &= (\sqrt{x+\frac{1}{4}})^2+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+(\frac{1}{2})^2 \\ &= (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 \\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}} &= 4 \\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2} &= 4 \\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 4 \\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 &= 4 \\ \sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 2 \\ \sqrt{x+\frac{1}{4}} &= \frac{3}{2} \\ x+\frac{1}{4} &= \frac{9}{4} \\ x &= 2 \\ \end{align} </math> </div></div> <ol start=5> <li>Berapa nilai x dari <math>\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^3}{\sqrt{8-x^2}}+x^2-8 &= 0 \\ \frac{x^3}{\sqrt{8-x^2}} &= 8-x^2 \\ x^3 &= (8-x^2)^{\frac{3}{2}} \\ x &= (8-x^2)^{\frac{1}{2}} \\ x^2 &= 8-x^2 \\ 2x^2-8 &= 0 \\ x^2-4 &= 0 \\ (x-2)(x+2) &= 0 \\ \text{membuktikan } \\ x=2 \text{ maka hasilnya 0 } \\ x=-2 \text{ maka hasilnya -8 } \\ \text{jadi } x=2 \\ \end{align} </math> </div></div> <ol start=6> <li>Berapa nilai x dari <math>\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} = 5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} &= 5 \\ \frac{x^{50}+x^{60}+x^{70}}{31}} &= 5^5 \\ x^{50}+x^{60}+x^{70} &= 5^5 \cdot 31 \\ x^{50}(1+x^{10}+x^{20}) &= 5^5 \cdot 31 \\ (x^{10}^5)(1+x^{10}+(x^{10}^2) &= 5^5 \cdot 31 \\ \text{ misalkan } x^{10} = a \\ a^5(1+a+a^2) &= 5^5 \cdot 31 \\ a &= 5 \\ x^{10} &= 5 \\ x &= ^5 log 10 \\ \end{align} </math> </div></div> <ol start=7> <li>Berapa nilai x dari <math>\sqrt{3x+5+\sqrt{4x+5}} = x</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{3x+5+\sqrt{4x+5}} &= x \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \text{misalkan } \sqrt{4x+5}=y \text{ dan } 4x+5=y^2 \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \sqrt{y^2+y-x} &= x \\ y^2+y &= x^2+x \\ y=x \\ 4x+5 &= y^2 \\ 4x+5 &= x^2 \\ x^2-4x-5 &= 0 \\ (x-5)(x+1) &= 0 \\ x=5 &\text{ atau } x=-1 \text{ (TM) } \\ \end{align} </math> </div></div> <ol start=8> <li>Berapa nilai x dari <math>\sqrt{1+\sqrt{1+x}} = \sqrt[3]{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{1+\sqrt{1+x}} &= \sqrt[3]{x} \\ \sqrt[3]{x} &= n \\ x &= n^3 \\ \sqrt{1+\sqrt{1+n^3}} &= n \\ 1+\sqrt{1+n^3} &= n^2 \\ \sqrt{1+n^3} &= n^2-1 \\ 1+n^3 &= n^4-2n^2+1 \\ n^4-n^3-2n^2 &= 0 \\ n^2(n^2-n-2) &= 0 \\ n^2(n-2)(n+1) &= 0 \\ n=0, n=2 \text{ atau } n=-1 \\ n &= 0 \\ x &= 0^3 \\ &= 0 \\ n &= 2 \\ x &= 2^3 \\ &= 8 \\ n &= -1 \\ x &= (-1)^3 \\ &= -1 \\ \text{yang paling mungkin untuk nilai x adalah } 8 \\ \end{align} </math> </div></div> <ol start=9> <li>Berapa nilai x dari <math>\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} &= \frac{\sqrt{1+x}}{\sqrt{x}} \\ \sqrt{x}(\sqrt{1+x}+\sqrt{x}) &= (\sqrt{1+x}-\sqrt{x})\sqrt{1+x} \\ \sqrt{x(1+x)}+x &= 1+x-\sqrt{x(1+x)} \\ 2\sqrt{x(1+x)} &= 1 \\ \sqrt{x(1+x)} &= \frac{1}{2} \\ x(1+x) &= \frac{1}{4} \\ x^2+x &= \frac{1}{4} \\ 4x^2+4x &= 1 \\ 4x^2+4x-1 &= 0 \\ x &= \frac{-4 \pm \sqrt{4^2-4(4)(-1)}}{2(4)} \\ &= \frac{-4 \pm \sqrt{32}}{8} \\ &= \frac{-4 \pm 4\sqrt{2}}{8} \\ &= \frac{-1 \pm \sqrt{2}}{2} \\ \text{karena akar x harus minimal nol jadi } x = \frac{-1+\sqrt{2}}{2} \\ \end{align} </math> </div></div> <ol start=10> <li>Berapa nilai x dari <math>\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x-\sqrt{x+1}}{x+\sqrt{x+1}} &= \frac{11}{19} \\ \text{misalkan } \sqrt{x+1}=y \text{ dan } x=y^2-1 \\ \frac{y^2-1-y}{y^2-1+y} &= \frac{11}{19} \\ 19(y^2-y-1) &= 11(y^2+y-1) \\ 19y^2-19y-19 &= 11y^2+11y-11 \\ 8y^2-30y-8 &= 0 \\ 4y^2-15y-4 &= 0 \\ (4y+1)(y-4) &= 0 \\ y=-\frac{1}{4} \text{ (TM) atau } & y=4 \\ x &= 4^2-1 \\ &= 15 \\ \end{align} </math> </div></div> <ol start=11> <li>Berapa nilai x dari <math>\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} &= 98 \\ \text{misalkan } \sqrt{x^2-1}=y \\ \frac{x+y}{x-y}+\frac{x-y}{x+y} &= 98 \\ \frac{(x+y)^2+(x-y)^2}{(x-y)(x+y)} &= 98 \\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} &= 98 \\ \frac{2(x^2+y^2)}{x^2-y^2} &= 98 \\ \frac{x^2+y^2}{x^2-y^2} &= 49 \\ x^2+y^2 &= 49(x^2-y^2) \\ x^2+y^2 &= 49x^2-49y^2 \\ 48x^2 &= 50y^2 \\ 24x^2 &= 25y^2 \\ 24x^2 &= 25(\sqrt{x^2-1})^2 \\ 24x^2 &= 25(x^2-1) \\ 24x^2 &= 25x^2-25 \\ x^2 &= 25 \\ x &= \pm 5 \\ \end{align} </math> </div></div> <ol start=12> <li>Berapa nilai x dari <math>\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \sqrt{x}=y \text{ dan } x=y^2 \\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} &= \frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\sqrt{\frac{y^2}{y^2+y}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\frac{y}{\sqrt{y^2+y}} \\ y^2+y-\sqrt{(y^2+y)(y^2-y)} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^4-y^2} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^2(y^2-1)} &= \frac{5}{4}y \\ y(y+1)-y\sqrt{y^2-1} &= \frac{5}{4}y \\ y+1-\sqrt{y^2-1} &= \frac{5}{4} \\ -\sqrt{y^2-1} &= \frac{1}{4}-y \\ y^2-1 &= (\frac{1}{4}-y)^2 \\ y^2-1 &= \frac{1}{16}-\frac{1}{2}y+y^2 \\ -1 &= \frac{1}{16}-\frac{1}{2}y \\ \frac{1}{2}y &= \frac{1}{16}+1 \\ \frac{1}{2}y &= \frac{17}{16} \\ y &= \frac{17}{8} \\ x &= (\frac{17}{8})^2 \\ &= \frac{289}{64} \\ \end{align} </math> </div></div> <ol start=13> <li>Berapa nilai x dari <math>\sqrt[4]{62+x}+\sqrt[4]{275-x}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ misalkan } \sqrt[4]{62+x}=a, 62+x=a^4, \sqrt[4]{275-x}=b \text{ dan } 275-x=b^4 \\ a+b &= 7 \\ (a+b)^2 &= 49 \\ a^2+b^2+2ab &= 49 \\ a^2+b^2 &= 49-2ab \\ a^4+b^4 &= 62+x+275-x \\ (a^2+b^2)^2-2(ab)^2 &= 337 \\ (49-2ab)^2-2(ab)^2 &= 337 \\ 2401-196ab+4(ab)^2-2(ab)^2 &= 337 \\ 2(ab)^2-196ab+2064 &= 0 \\ (ab)^2-98ab+1032 &= 0 \\ (ab-12)(ab-86) &= 0 \\ ab = 12 \text{ atau } & ab = 86 \text{ (TM) karena hasil kali maksimum yaitu 12 } \\ ab =12 \text{ dan } a+b=7 \\ a+b &= 7 \\ b &= 7-a \\ ab &= 12 \\ a(7-a) &= 12 \\ -a^2+7a &= 12 \\ a^2-7a+12 &= 0 \\ (a-3)(a-4) &= 0 \\ a=3 \text{ atau } & a=4 \\ a=3, b=4 \\ 62+x &= a^4 \\ 62+x &= (3)^4 \\ 62+x &= 81 \\ x &= 19 \\ a=4, b=3 \\ 62+x &= a^4 \\ 62+x &= (4)^4 \\ 62+x &= 256 \\ x &= 194 \\ \end{align} </math> </div></div> <ol start=14> <li>Berapa nilai x dari <math>\sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2} &= 7 \\ (\sqrt[3]{8+x})^2-\sqrt[3]{8+x} \sqrt[3]{27-x}+(\sqrt[3]{27-x})^2 &= 7 \\ \text{misalkan } \sqrt[3]{8+x}=a, 8+x=a^3, \sqrt[3]{27-x}=b \text{ dan } 27-x=b^3 \\ a^2-ab+b^2 &= 7 \\ a^3+b^3 &= 8+x+27-x \\ &= 35 \\ a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ 35 &= (a+b)(7) \\ a+b &= 5 \\ b &= 5-a \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ 5^3 &= 35+3ab(5) \\ 125 &= 35+15ab \\ 80 &= 15ab \\ ab &= 6 \\ a(5-a) &= 6 \\ 5a-a^2 &= 6 \\ a^2-5a+6 &= 6 \\ (a-2)(a-3) &= 6 \\ a=2 &\text{ atau } a=3 \\ a=2, b=3 \text{ dan } a=3,b=2 \\ 8+x &= a^3 \\ &= 2^3 \\ &= 8 \\ x &= 0 \\ 8+x &= a^3 \\ &= 3^3 \\ &= 27 \\ x &= 19 \\ \end{align} </math> </div></div> <ol start=15> <li>Berapa nilai x dari <math>3^x+5^x-9^x+15^x-25^x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3^x+5^x-9^x+15^x-25^x &= 1 \\ 3^x+5^x-(3^2)^x+(3 \cdot 5)^x-(5^2)^x &= 1 \\ 3^x+5^x-(3^x)^2+(3^x \cdot 5^x)-(5^x)^2 &= 1 \\ \text{misalkan } 3^x=a \text{ dan } 5^x=b \\ a+b-a^2+ab-b^2 &= 1 \\ a^2-ab+b^2-a-b+1 &= 0 \\ 2a^2-2ab+2b^2-2a-2b+2 &= 0 \\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1 &= 0 \\ (a-b)^2+(a-1)^2+(b-1)^2 &= 0 \\ a-b=0; a-1=0; b-1 &= 0 \\ a=b &= 1 \\ 3^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> <ol start=16> <li>Berapa nilai x dari <math>^6log x^2+^{6x}log \frac{6}{x}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ \text{misalkan } 6x=a \text{ maka } x=\frac{a}{6} \\ ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ ^6log (\frac{a}{6})^2+^{6 \frac{a}{6}}log \frac{6}{\frac{a}{6}} &= 1 \\ ^6log \frac{a^2}{6^2}+^alog \frac{6^2}{a} &= 1 \\ ^6log a^2-^6log 6^2+^alog 6^2-^alog a &= 1 \\ 2 ^6log a-2 ^6log 6+2 ^alog 6-^alog a &= 1 \\ 2 ^6log a-2+2 \frac{1}{^6log a}-1 &= 1 \\ 2 ^6log a+2 \frac{1}{^6log a}-4 &= 0 \\ 2 ^6log^2 a-4 ^6log a+2 &= 0 \\ ^6log^2 a-2 ^6log a+1 &= 0 \\ (^6log a-1)^2 &= 0 \\ ^6log a &= 1 \\ a &= 6 \\ x &= \frac{a}{6} \\ &= \frac{6}{6} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=17> <li>Berapa nilai x dari (x+500)³+x=20?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+500)^3+x &= 20 \\ \text{misalkan } a=x+500 \text{ maka } x=a-500 \\ a^3+a-500 &= 20 \\ a^3+a &= 520 \\ a(a^2+1) &= 8 \cdot 65 \\ a(a^2+1) &= 8(64+1) \\ a(a^2+1) &= 8(8^2+1) \\ a &= 8 \\ x &= 8-500 \\ &= -492 \\ \end{align} </math> </div></div> <ol start=18> <li>Berapa nilai x dari <math>\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2} &= 0 \\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}} &= \frac{1}{2} \\ \frac{x^n+4^n}{x^n+16^n} &= (\frac{1}{2})^n \\ \frac{x^n+4^n}{x^n+16^n} &= \frac{1}{2^n} \\ 2^n(x^n+4^n) &= x^n+16^n \\ 2^n(x^n+2^{2n}) &= x^n+2^{4n} \\ 2^n \cdot x^n+2^{3n} &= x^n+2^{4n} \\ 2^n \cdot x^n-x^n &= 2^{4n}-2^{3n} \\ x^n(2^n-1) &= 2^{3n}(2^n-1) \\ x^n &= 2^{3n} \\ x^n &= (2^3)^n \\ x^n &= 8^n \\ x &= 8 \\ \end{align} </math> </div></div> <ol start=19> <li>Berapa hasil dari <math>\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ x &= \frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5 \cdot 6}+\sqrt{5 \cdot 5}+\sqrt{6 \cdot 4}+\sqrt{5 \cdot 4}}{\sqrt{5 \cdot 4}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5} \cdot \sqrt{6}+\sqrt{5} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{4}}{2 \cdot \sqrt{5}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{6} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{5}+\sqrt{5} \cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}} \\ &= \frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ &= \frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ \frac{1}{x} &= \frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}} \\ &= \frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ &= \frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1} \\ &= \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5} \\ &= \sqrt{6}-\sqrt{4} \\ &= \sqrt{6}-2 \\ x &= \frac{1}{\sqrt{6}-2} \\ &= \frac{\sqrt{6}+2}{6-4} \\ &= \frac{\sqrt{6}+2}{2} \\ &= 1+\frac{\sqrt{6}}{2} \\ \end{align} </math> </div></div> <ol start=20> <li>Berapa hasil dari <math>(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5 \\ \text{misalkan } x=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ x &= \frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ &= \frac{\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}} \\ &= \frac{\sqrt{3}+1}{4} \\ 4x &= \sqrt{3}+1 \\ 4x-1 &= \sqrt{3} \\ (4x-1)^2 &= 3 \\ 16x^2-8x+1 &= 3 \\ 16x^2 &= 8x+2 \\ 8x^2 &= 4x+1 \\ x^2 &= \frac{4x+1}{8} \\ *cara 1 \\ x^3 &= x \cdot x^2 \\ &= x(\frac{4x+1}{8}) \\ &= \frac{4x^2+x}{8} \\ &= \frac{4x^2}{8}+\frac{x}{8} \\ &= \frac{4(\frac{4x+1}{8})}{8}+\frac{x}{8} \\ &= \frac{16x+4}{64}+\frac{x}{8} \\ &= \frac{4x+1}{16}+\frac{x}{8} \\ &= \frac{4x+1+2x}{16} \\ &= \frac{6x+1}{16} \\ x^5 &= x^2 \cdot x^3 \\ &= (\frac{4x+1}{8})(\frac{6x+1}{16}) \\ &= \frac{24x^2+10x+1}{128} \\ &= \frac{24x^2}{128}+\frac{10x+1}{128} \\ &= \frac{24(\frac{4x+1}{8})}{128}+\frac{10x+1}{128} \\ &= \frac{96x+24}{1024}+\frac{10x+1}{128} \\ &= \frac{96x+24+80x+8}{1024} \\ &= \frac{176x+32}{1024} \\ &= \frac{176x}{1024}+\frac{32}{1024} \\ &= \frac{176}{1024}(\frac{\sqrt{3}+1}{4})+\frac{32}{1024} \\ &= \frac{44(\sqrt{3}+1)}{1024}+\frac{32}{1024} \\ &= \frac{44\sqrt{3}+44}{1024}+\frac{32}{1024} \\ &= \frac{76+44\sqrt{3}}{1024} \\ &= \frac{19+11\sqrt{3}}{256} \\ *cara 2 \\ x^4 &= (x^2)^2 \\ &= (\frac{4x+1}{8})^2 \\ &= \frac{16x^2+8x+1}{64} \\ &= \frac{16x^2}{64}+\frac{8x}{64}+\frac{1}{64} \\ &= \frac{x^2}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{\frac{4x+1}{8}}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{4x}{32}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{8}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{4}+\frac{3}{64} \\ x^5 &= x \cdot x^4 \\ &= (\frac{\sqrt{3}+1}{4})(\frac{x}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\frac{\sqrt{3}+1}{4}}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\sqrt{3}+1}{16}+\frac{3}{64}) \\ &= \frac{(\sqrt{3}+1)^2}{64}+(\frac{\sqrt{3}+1}{4})\frac{3}{64} \\ &= \frac{3+2\sqrt{3}+1}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{4+2\sqrt{3}}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{16+8\sqrt{3}}{256}+\frac{3\sqrt{3}+3}{256} \\ &= \frac{19+11\sqrt{3}}{256} \\ \end{align} </math> </div></div> <ol start=21> <li>Berapa hasil dari <math>\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4} &= \frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots) \\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots &= \frac{1}{1-\frac{1}{4}} \\ &= \frac{4}{3} \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{4}{3}) \\ &= \frac{1}{4}+\frac{16}{3} \\ &= \frac{67}{12} \\ x &= \frac{67}{9} \\ \end{align} </math> </div></div> <ol start=22> <li>Berapa nilai y-x jika <math>\frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} = \frac{x}{y}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} &= \frac{x}{y} \\ \frac{\frac{106 \times 107}{2}}{\frac{106}{2}(4+109)} &= \frac{x}{y} \\ \frac{53 \times 107}{53 \times 113} &= \frac{x}{y} \\ y-x &= 113-107 = 6 \\ \end{align} </math> </div></div> <ol start=23> <li>Berapa angka satuan dari hasil 17²⁰²⁴?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan angka satuannya} \\ 17^1 &= 7 \\ 17^2 &= 9 \\ 17^3 &= 3 \\ 17^4 &= 1 \\ 17^5 &= 7 \\ 17^6 &= 9 \\ 17^7 &= 3 \\ 17^8 &= 1 \\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1} \end{align} </math> </div></div> <ol start=24> <li>Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ 1! + 2! + 3! + 4! + \dots + 2024! &= 1 + (1x2) + (1x2x3) + (1x2x3x4) + \dots + 2024! \\ &= 1 + 2 + 6 + 24 + 120 + 720 + \dots + 2024! \\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka } 1+2+6+24 = 33 \text{ jadi angka satuannya adalah } 3 \end{align} </math> </div></div> <ol start=25> <li>Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ \frac{1! + 2! + 3! + 4! + \dots + 2024!}{12} &= \frac{1 + 1x2 + 1x2x3 + 1x2x3x4 + \dots + 2024!}{12} \\ &= \frac{1 + 2 + 6 + 24 + \dots + 2024!}{12} \\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi } 1+2+6 = 9 \end{align} </math> </div></div> <ol start=26> <li>Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{penjumlahan} \\ 1+1+1+1+ \dots \text{ (sebanyak 88 buah) }+x+y &= A \\ 88+x+y &= A \\ \text{perkalian} \\ 1 \times 1 \times 1 \times \dots \text{ (sebanyak 88 buah) }\times x \times y &= A \\ x \times y &= A \\ 88+x+y &= xy \\ xy-y &= 88+x \\ y(x-1) &= 88+x \\ y &= \frac{88+x}{x-1} \\ \text{uji selidiki untuk x=2} \\ y &= \frac{88+2}{2-1} \\ &= 90 \\ \text{buktikan} \\ 88+x+y &= xy \\ 88+2+90 &= 2(90) \\ 180 &= 180 \\ \text{terbukti} \\ \text{nilai A adalah } 180 \\ \end{align} </math> </div></div> <ol start=27> <li>Berapakah nilai x, y dan z dari <math>x+y-z=1, x^2+y^2-z^2=-5 \text{ dan } x^3+y^3-z^3=-53</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+y-z &= 1 \\ x+y &= z+1 \\ x^2+2xy+y^2 &= z^2+2z+1 \\ x^2+y^2-z^2 &= 2z+1-2xy \\ -5 &= 2z+1-2xy \\ 2xy &= 2z+6 \\ xy &= z+3 \\ x^2+y^2-z^2 &= -5 \\ x^2+y^2 &= z^2-5 \\ x^3+y^3-z^3 &= -53 \\ (x+y)(x^2-xy+y^2)-z^3+53 &= 0 \\ (x+y)(x^2+y^2-xy)-z^3+53 &= 0 \\ (z+1)(z^2-5-(z+3))-z^3+53 &= 0 \\ (z+1)(z^2-z-8)-z^3+53 &= 0 \\ z^3-z^2-8z+z^2-z-8-z^3+53 &= 0 \\ -9z+45 &= 0 \\ -9z &= -45 \\ z &= 5 \\ x+y &= 5+1 \\ x+y &= 6 \\ x &= 6-y \\ xy &= 5+3 \\ xy &= 8 \\ (6-y)y &= 8 \\ 6y-y^2 &= 8 \\ y^2-6y+8 &= 0 \\ (y-4)(y-2) &= 0 \\ y=4 \text{ atau } y=2 \\ \text{jika } y=4 \\ x+y &= z+1 \\ x+4 &= 5+1 \\ x &= 2 \\ \text{jika } y=2 \\ x+y &= z+1 \\ x+2 &= 5+1 \\ x &= 4 \\ \end{align} </math> </div></div> <ol start=28> <li>Berapakah nilai titik koordinat (x,y) dari <math>\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}</math> dan <math>\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x+y}+\sqrt{x-y} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+y}-\sqrt{x-y} &= \sqrt{\frac{52y}{3x}} \\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y}) &= \sqrt{\frac{432x}{13y}} \cdot \sqrt{\frac{52y}{3x}} \\ x+y-x+y &= \sqrt{\frac{432x \cdot 52y}{13y \cdot 3x}} \\ 2y &= \sqrt{144 \cdot 4} \\ 2y &= \sqrt{576} \\ 2y &= 24 \\ y &= 12 \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13(12)}} \\ x+12+x-12+2 \cdot \sqrt{x+12} \cdot \sqrt{x-12} &= \frac{36x}{13} \\ 2x+2 \sqrt{x^2-144} &= \frac{36x}{13} \\ 2(x+\sqrt{x^2-144}) &= \frac{36x}{13} \\ x+\sqrt{x^2-144} &= \frac{18x}{13} \\ \sqrt{x^2-144} &= \frac{5x}{13} \\ x^2-144 &= \frac{25x^2}{169} \\ \frac{144x^2}{169}-144 &= 0 \\ \frac{x^2}{169}-1 &= 0 \\ x^2-169 &= 0 \\ (x-13)(x+13) &= 0 \\ x_1=13 &\text{ atau } x_2=-13 \text{ (TM) karena } x>y \\ \end{align} </math> jadi titik koordinat (13,12) </div></div> <ol start=29> <li>Berapakah nilai dari <math>x^2-7x</math> jika <math>(x-2)^2+\frac{1}{(x-2)^2} = 11</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x-2)^2+\frac{1}{(x-2)^2} &= 11 \\ (x-2)^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2)^2} &= 11-2 \\ (x-2-\frac{1}{x-2})^2 &= 9 \\ x-2-\frac{1}{x-2} &= 3 \\ (x-2)^2-1 &= 3(x-2) \\ x^2-4x+4-1 &= 3x-6 \\ x^2-7x &= -9 \\ \end{align} </math> </div></div> <ol start=30> <li>Berapakah nilai dari <math>\frac{(x+y)^2(x+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)}</math> jika xy+yz+xz=1?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy+yz+xz &= 1 \\ x^2+xy+yz+xz &= x^2+1 \\ x(x+y)+z(x+y) &= x^2+1 \\ (x+y)(x+z) &= x^2+1 \\ \text{dengan pola yang sama } \\ (y+x)(y+z) &= y^2+1 \\ (x+z)(y+z) &= z^2+1 \\ \frac{(x+y)^2(y+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)} &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)} \\ &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)^2(y+z)^2(x+z)^2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=31> <li>Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} w+5 &= w+x+y+z+5 \\ x+4 &= w+x+y+z+5 \\ y+3 &= w+x+y+z+5 \\ z+2 &= w+x+y+z+5 \\ \text{jumlahkan keempat persamaan } \\ w+x+y+z+14 &= 4(w+x+y+z+5) \\ w+x+y+z+14 &= 4(w+x+y+z)+20 \\ 3(w+x+y+z) &= -6 \\ w+x+y+z &= -2 \\ \end{align} </math> </div></div> <ol start=32> <li>Berapakah nilai dari <math>\frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2}</math> jika <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3</math> dan x+y+z=xyz?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2} &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2} \\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) \\ 3^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz}) \\ 9 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz}) \\ &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2 \\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} &= 7 \\ \end{align} </math> </div></div> <ol start=33> <li>Berapakah nilai dari <math>\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}</math> jika <math>x^2+y^2+z^2 = -2(ab+bc+ac)</math>?>/li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+z^2 &= -2(xy+yz+xz) \\ x^2+y^2+z^2+2(xy+yz+xz) &= 0 \\ (x+y+z)^2 &= 0 \\ x+y+z &= 0 \\ x+y &= -z \\ x+z &= -y \\ y+z &= -x \\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z} &= \frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x} \\ &= -2-(-5)-(-7) \\ &= 10 \\ \end{align} </math> </div></div> <ol start=34> <li>Berapakah nilai dari <math>\frac{20xyz}{xy+yz+xz}</math> jika <math>16^x = 256^y = 625^z = 40</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 16^x = 256^y = 625^z &= 40 \\ 2^{4x} = 4^{4y} = 5^{4z} &= 40 \\ 2^{4x} &= 40 \\ 2 &= 40^{\frac{1}{4x}} \\ 4^{4y} &= 40 \\ 4 &= 40^{\frac{1}{4y}} \\ 5^{4z} &= 40 \\ 5 &= 40^{\frac{1}{4z}} \\ 2 \cdot 4 \cdot 5 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z}} \\ 1 &= \frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z} \\ 4 &= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{20xyz}{xy+yz+xz} &= 20 \cdot \frac{xyz}{xy+yz+xz} \\ &= 20 \cdot (\frac{xy+yz+xz}{xyz})^{-1} \\ &= 20 \cdot (\frac{1}{z} + \frac{1}{x} + \frac{1}{y})^{-1} \\ &= 20 \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{-1} \\ &= 20 \cdot (4)^{-1} \\ &= 20 \cdot \frac{1}{4} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=35> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+3x^2+1}</math> jika <math>6x^2+25x+6=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 6x^2+25x+6 &= 0 \\ 6x+25+\frac{6}{x} &= 0 \\ 6(x+\frac{1}{x}) &= -25 \\ x+\frac{1}{x} &= \frac{-25}{6} \\ (c+\frac{1}{x})^2 &= (\frac{-25}{6})^2 \\ x^2+2+\frac{1}{x^2} &= \frac{625}{36} \\ x^2+\frac{1}{x^2} &= \frac{625}{36}-2 \\ x^2+\frac{1}{x^2} &= \frac{553}{36} \\ \frac{x^2}{x^4+3x^2+1} &= \frac{1}{x^2+3+\frac{1}{x^2}} \\ &= \frac{1}{a^2+\frac{1}{x^2}+3} \\ &= \frac{1}{\frac{553}{36}+3} \\ &= \frac{1}{\frac{661}{36}} \\ &= \frac{36}{661} \\ \end{align} </math> </div></div> <ol start=36> <li>Berapakah nilai dari <math>\frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5} &= \frac{(9+2\sqrt{20})^{1013}}{((2)^3+3(2)^2(\sqrt{5})+3(2)(\sqrt{5})^2+(\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{((2+\sqrt{5})^2)^{1013}}{((2+\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{(2+\sqrt{5})^{2026}}{(2+\sqrt{5})^{2025}}+6-\sqrt{5} \\ &= 2+\sqrt{5}+6-\sqrt{5} \\ &= 8 \\ \end{align} </math> </div></div> <ol start=37> <li>Berapakah nilai dari <math>27x^3+\frac{8}{x^3}</math> jika <math>3x+\frac{2}{x}=6</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3x+\frac{2}{x} &= 6 \\ (3x+\frac{2}{x})^3 &= 6^3 \\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3} &= 216 \\ 27x^3+18(6)+\frac{8}{x^3} &= 216 \\ 27x^3+108+\frac{8}{x^3} &= 216 \\ 27x^3+\frac{8}{x^3} &= 108 \\ \end{align} </math> </div></div> <ol start=38> <li>Berapakah nilai dari <math>x^6+\frac{8}{x^3}</math> jika <math>x^3+\frac{1}{x^3}=8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3+\frac{1}{x^3} &= 8 \\ x^3 &= 8-\frac{1}{x^3} \\ x^6 &= 8x^3-1 \\ x^6+\frac{8}{x^3} &= 8x^3-1+\frac{8}{x^3} \\ &= 8x^3+\frac{8}{x^3}-1 \\ &= 8(x^3+\frac{1}{x^3})-1 \\ &= 8(8)-1 \\ &= 63 \\ \end{align} </math> </div></div> <ol start=39> <li>Berapakah nilai dari <math>4x+\frac{25}{x}</math> jika <math>2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2\sqrt{x}+\frac{5}{\sqrt{x}} &= 4x-\frac{25}{x} \\ 2\sqrt{x}+\frac{5}{\sqrt{x}} &= (2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}}) \\ 1 &= 2\sqrt{x}-\frac{5}{\sqrt{x}} \\ 1^2 &= (2\sqrt{x}-\frac{5}{\sqrt{x}})^2 \\ 1 &= 4x-20+\frac{25}{x} \\ 4x+\frac{25}{x} &= 21 \\ \end{align} </math> </div></div> <ol start=40> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2-x+1}{x^2+x+1} &= \frac{5}{6} \\ \frac{x^2+1-x}{x^2+1+x} &= \frac{5}{6} \\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} &= \frac{5}{6} \\ \text{ misalkan } x+\frac{1}{x} &= y \\ \frac{y-1}{y+1} &= \frac{5}{6} \\ 6(y-1) &= 5(y+1) \\ 6y-6 &= 5y+5 \\ y &= 11 \\ x+\frac{1}{x} &= 11 \\ \end{align} </math> </div></div> <ol start=41> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt{x}+x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x}+x &= 1 \\ x-1 &= -\sqrt{x} \\ (x-1)^2 &= (-\sqrt{x})^2 \\ x^2-2x+1 &= x \\ x^2-3x+1 &= 0 \\ x-3+\frac{1}{x} &= 0 \\ x+\frac{1}{x} &= 3 \\ \end{align} </math> </div></div> <ol start=42> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt[3]{x}-\sqrt[3]{x-36}=3</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x}-\sqrt[3]{x-36} &= 3 \\ (\sqrt[3]{x}-\sqrt[3]{x-36})^3 &= 3^3 \\ x-(x-36)-3 \sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36}) &= 27 \\ 36-3 \sqrt[3]{x(x-36)}3 &= 27 \\ -9 \sqrt[3]{x(x-36)} &= -9 \\ \sqrt[3]{x(x-36)} &= 1 \\ x(x-36) &= 1 \\ x^2-36x-1 &= 0 \\ x-36-\frac{1}{x} &= 0 \\ x-\frac{1}{x} &= 36 \\ \end{align} </math> </div></div> <ol start=43> <li>Berapakah nilai dari <math>x+\frac{16}{x}</math> jika <math>x-3\sqrt{x}=4</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x-3\sqrt{x} &= 4 \\ x-4 &= 3\sqrt{x} \\ x^2-8x+16 &= 9x \\ x^2-17x+16 &= 0 \\ x-17+\frac{16}{x} &= 0 \\ x+\frac{16}{x} &= 17 \\ \end{align} </math> </div></div> <ol start=44> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+4}</math> jika <math>x^2-7x+2=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2-7x+2 &= 0 \\ x^2+2 &= 7x \\ x+\frac{2}{x} &= 7 \\ x^2+4+\frac{4}{x^2} &= 49 \\ x^2+\frac{4}{x^2} &= 45 \\ \frac{x^4+4}{x^2} &= 45 \\ \frac{x^2}{x^4+4} &= \frac{1}{45} \\ \end{align} </math> </div></div> <ol start=45> <li>Berapakah nilai dari <math>x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}</math> jika <math>x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}} &= 25 \\ x^{\frac{1}{2}}+x^{-\frac{1}{2}} &= 23 \\ x+2+x^{-1} &= 529 \\ x+x^{-1} &= 527 \\ x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+x^{-\frac{3}{4}} &= 110 \\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1} &= x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}} \\ &= 527+110 \\ &= 637 \\ \end{align} </math> </div></div> <ol start=46> <li>Berapakah nilai dari <math>\sqrt{8x^6+x^5+x^4+5x^3+1}</math> jika <math>\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5} &= 0 \\ \frac{x^2+x+1}{x^5} &= 0 \\ x^2+x+1 &= 0 \\ x^2+x+1 &= 0 \\ (x-1)(x^2+x+1) &= 0(x-1) \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ \sqrt{8x^6+x^5+x^4+5x^3+1} &= \sqrt{(2x^3)^2+x^3x^2+x^3x+5x^3+1} \\ &= \sqrt{(2(1))^2+(1)x^2+(1)x+5(1)+1} \\ &= \sqrt{(2)^2+x^2+x+5+1} \\ &= \sqrt{4+x^2+x+1+5} \\ &= \sqrt{4+0+5} \\ &= \sqrt{9} \\ &= 3 \\ \end{align} </math> </div></div> <ol start=47> <li>Berapakah nilai dari <math>f(1)+f(2)+f(3)+ \dots + f(99)</math> jika <math>f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{1}{\sqrt{x+1}+\sqrt{x}} \\ &= \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} \\ &= \sqrt{x+1}-\sqrt{x} \\ f(1)+f(2)+f(3)+ \dots + f(98)+f(99) &= \sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+ \cdot + \sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99} \\ &= \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+ \cdot + \sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ &= \sqrt{100}-\sqrt{1} \\ &= 10-1 \\ &= 9 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots + \frac{2024}{2025})</math> jika <math>h(x)=\frac{3}{3+9^x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} h(x) &= \frac{3}{3+9^x} \\ h(1-x) &= \frac{3}{3+9^{1-x}} \\ &= \frac{3}{3+\frac{9}{9^x}} \\ &= \frac{9^x}{3+9^x} \\ h(x)+h(1-x) &= \frac{3}{3+9^x}+\frac{9^x}{3+9^x} \\ &= \frac{3+9^x}{3+9^x} \\ &= 1 \\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots +(1-\frac{2}{2025})+(1-\frac{1}{2025})) \\ & 5(1+1+1+ \dots +1+1) \text{ sebanyak 1012 kali } \\ & 5(1012) \\ & 5060 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>\frac{7^{2025} - 7^{2023} + 432}{7^{2024} + 7^{2023} + 72}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72} &= \frac{7^{2023}7^{2}-7^{2023} + 48 \times 9}{7^{2023}7^1+7^{2023}+8 \times 9} \\ &= \frac{7^{2023}(7^{2}-1)+48 \times 9}{7^{2023}(7^1+1)+8 \times 9} \\ &= \frac{7^{2023}(49-1)+48 \times 9}{7^{2023}(7+1) + 8 \times 9} \\ &= \frac{7^{2023} \times 48+48 \times 9}{7^{2023} \times 8+8 \times 9} \\ &= \frac{48(7^{2023}+9)}{8(7^{2023}+9)} \\ &= \frac{48}{8} \\ &= 6 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>tan (x+\frac{\pi}{4})</math> jika <math>\frac{1}{cos x}-tan x = \frac{4}{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{cos x}-tan x &= \frac{4}{5} \\ sec x-tan x &= \frac{4}{5} \\ sec^2 x-tan^2 x &= 1 \\ (sec x+tan x)(sec x-tan x) &= 1 \\ (sec x+tan x)\frac{4}{5} &= 1 \\ sec x+tan x &= \frac{5}{4} \\ \text{kedua persamaan dengan cara metode eliminasi } \\ 2 tan x &= \frac{5}{4}-\frac{4}{5} \\ 2 tan x &= \frac{9}{20} \\ tan x &= \frac{9}{40} \\ tan (x+\frac{\pi}{4}) &= \frac{tan x+tan \frac{\pi}{4}}{1-tan x \cdot tan \frac{\pi}{4}} \\ &= \frac{\frac{9}{40}+1}{1-\frac{9}{40} \cdot 1} \\ &= \frac{\frac{49}{40}}{\frac{31}{40}} \\ &= \frac{49}{31} \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>sin^3 x+csc^3 x</math> jika <math>sin x-csc x = 8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ Dengan menggunakan rumus: } (a-b)^3 &= a^3-b^3-3ab(a-b) \\ (sin x-csc x)^3 &= sin^3 x-csc^3 x-3sin x csc x(sin x-csc x) \\ 8^3 &= sin^3 x-csc^3 x-3sin x (\frac{1}{sin x})(8) \\ 512 &= sin^3 x-csc^3 x-24 \\ sin^3 x-csc^3 x &= 512+24 \\ sin^3 x-csc^3 x &= 536 \\ \end{align} </math> </div></div> <ol start=52> <li>Berapakah nilai dari <math>(sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2</math> jika <math>\frac{1}{sin x}+\frac{1}{cos x} = 10</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{sin x}+\frac{1}{cos x} &= 10 \\ \frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} &= 100 \\ (sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2 &= sin^2 x+\frac{2sin x}{cos x}+\frac{1}{cos^2 x}+cos^2 x+\frac{2cos x}{sin x}+\frac{1}{sin^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2(sin^2 x+cos^2 x)}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+100 \\ &= 101 \\ \end{align} </math> </div></div> <ol start=53> <li>Berapakah nilai dari (x-1)⁶ jika <math>x=\frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin 80^\circ &= cos 10^\circ \\ sin 80^\circ-cos 10^\circ &= 0 \\ x &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+2 sin 20^\circ cos 20^\circ}{cos 10^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+4 sin 10^\circ cos 10^\circ cos 20^\circ}{cos 10^\circ} \\ &= 4 cos 55^\circ cos 25^\circ+4 sin 10^\circ cos 20^\circ \\ &= 2(2 cos 55^\circ cos 25^\circ+2 sin 10^\circ cos 20^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ+sin (-10)^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ-sin 10^\circ) \\ &= 2(cos 80^\circ-sin 10^\circ+cos 30^\circ+sin 30^\circ) \\ &= 2(cos 80^\circ-sin (90^\circ-80^\circ)+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(cos 80^\circ-cos 80^\circ+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= \sqrt{3}+1 \\ x-1 &= \sqrt{3} \\ (x-1)^6 &= (\sqrt{3})^6 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=54> <li>Berapakah nilai dari x jika <math>x=\frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ} \\ 2x sin 20^\circ-x sin 40 ^\circ &= x sin 20^\circ-x^2 sin 10^\circ \\ x^2 sin 10^\circ+x sin 20^\circ-x sin 40 ^\circ &= 0 \\ x(x sin 10^\circ+sin 20^\circ-sin 40 ^\circ) &= 0 \\ x = 0 &\text{ atau } x sin 10^\circ+sin 20^\circ-sin 40 ^\circ = 0 \\ x sin 10^\circ+sin 20^\circ-sin 40 ^\circ &= 0 \\ x sin 10^\circ &= sin 40 ^\circ-sin 20^\circ \\ x &= \frac{sin 40 ^\circ-sin 20^\circ}{sin 10^\circ} \\ &= \frac{2 cos 30 ^\circ sin 10^\circ}{sin 10^\circ} \\ &= 2 cos 30 ^\circ \\ &= \frac{2 \sqrt{3}}{2} \\ &= \sqrt{3} \\ \end{align} </math> </div></div> <ol start=55> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x^2}{x^2-16y^2} = \frac{625}{49}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2}{x^2-16y^2} &= \frac{625}{49} \\ \frac{x^2-16y^2}{x^2} &= \frac{49}{625} \text{ (terbalik posisinya)} \\ 1-\frac{16y^2}{x^2} &= \frac{49}{625} \\ \frac{16y^2}{x^2} &= 1 - \frac{49}{625} \\ (\frac{4y}{x})^2 &= \frac{576}{625} \\ (\frac{4y}{x})^2 &= (\frac{24}{25})^2 \\ \frac{4y}{x} &= \frac{24}{25} \\ \frac{y}{x} &= \frac{6}{25} \\ \frac{x}{y} &= \frac{25}{6} \\ \end{align} </math> </div></div> <ol start=56> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x}{y}+\frac{x+10y}{y+10x} = 2</math> serta bilangan real untuk x dan y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x}{y}+\frac{x+10y}{y+10x} &= 2 \\ \frac{x}{y}+\frac{\frac{x}{y}+10}{1+10\frac{x}{y}} &= 2 \\ \text{misalkan } \frac{x}{y} = a \\ a+\frac{a+10}{1+10a} &= 2 \\ a(1+10a)+a+10 &= 2(1+10a) \\ 10a^2+a+a+10 &= 2+20a \\ 10a^2-18a+8 &= 0 \\ 5a^2-9a+4 &= 0 \\ (5a-4)(a-1) &= 0 \\ a = \frac{4}{5} &\text{ atau } a = 1 \\ \text{jadi } \frac{x}{y} = {\frac{4}{5}, 1} \\ \end{align} </math> </div></div> <ol start=57> <li>Berapakah nilai dari xy jika <math>x^4+y^4+x^2y^2=15 \text{ dan } x^2+y^2+xy=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+xy &= 5 \\ x^2+y^2 &= 5-xy \\ x^4+y^4+x^2y^2 &= 15 \\ (x^2)^2+(y^2)^2+2x^2y^2-x^2y^2 &= 15 \\ (x^2+y^2)^2-x^2y^2 &= 15 \\ (5-xy)^2-x^2y^2 &= 15 \\ 25-10xy+x^2y^2-x^2y^2 &= 15 \\ 25-10xy &= 15 \\ 10xy &= 10 \\ xy &= 1 \\ \end{align} </math> </div></div> <ol start=58> <li>Berapakah nilai dari x jika <math>4^x = 63(4^3+1)(4^6+1)(4^{12}+1)+1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 4^x &= 63(4^3+1)(4^6+1)(4^{12}+1)+1 \\ 4^x-1 &= 63(4^3+1)(4^6+1)(4^{12}+1) \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{4^3-1} \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{63} \\ &= (4^3+1)(4^6+1)(4^{12}+1)(4^3-1) \\ &= (4^3-1)(4^3+1)(4^6+1)(4^{12}+1) \\ &= (4^6-1)(4^6+1)(4^{12}+1) \\ &= (4^{12}-1)(4^{12}+1) \\ &= 4^{24}-1 \\ 4^x &= 4^{24} \\ x &= 24 \\ \end{align} </math> </div></div> <ol start=59> <li>Berapakah nilai dari <math>\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}</math> jika <math>x=\sqrt{9+4\sqrt{5}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \sqrt{9+4\sqrt{5}} \\ x &= 2+\sqrt{5} \\ x^2 &= 9+4\sqrt{5} \\ x^2-4x &= 9+4\sqrt{5}-4(2+\sqrt{5}) \\ x^2-4x &= 1 \\ x^2 &= 4x+1 \\ x^3 &= x \cdot x^2 \\ &= x(4x+1) \\ &= 4x^2+x \\ &= 4(4x+1)+x \\ &= 16x+4+x \\ &= 17x+4 \\ x^4 &= x \cdot x^3 \\ &= x(17x+4) \\ &= 17x^2+4x \\ &= 17(4x+1)+4x \\ &= 68x+17+4x \\ &= 72x+17 \\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1} &= \frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1} \\ &= \frac{72x+17-85x-20+8x+2+5x+3}{2} \\ &= \frac{2}{2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=60> <li>Berapakah nilai dari <math>\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}</math> jika 2x-1=<math>\sqrt{61}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \frac{x^3+1}{x^5-x^4-x^3+x^2} = p \\ p &= \frac{x^3+1}{x^5-x^4-x^3+x^2} \\ &= \frac{x^3+1}{x^5-x^4-(x^3-x^2)} \\ &= \frac{x^3+1}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)} \\ &= \frac{x^2-x+1}{x^2(x-1)^2} \\ &= \frac{x^2-x+1}{(x(x-1))^2} \\ &= \frac{x(x-1)+1}{(x(x-1))^2} \\ 2x-1 &= \sqrt{61} \\ x &= \frac{\sqrt{61}+1}{2} \\ x-1 &= \frac{\sqrt{61}-1}{2} \\ x(x-1) &= (\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2}) \\ &= \frac{61-1}{4} \\ &= \frac{60}{4} \\ &= 15 \\ p &= \frac{x(x-1)+1}{(x(x-1))^2} \\ &= \frac{15+1}{15^2} \\ &= \frac{16}{15^2} \\ \sqrt{p} &= \sqrt{\frac{16}{15^2}} \\ &= \frac{4}{15} \\ \end{align} </math> </div></div> <ol start=61> <li>Berapakah nilai dari <math>(\frac{x-3}{x})^{25}</math> jika <math>x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\sqrt[5]{8}+\sqrt[5]{2} &= 1+\sqrt[5]{16}+\sqrt[5]{4} \\ x+(\sqrt[5]{2})^3+\sqrt[5]{2} &= 1+(\sqrt[5]{2})^4+(\sqrt[5]{2})^2 \\ x &= (\sqrt[5]{2})^4-(\sqrt[5]{2})^3+(\sqrt[5]{2})^2-\sqrt[5]{2}+1 \\ \text{misalkan } \sqrt[5]{2} = p \\ x &= p^4-p^3+p^2-p+1 \\ x &= \frac{p^5+1}{p+1} \\ (\frac{x-3}{x})^{25} &= (1-\frac{3}{x})^{25} \\ &= (1-\frac{3}{\frac{p^5+1}{p+1}})^{25} \\ &= (1-\frac{3(p+1)}{p^5+1})^{25} \\ &= (1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2})^5+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{2+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-3\sqrt[5]{2}-3)}{3})^{25} \\ &= (-\sqrt[5]{2})^{25} \\ &= (-2)^5 \\ &= -32 \\ \end{align} </math> </div></div> <ol start=62> <li>Berapakah nilai dari <math>x^{50}+x^{49}+x^{48}+x^{47}+x^{46}</math> jika <math>x^2+x+1=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+x+1 &= 0 \\ x^2+x &= -1 \\ \frac{x^3-1}{x-1} &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46} &= x^{48}(x^2+x+1)+x^{45}(x^2+x) \\ &= x^{48}(0)+(x^3)^{15}(-1) \\ &= 0+(1)^{15}(-1) \\ &= -1 \\ \end{align} </math> </div></div> <ol start=63> <li>Berapakah 2²⁴ dari <math>8^7+8^6+8^5+8^4+8^3+8^2+8+1=A</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= A \\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1) &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8 &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= 8A+1 \\ 8^8+A &= 8A+1 \\ 8^8 &= 7A+1 \\ (2^3)^8 &= 7A+1 \\ 2^{24} &= 7A+1 \\ \end{align} </math> </div></div> <ol start=64> <li>Berapakah nilai dari <math>x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1</math> jika <math>x+\frac{1}{x}=\sqrt{3}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{x} &= \sqrt{3} \\ x^2+2+\frac{1}{x^2} &= 3 \\ x^2-1+\frac{1}{x^2} &= 0 \\ x^2(x^2-1+\frac{1}{x^2}) &= x^2(0) \\ x^4-x^2+1 &= 0 \\ (x^2+1)(x^4-x^2+1) &= (x^2+1)0 \\ x^6-x^4+x^2+x^4-x^2+1 &= 0 \\ x^6+1 &= 0 \\ x^6 &= -1 \\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1 &= {x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1 \\ &= (-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2-1+1 \\ &= -1+1-1+1-1+1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=65> <li>Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari <math>\frac{c-b}{a-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= ax^2+bx+c \\ f(2) &= a(2)^2+2b+c = 4 \\ &= 4a+2b+c = 4 \\ f(7) &= a(7)^2+7b+c = 49 \\ &= 49a+7b+c = 49 \\ 49a+7b+c &= 49 \\ 4a+2b+c &= 4 \\ 45a+5b &= 45 \text{ (f(7) dikurangi f(2)) } \\ 9a+b &= 9 \\ b &= -9a+9 \\ 4a+2b+c &= 4 \\ 4a+2(-9a+9)+c &= 4 \\ 4a-18a+18+c &= 4 \\ -14a+18+c &= 4 \\ c &= 14a-14 \\ \frac{c-b}{a-1} &= \frac{14a-14-(-9a+9)}{a-1} \\ &= \frac{14(a-1)+9(a-1)}{a-1} \\ &= \frac{(14+9)(a-1)}{a-1} \\ &= 23 \\ \end{align} </math> </div></div> <ol start=66> <li>Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+y)^3 &= x^3+y^3+3xy(x+y) \\ 11^3 &= 242+3xy(11) \text{ (dibagi 11)} \\ 11^2 &= 22+3xy \\ 121 &= 22+3xy \\ 99 &= 3xy \\ xy &= 33 \\ (x-y)^2 &= x^2+y^2-2xy \\ &= ((x+y)^2-2xy)-2xy \\ &= (x+y)^2-4xy \\ &= 11^2-4(33) \\ &= 121-132 \\ &= -11 \\ \end{align} </math> </div></div> <ol start=67> <li>Berapa f(1)+f(-1) jika <math>f(\frac{ax-b}{bx-a})</math>=x²-5x+6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ jika} f(1) = f(\frac{ax-b}{bx-a}) \\ 1 &= \frac{ax-b}{bx-a} \\ bx-a &= ax-b \\ (b-a)x &= -b+a \\ &= -(b-a) \\ &= -1 \\ f(1) &= x^2-5x+6 \\ &= (-1)^2-5(-1)+6 \\ &= 12 \\ \text{ jika} f(-1) = f(\frac{ax-b}{bx-a}) \\ -1 &= \frac{ax-b}{bx-a} \\ -(bx-a) &= ax-b \\ -bx+a &= ax-b \\ (-b-a)x &= -b-a \\ &= 1 \\ f(-1) &= x^2-5x+6 \\ &= (1)^2-5(1)+6 \\ &= 2 \\ f(1)+f(-1) &= 12+2 \\ &= 14 \\ \end{align} </math> </div></div> <ol start=68> <li>berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)-x &= f(x-1) \\ f(x)-f(x-1) &= x \\ x=1 ; f(1)-f(0) &= 1 \\ x=2 ; f(2)-f(1) &= 2 \\ x=3 ; f(3)-f(2) &= 3 \\ x=4 ; f(4)-f(3) &= 4 \\ \dots \\ x=200 ; f(200)-f(199) &= 200 \\ \text{ jumlahkan tersebut menjadi } \\ f(200)-f(0) &= 1+2+3+4+\dots+200 \\ &= \frac{200 \cdot 201}{2} \\ &= 20.100 \\ f(200)-1 &= 20.100 \\ &= 20.101 \\ \end{align} </math> </div></div> <ol start=69> <li>Misalkan f(x) adalah fungsi rekursif yang berlaku ∀x ∈ R sebagai berikut: : f(x)+f(15-x) = 2024 : f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)+f(15-x) &= 2024 \\ f(15+x) &= f(x)+2020 \\ *cara 1 \\ \text{ganti x dengan 15+x } \\ f(15+x)+f(-x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ *cara 2 \\ \text{ganti x dengan -x } \\ f(-x)+f(15+x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ \end{align} </math> </div></div> <ol start=70> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>2f(\frac{2002}{x}) + f(x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2f(\frac{2002}{x}) + f(x) &= 3x \\ \text{ganti x dengan 2 } \\ 2f(\frac{2002}{2}) + f(2) &= 3(2) \\ 2f(1001) + f(2) &= 6 \\ \text{ganti x dengan 1001 } \\ 2f(\frac{2002}{1001}) + f(1001) &= 3(1001) \\ 2f(2) + f(1001) &= 3003 \\ 2f(2) + f(1001) &= 3003 \\ f(1001) &= 3003 - 2f(2) \\ 2f(1001) + f(2) &= 6 \\ 2(3003 - 2f(2)) + f(2) &= 6 \\ 6006 - 4f(2) + f(2) &= 6 \\ 3f(2) &= 6000 \\ f(2) &= 2000 \\ \end{align} </math> </div></div> <ol start=71> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>f(\frac{1}{x}) + \frac{1}{x}f(-x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{1}{x})+\frac{1}{x}f(-x) &= 3x \\ \text{ganti x dengan 1/3 } \\ f(3)+3f(-\frac{1}{3}) &= 1 \\ \text{ganti x dengan -3 } \\ f(-\frac{1}{3}) - \frac{1}{3}f(3) &= -9 \\ \text{dikalikan 3 } \\ 3f(-\frac{1}{3})-f(3) &= -27 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ 2f(3) &= 28 \\ f(3) &= 14 \\ \end{align} </math> </div></div> <ol start=72> <li>Diketahui polinom <math>f(7^b-1)=7^{3b}-10</math>. tentukan nilai f(5)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(5) &= f(7^b-1) \\ 5 &= 7^b-1 \\ 7^b &= 6 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(6-1) &= 6^3-10 \\ f(5) &= 216-10 \\ &= 206 \\ *cara 2 \\ \text{misalkan } 7^b-1=a \text{ maka } 7^b=a+1 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(a) &= (a+1)^3-10 \\ f(5) &= (5+1)^3-10 \\ &= 6^3-10 \\ &= 216-10 \\ &= 206 \\ \end{align} </math> </div></div> <ol start=73> <li>Diketahui polinom <math>f(6^b-7)=6^{3b}-2 \cdot 6^{2b}-4</math>. tentukan nilai f(-2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(-2) &= f(6^b-7) \\ -2 &= 6^b-7 \\ 6^b &= 5 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(5-7) &= 5^3-2 \cdot 5^2-4 \\ f(-2) &= 125-50-4 \\ &= 71 \\ *cara 2 \\ \text{misalkan } 6^b-7=a \text{ maka } 6^b=a+7 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(a) &= (a+7)^3-2(a+7)^2-4 \\ f(-2) &= (-2+7)^3-2(-2+7)^2-4 \\ &= 5^3-2(5)^2-4 \\ &= 125-50-4 \\ &= 71 \\ \end{align} </math> </div></div> <ol start=74> <li>Jika <math>f(xy)=\frac{f(x)}{y}</math> dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(10) &= 7 \\ f(2 \cdot 5) &= 7 \\ f(xy) &= \frac{f(x)}{y} \\ f(2 \cdot 5) &= \frac{f(2)}{5} \\ 7 &= \frac{f(2)}{5} \\ f(2) &= 35 \\ \end{align} </math> </div></div> <ol start=75> <li>Jika <math>f(xy)=\frac{f(x+y)}{xy}</math> dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(15) &= 16 \\ f(3 \cdot 5) &= 16 \\ f(xy) &= \frac{f(x+y)}{xy} \\ f(3 \cdot 5) &= \frac{f(3+5)}{3 \cdot 5} \\ f(15) &= \frac{f(8)}{15} \\ 16 &= \frac{f(8)}{15} \\ f(8) &= 240 \\ \end{align} </math> </div></div> <ol start=76> <li>Jika <math>f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15</math> maka tentukan nilai f(16)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x+\frac{1}{x}+6) &= x^2+\frac{1}{x^2}+15 \\ &= (x+\frac{1}{x})^2-2+15 \\ &= (x+\frac{1}{x})^2+13 \\ \text{misalkan } x+\frac{1}{x} &= p \\ f(x+\frac{1}{x}+6) &= (x+\frac{1}{x})^2+13 \\ f(p+6) &= p^2+13 \\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 } \\ f(p+6) &= p^2+13 \\ f(10+6) &= 10^2+13 \\ f(16) &= 100+13 \\ &= 113 \\ \end{align} </math> </div></div> <ol start=77> <li>Tentukan nilai x jika <math>f(x)=\frac{4}{4-x}</math> dan <math>f(x \cdot f(x))^{\frac{f(4x)}{f(x)}}=256</math>!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{4}{4-x} \\ f(4x) &= \frac{4}{4-4x} \\ \frac{f(4x)}{f(x)} &= \frac{\frac{4}{4-4x}}{\frac{4}{4-x}} \\ &= \frac{4-x}{4-4x} \\ f(x \cdot f(x)) &= f(x(\frac{4}{4-x})) \\ &= f(\frac{4x}{4-x}) \\ &= \frac{4}{4-(\frac{4x}{4-x})} \\ &= \frac{4}{\frac{16-4x-4x}{4-x}} \\ &= \frac{4}{\frac{16-8x}{4-x}} \\ &= \frac{4(4-x)}{4(4-4x)} \\ &= \frac{4-x}{4-4x} \\ \text{misalkan } \frac{4-x}{4-4x} &= a \\ f(x \cdot f(x))^{\frac{f(4x)}{f(x)}} &= 256 \\ a^a &= 256 \\ a^a &= 4^4 \\ a &= 4 \\ \frac{4-x}{4-4x} &= 4 \\ 4-x &= 16-16x \\ 15x &= 12 \\ x &= \frac{4}{5} \\ \end{align} </math> </div></div> <ol start=78> <li>Fungsi <math>f(x) = \frac{kx}{2x+1} \text{dengan } x \neq -\frac{1}{2}</math>. Dengan f(f(x)) = x maka tentukan nilai k!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{kx}{2x+1} \\ f(f(x)) &= x \\ f(\frac{kx}{2x+1}) &= x \\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1} &= x \\ \frac{\frac{k^2x}{2x+1}}{\frac{2kx+2x+1}{2x+1}} &= x \\ \frac{k^2x}{2kx+2x+1} &= x \\ \frac{k^2}{2kx+2x+1} &= 1 \\ k^2 &= 2kx+2x+1 \\ k^2-2kx &= 2x+1 \\ k^2-2kx+x^2 &= x^2+2x+1 \\ (k-x)^2 &= (x+1)^2 \\ (k-x)^2-(x+1)^2 &= 0 \\ (k-x+x+1)(k-x-(x+1)) &= 0 \\ k=-1 &\text{ atau } k=2x+1 &\text{ (TM) } \\ \end{align} </math> </div></div> <ol start=79> <li>Jika n = 2023²+2024² maka berapa hasil dari <math>\sqrt{2n-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n &= 2023^2+2024^2 \\ &= 2023^2+(2023+1)^2 \\ \text{misalkan 2023 = p} \\ n &= p^2+(p+1)^2 \\ &= p^2+p^2+2p+1 \\ &= 2p^2+2p+1 \\ \sqrt{2n-1} &= \sqrt{2(2p^2+2p+1)-1} \\ &= \sqrt{4p^2+4p+2-1} \\ &= \sqrt{4p^2+4p+1} \\ &= \sqrt{(2p+1)^2} \\ &= 2p+1 \\ &= 2(2023)+1 \\ &= 4046+1 \\ &= 4047 \\ \end{align} </math> </div></div> <ol start=80> <li>Tentukan nilai dari a+b+c merupakan bilangan bulat positif jika ab = 2, bc = 3 dan ac = 6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab \cdot bc \cdot ac &= 2 \cdot 3 \cdot 6 \\ (abc)^2 &= 36 \\ abc &= \pm 6 \\ abc &= 6 \\ \frac{abc}{ab} &= c = \frac{6}{2} = 3 \\ \frac{abc}{bc} &= a = \frac{6}{3} = 2 \\ \frac{abc}{ac} &= b = \frac{6}{6} = 1 \\ a+b+c &= 6 \\ \end{align} </math> </div></div> <ol start=81> <li>Tentukan nilai dari (a-c)^b jika <math>\frac{ab}{a+b} = \frac{1}{3}</math>, <math>\frac{bc}{b+c} = \frac{1}{4}</math> dan <math>\frac{ac}{a+c} = \frac{1}{9}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{ab}{a+b} &= \frac{1}{3} \\ \frac{a+b}{ab} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{b} + \frac{1}{a} &= 3 \\ \frac{bc}{b+c} &= \frac{1}{4} \\ \frac{b+c}{bc} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{b} &= 4 \\ \frac{ac}{a+c} &= \frac{1}{9} \\ \frac{a+c}{ac} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{a} &= 9 \\ \text{Misalkan 1/a = x, 1/b = y dan 1/c = z} \\ x+y &= 3 \\ y+z &= 4 \\ x+z &= 9 \\ x+y &= 3 \\ y+z &= 4 \\ x-z &= -1 \\ x-z &= -1 \\ x+z &= 9 \\ 2x &= 8 \\ x &= 4 \\ x-z &= -1 \\ 4-z &= -1 \\ z &= 5 \\ x+y &= 3 \\ 4+y &= 3 \\ y &= -1 \\ \frac{1}{a} &= 4 \\ a &= \frac{1}{4} \\ \frac{1}{b} &= -1 \\ b &= -1 \\ \frac{1}{c} &= 5 \\ c &= \frac{1}{5} \\ (a-c)^b &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=82> <li>Tentukan nilai dari a, b dan c jika <math>\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}</math> dan a+2b+3c=28?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan k untuk semua ketiga persamaan tersebut } \\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5} &= k \\ a+b &= 2k \\ a+c &= 4k \\ b+c &= 5k \\ 2a+b+c &= 6k \\ 2a+5k &= 6k \\ k &= 2a \\ a &= \frac{k}{2} \\ b &= \frac{3k}{2} \\ c &= \frac{7k}{2} \\ a+2b+3c &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ a &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ b &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ c &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=83> <li>Tentukan nilai dari (b+c)^a jika <math>\frac{a+b+c}{2} = \sqrt{a-2}+\sqrt{b-1}+\sqrt{c}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{a+b+c}{2} &= \sqrt{a-2}+\sqrt{b-1}+\sqrt{c} \\ a+b+c &= 2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}) \\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c} &= 0 \\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1 &= 0 \\ (\sqrt{a-2}-1)^2+(\sqrt{b-1}-1)^2+(\sqrt{c}-1)^2 &= 0 \\ (\sqrt{a-2}-1)^2 &= 0 \\ \sqrt{a-2}-1 &= 0 \\ \sqrt{a-2} &= 1 \\ a-2 &= 1 \\ a &= 3 \\ (\sqrt{b-1}-1)^2 &= 0 \\ \sqrt{b-1}-1 &= 0 \\ \sqrt{b-1} &= 1 \\ b-1 &= 1 \\ b &= 1 \\ (\sqrt{c}-1)^2 &= 0 \\ \sqrt{c}-1 &= 0 \\ \sqrt{c} &= 1 \\ c &= 1 \\ (b+c)^a &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=84> <li>x dan y merupakan bilangan tak nol. Jika xy = <math>\frac{x}{y}</math> = x-y maka berapa nilai x+y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy &= \frac{x}{y} \\ y^2 &= 1 \\ y^2 - 1 &= 0 \\ (y-1)(y+1) &= 0 \\ y = 1 &\text{ atau } y = -1 \\ \frac{x}{y} &= x-y \\ x &= xy-y^2 \\ x-xy &= -y^2 \\ x(1-y) &= -y^2 \\ x &= \frac{-y^2}{1-y} \\ \text{cek y=1 } \\ x &= \frac{-1^2}{1-1} \\ \text{tidak memenuhi syarat } \\ \text{cek y=-1 } \\ x &= \frac{-(-1)^2}{1-(-1)} \\ &= \frac{-1}{2} \\ x+y &= -1-\frac{1}{2} \\ &= -\frac{3}{2} \\ \end{align} </math> </div></div> <ol start=86> <li>Berapa nilai x dari <math>(\frac{a}{b})^3+(\frac{b}{a})^3 = 2\sqrt{x}</math> jika <math>\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b} \\ \frac{a+b}{ab} &= \frac{1}{a+b} \\ (a+b)^2 &= ab \\ a^2+2ab+b^2 &= ab \\ a^2+b^2 &= -ab \\ \text{misalkan } \frac{a}{b}+\frac{b}{a} = n \\ \frac{a}{b}+\frac{b}{a} &= n \\ \frac{a^2+b^2}{ab} &= n \\ a^2+b^2 &= nab \\ n &= -1 \\ \frac{a}{b}+\frac{b}{a} &= n \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3(\frac{a}{b}+\frac{b}{a}) &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3n &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3 &= n^3-3n \\ &= (-1)^3-3(-1) \\ &= 2 \\ 2\sqrt{x} &= 2 \\ \sqrt{x} &= 1 \\ x &= 1 \\ \end{align} </math> </div></div> <ol start=87> <li>Berapa nilai m dari <math>x^2-mx-1=0</math> jika <math>\sqrt[3]{x_1}+\sqrt[3]{x_2}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x_1} &= a \\ x_1 &= a^3 \\ \sqrt[3]{x_2} &= b \\ x_2 &= b^3 \\ \sqrt[3]{x_1}+\sqrt[3]{x_2} &= 1 \\ a+b &= 1 \\ x^2-mx-1 &= 0 \\ x_1+x_2 &= m \\ x_1 \cdot x_2 &= -1 \\ x_1+x_2 &= m \\ a^3+b^3 &= m \\ x_1 \cdot x_2 &= -1 \\ a^3 \cdot b^3 &= -1 \\ (ab)^2 &= (-1)^3 \\ ab &= -1 \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ (1)^3 &= m+3(-1)(1) \\ 1 &= m-3 \\ m &= 4 \\ \end{align} </math> </div></div> <ol start=88> <li>Berapa nilai <math>\frac{x_1}{x_2}</math> dari <math>ax^2-18x-b=0</math> jika <math>ab=45</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab &= 45 \\ b &= \frac{45}{a} \\ ax^2-18x-b &= 0 \\ ax^2-18x-\frac{45}{a} &= 0 \\ a^2x^2-18ax-45 &= 0 \\ (ax-3)(ax-15) &= 0 \\ ax-3 &= 0 \\ x &= \frac{3}{a} \\ ax-15 &= 0 \\ x &= \frac{15}{a} \\ \frac{x_1}{x_2} &= \frac{\frac{3}{a}}{\frac{15}{a}} \\ &= \frac{3}{15} \\ &= \frac{1}{5} \\ \frac{x_1}{x_2} &= \frac{\frac{15}{a}}{\frac{3}{a}} \\ &= \frac{15}{3} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=89> <li>Jika <math>\frac{u_3}{u_1+u_2} = \frac{7}{8}</math> merupakan barisan aritmetika maka berapa dari <math>\frac{u_2+u_3}{u_1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{u_3}{u_1+u_2} &= \frac{7}{8} \\ \frac{a+2b}{a+a+b} &= \frac{7}{8} \\ \frac{a+2b}{2a+b} &= \frac{7}{8} \\ 8(a+2b) &= 7(2a+b) \\ 8a+16b &= 14a+7b \\ 9b &= 6a \\ b &= \frac{2a}{3} \\ \frac{u_2+u_3}{u_1} &= \frac{a+b+a+2b}{a} \\ &= \frac{2a+3b}{a} \\ &= \frac{2a+3(\frac{2a}{3})}{a} \\ &= \frac{2a+2a}{a} \\ &= \frac{4a}{a} \\ &= 4 \\ \end{align} </math> </div></div> <ol start=90> <li>Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7p+q}{2p+q} &= \frac{17p+q}{7p+q} \\ (7p+q)^2 &= (17p+q)(2p+q) \\ 49p^2+14pq+q^2 &= 34p^2+19pq+q^2 \\ 15p^2 &= 5pq \\ 3p &= q \\ \frac{7p+q}{2p+q} &= \frac{7p+3p}{2p+3p} \\ &= \frac{10p}{5p} \\ &= 2 \\ \end{align} </math> </div></div> <ol start=91> <li>Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{rataan geometris } \\ \sqrt{a \cdot b} &= b-24 \\ a \cdot b &= (b-24)^2 \\ \text{rataan aritmatik } \\ \frac{a+b}{2} &= a+15 \\ a+b &= 2(a+15) \\ a+b &= 2a+30 \\ a &= b-30 \\ a \cdot b &= (b-24)^2 \\ (b-30)b &= (b-24)^2 \\ b^2-30b &= b^2-48b+576 \\ 18b &= 576 \\ b &= 32 \\ a &= b-30 \\ &= 32-30 \\ &= 2 \\ a+b &= 32+2 \\ &= 34 \\ \end{align} </math> </div></div> <ol start=91> <li>Segitiga lancip ABC dengan <math>\frac{a^4+b^4+c^4+a^2b^2}{c^2(a^2+b^2)}=2</math>. tentukan nilai sudut C?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{syarat segitiga lancip semua sudut masing-masing kurang dari } 90^\circ \\ c^2 &= a^2+b^2-2ab cos C \\ cos C &= \frac{a^2+b^2-c^2}{2ab} \\ a^4+b^4+c^4+a^2b^2 &= 2c^2(a^2+b^2) \\ a^4+b^4+a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-2c^2(a^2+b^2)+(c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= (ab)^2 \\ a^2+b^2-c^2 &= \pm ab \\ cos C &= \pm \frac{ab}{2ab} \\ &= \pm \frac{1}{2} \\ &= \frac{1}{2} \text{ (karena sudut harus kurang dari } 90^\circ) \\ C &= 60^\circ \\ \end{align} </math> </div></div> <ol start=92> <li>Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan panjang DA dan EA } = x \text{ dan panjang AB } = y \\ \text{luas segitiga CAB } &= \frac{CA \cdot AB}{2} \\ 45 &= \frac{(x+9)(x+5)}{2} \\ 90 &= x^2+14x+45 \\ x^2+14x &= 45 \\ y^2 &= (x+9)^2+(x+5)^2 \\ &= x^2+18x+81+x^2+10x+25 \\ &= 2x^2+28x+106 \\ &= 2(x^2+14x)+106 \\ &= 2(45)+106 \\ &= 196 \\ y &= 14 \\ \end{align} </math> jadi panjang BC adalah 14 cm </div></div> <ol start=93> <li>Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kita cari ukuran GC yaitu } \\ \frac{GC}{AD} &= \frac{CE}{DE} \\ \frac{GC}{15} &= \frac{6}{18} \\ GC &= 5 \\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC } \\ &= \frac{1}{2} \cdot BC \cdot CF - \frac{1}{2} \cdot GC \cdot CE \\ &= \frac{1}{2} \cdot 15 \cdot 18 - \frac{1}{2} \cdot 5 \cdot 6 \\ &= 135 - 15 \\ &= 120 \\ \end{align} </math> jadi luas daerah BFEG adalah 120 cm² </div></div> <ol start=94> <li>Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ AI &= \frac{1}{5} AF \\ &= \frac{1}{5} 10 \\ &= 2 \\ IF &= AF-AI \\ &= 10-2 \\ &= 8 \\ \text{luas trapesium AFGD } &= \frac{(AD+EF) \cdot AF}{2} \\ &= \frac{(4+6)10}{2} \\ &= 50 \\ \text{luas segitiga AID } &= \frac{AI \cdot AF}{2} \\ &= \frac{(2)4}{2} \\ &= 4 \\ \text{luas segitiga IFG } &= \frac{IF \cdot FG}{2} \\ &= \frac{(8)6}{2} \\ &= 24 \\ \text{luas daerah segitiga IGD } &= \text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG } \\ &= 50-4-24 \\ &= 22 \\ \end{align} </math> jadi luas daerah segitiga IGD adalah 22 cm² </div></div> <ol start=96> <li>Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-<math>\pi</math>) cm³ maka berapa luas permukaan kerucut tersebut?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ \text{volume balok} \\ V_b &= x^2(12) \\ \text{volume kerucut} \\ V_b &= \frac{1}{3}\pi x^2(12) \\ &= 4\pi x^2 \\ V_{b-k} &= Vb-Vk \\ 100(3-\pi) &= 12x^2-4\pi x^2 \\ 100(3-\pi) &= 4x^2(3-\pi) \\ x^2 &= 25 \\ x &= 5 \\ s &= \sqrt{12^2+5^2} \\ &= \sqrt{144+25} \\ &= \sqrt{169} \\ &= 13 \\ \text{luas permukaan kerucut } &= \pi r(r+s) \\ &= \pi(5)(5+13) \\ &= 90\pi \\ \end{align} </math> jadi luas daerah permukaan kerucut adalah 90<math>\pi</math> cm² </div></div> <ol start=97> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 3a+1 \\ B &= 3b+2 \\ A(A+1)+3B \\ (3a+1)(3a+1+1)+3(3b+2) \\ (3a+1)(3a+2)+9b+6 \\ 9a^2+9a+2+9b+6 \\ 9a^2+9a+9b+8 \\ 9(a^2+a+b)+8 \\ \text{sisa pembagiannya adalah } 8 \\ \end{align} </math> </div></div> <ol start=98> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 9a+7 \\ B &= 9b+8 \\ A(A-5)+9B \\ (9a+7)(9a+7-5)+9(9b+8) \\ (9a+7)(9a+2)+81b+72 \\ 81a^2+81a+14+81b+72 \\ 81a^2+81a+81b+86 \\ 81a^2+81a+81b+81+5 \\ 81(a^2+a+b+1)+5 \\ \text{sisa pembagiannya adalah } 5 \\ \end{align} </math> </div></div> <ol start=99> <li>Jika <math>\begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix}</math> maka berapa hasil dari A²¹+A²⁵+A⁴⁶?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A^2 &= A \cdot A \\ &= \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \\ A^3 &= A^2 \cdot A \\ &= \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ &= - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= -I \\ A^{21}+A^{25}+A^{46} &= A^{21} \cdot (I+A^4+A^{25}) \\ &= A^{21} \cdot (I+A^3 \cdot A +A^{24} \cdot A) \\ &= (A^3)^7 \cdot (I+A^3 \cdot A +(A^3)^8 \cdot A) \\ &= (-I)^7 \cdot (I-I \cdot A +(-I)^8 \cdot A) \\ &= -I \cdot (I-A+A) \\ &= -I \cdot I \\ &= -I \\ &= -\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ \end{align} </math> </div></div> <ol start=100> <li>Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida?</li></ol> : bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15 : ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9). : ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8) : maka Ida menulis bilangan 8. <ol start=101> <li>Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9?</li></ol> : Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu. : untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja. : untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja. : Jadi banyaknya bilangan mungkin 2. <ol start=102> <li>Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli!</li></ol> ;cara 1 # 8¹ = 1 # 8² = 1 (8²=8¹x8¹ sama dengan 1x1) # 8³ = 1 (8³=8¹x8² sama dengan 1x1) # 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2) # 8⁵ = 1 # 8ⁿ = 1 (semua n untuk bilangan asli) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli ;cara 2 # 8ⁿ = b mod 7 # 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil) # (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya) # 8ⁿ = 1ⁿ mod 7 # 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli <ol start=103> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 5?</li></ol> ;cara 1 # 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 # 7¹ = 7 (sisa 1) # 7² = 49 (sisa 2) # 7³ = 343 (sisa 3) # 7⁴ = 2,401 (sisa 0) # 7⁵ = 16,807 # 7⁶ = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3 ;cara 2 :17¹ = 2 :17² = 4 :17³ = 3 :17⁴ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3 :17⁹⁹ = (17⁴)²⁴ x 17³ Untuk 17⁴ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 3. Jadi 17⁹⁹ dibagi 7 bersisa 3 ;cara 3 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 17⁴ ::17⁴ = 1 mod 5 ::(17⁴)²⁴ = 1²⁴ mod 5 ::17⁹⁶ = 1²⁴ mod 5 ::17⁹⁶ = 1 mod 5 ::17⁹⁶ x 17³ = 1 x 17³ mod 5 ::17⁹⁹ = 17³ mod 5 ::17⁹⁹ = 17 x 17 x 17 mod 5 ::17⁹⁹ = 2 x 2 x 2 mod 5 ::17⁹⁹ = 8 mod 5 ::17⁹⁹ = 3 mod 5 Jadi 17⁹⁹ dibagi 5 bersisa 3 <ol start=104> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 7?</li></ol> ;cara 1 :17¹ = 3 :17² = 2 :17³ = 6 :17⁴ = 4 :17⁵ = 5 :17⁶ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3 :17⁹⁹ = (17⁶)¹⁶ x 17³ Untuk 17⁶ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 6. Jadi 17⁹⁹ dibagi 7 bersisa 6 ;cara 2 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 17⁶ ::17⁶ = 1 mod 7 ::(17⁶)¹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1 mod 7 ::17⁹⁶ x 17³ = 1 x 17³ mod 7 ::17⁹⁹ = 17³ mod 7 ::17⁹⁹ = 17 x 17 x 17 mod 7 ::17⁹⁹ = 3 x 3 x 3 mod 7 ::17⁹⁹ = 27 mod 7 ::17⁹⁹ = 6 mod 7 Jadi 17⁹⁹ dibagi 7 bersisa 6 <ol start=105> <li>Berapa hasil sisa dari 41²⁰²⁴ dibagi 33?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 41^{2024} &= 41^{2024} \text{ mod } 33 \\ &= (33 \times 3 + 2)^{2024} \text{ mod } 33 \\ &= 2^{2024} \text{ mod } 33 \\ &= 2^{2020} 2^4 \text{ mod } 33 \\ &= (2^5)^{404} 2^4 \text{ mod } 33 \\ &= (33 - 1)^{404} 2^4 \text{ mod } 33 \\ &= (-1)^{404} 2^4 \text{ mod } 33 \\ &= 2^4 \text{ mod } 33 \\ &= 16 \text{ mod } 33 \\ \text{Jadi hasil sisa adalah } 16 \\ \end{align} </math> </div></div> <ol start=106> <li>Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 99^{99} &= (3^2 \times 11)^{99} \\ &= 3^{198} \times 11^{99} \\ 243^n &= (3^5)^n \\ &= 3^{5n} \\ \text{agar bisa membagi, maka} \\ 5n &= 198 \\ n &= 39.6 \\ \text{jadi bilangan n terbesar adalah } 39 \\ \end{align} </math> </div></div> <ol start=107> <li>Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸?</lu></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 88^{88} &= (8 \times 11)^{88} \\ &= 8^{88} \times 11^{88} \\ &= 8^{87} \times 8 \times 11^{88} \\ &= (8^3)^{29} \times 8 \times 11^{88} \\ &= 512^{29} \times 8 \times 11^{88} \\ 512^n &= 512^{29} \\ \text{jadi bilangan n terbesar adalah } 29 \\ \end{align} </math> </div></div> <ol start=108> <li>Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6!</li></ol> ; cara 1 : KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. : N dibagi 3 sisa 1 : N dibagi 5 sisa 2 : N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a : KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70 : KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21 : KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97 ; cara 2 : Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35 : kemudian ditambahkan sisa masing-masing sesuai dengan KPK. : KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, <b>97</b> : KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, <b>97</b> : KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, <b>97</b> Jadi bilangan bulat positif adalah 97 :: NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 <ol start=109> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |- | 4 || 0 || Semua isi ember B dibuang |- | 1 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. <ol start=110> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |} nah ada ember A berisi 4 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |- | 1 || 0 || Semua isi ember B dibuang |- | 0 || 1 || Tuangkan 1 l dari ember A ke B sehingga ember A kosong |- | 3 || 1 || Isikan 3 l ke ember A |- | 0 || 4 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |} nah ada ember B berisi 4 liter. # Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kemungkinan 1} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 +3 \cdot ((\frac{4}{3}^x))^2 &= 7 \cdot (\frac{4}{3}^x) \\ \text{misalkan } \frac{4}{3}^x = y \\ 4 +3y^2 &= 7y \\ 3y^2-7y+4 &= 0 \\ (3y-4)(y-1) &= 0 \\ y_1 = \frac{4}{3} &\text{ atau } y_2 = 1 \\ : y_1 = \frac{4}{3} \\ \frac{4}{3}^x &= \frac{4}{3} \\ x &= 1 \\ : y_2 = 1 \\ \frac{4}{3}^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-6}></math>? [[Kategori:Soal-Soal Matematika]] m3qpuwr8g8kxqzpeiy7ap1cavojbiq5 117485 117484 2026-07-13T00:58:14Z Akuindo 8654 117485 wikitext text/x-wiki contoh soal <ol start=1> <li>Berapa hasil dari <math>\sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Misalkan 2020 = p} \\ \sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64} &= \sqrt{(2020-5) \cdot (2020-3) \cdot (2020+3) \cdot (2020+5) + 64} \\ &= \sqrt{(p-5) \cdot (p-3) \cdot (p+3) \cdot (p+5) + 64} \\ &= \sqrt{(p-5) \cdot (p+5) \cdot (p-3) \cdot (p+3) + 64} \\ &= \sqrt{(p^2-25) \cdot (p^2-9) + 64} \\ &= \sqrt{p^4-34p^2+ 225 + 64} \\ &= \sqrt{p^4-34p^2+ 289} \\ &= \sqrt{(p^2-17)^2} \\ &= p^2-17 \\ &= 2020^2-17 \\ &= (2000+20)^2-17 \\ &= 4.000.000+80.000+400-17 \\ &= 4.080.383 \\ \end{align} </math> </div></div> <ol start=2> <li>Berapa nilai x dari <math>\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} = \frac{9}{10}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} &= \frac{9}{10} \\ \text{misalkan untuk } \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} = p \\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} &= p \\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}} &= p^2 \\ x^2-x-p &= p^2 \\ x^2-2x+1+x-1 &= p^2+p \\ (x-1)^2+(x-1) &= p^2+p \\ x-1 &= p \\ \text{misalkan untuk } \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}} &= q^3 \\ x^2 q &= q^3 \\ x^2 &= q^2 \\ x &= q \\ \frac{x-1}{x} &= \frac{9}{10} \\ x &= 10 \\ \end{align} </math> </div></div> <ol start=3> <li>Berapa nilai x dari <math>(\frac{x}{x+10})^{x+10}=\frac{1}{1024}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{x+10}{x})^{-(x+10)} &= (1024)^{-1} \\ (\frac{x+10}{x})^{x+10} &= 1024 \\ (\frac{x+10}{x})^{x+10} &= 2^{10} \\ (\frac{x+10}{x})^{\frac{x+10}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (\frac{1}{2})^{-1} \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (1+(-\frac{1}{2}))^{(1+(-\frac{2}{1}))} \\ \frac{10}{x} &= -\frac{1}{2} \\ x &= -20 \\ \end{align} </math> </div></div> <ol start=4> <li>Berapa nilai x dari <math>x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{2}+\sqrt{x+\frac{1}{4}} &= (\sqrt{x+\frac{1}{4}})^2+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+(\frac{1}{2})^2 \\ &= (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 \\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}} &= 4 \\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2} &= 4 \\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 4 \\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 &= 4 \\ \sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 2 \\ \sqrt{x+\frac{1}{4}} &= \frac{3}{2} \\ x+\frac{1}{4} &= \frac{9}{4} \\ x &= 2 \\ \end{align} </math> </div></div> <ol start=5> <li>Berapa nilai x dari <math>\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^3}{\sqrt{8-x^2}}+x^2-8 &= 0 \\ \frac{x^3}{\sqrt{8-x^2}} &= 8-x^2 \\ x^3 &= (8-x^2)^{\frac{3}{2}} \\ x &= (8-x^2)^{\frac{1}{2}} \\ x^2 &= 8-x^2 \\ 2x^2-8 &= 0 \\ x^2-4 &= 0 \\ (x-2)(x+2) &= 0 \\ \text{membuktikan } \\ x=2 \text{ maka hasilnya 0 } \\ x=-2 \text{ maka hasilnya -8 } \\ \text{jadi } x=2 \\ \end{align} </math> </div></div> <ol start=6> <li>Berapa nilai x dari <math>\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} = 5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} &= 5 \\ \frac{x^{50}+x^{60}+x^{70}}{31}} &= 5^5 \\ x^{50}+x^{60}+x^{70} &= 5^5 \cdot 31 \\ x^{50}(1+x^{10}+x^{20}) &= 5^5 \cdot 31 \\ (x^{10}^5)(1+x^{10}+(x^{10}^2) &= 5^5 \cdot 31 \\ \text{ misalkan } x^{10} = a \\ a^5(1+a+a^2) &= 5^5 \cdot 31 \\ a &= 5 \\ x^{10} &= 5 \\ x &= ^5 log 10 \\ \end{align} </math> </div></div> <ol start=7> <li>Berapa nilai x dari <math>\sqrt{3x+5+\sqrt{4x+5}} = x</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{3x+5+\sqrt{4x+5}} &= x \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \text{misalkan } \sqrt{4x+5}=y \text{ dan } 4x+5=y^2 \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \sqrt{y^2+y-x} &= x \\ y^2+y &= x^2+x \\ y=x \\ 4x+5 &= y^2 \\ 4x+5 &= x^2 \\ x^2-4x-5 &= 0 \\ (x-5)(x+1) &= 0 \\ x=5 &\text{ atau } x=-1 \text{ (TM) } \\ \end{align} </math> </div></div> <ol start=8> <li>Berapa nilai x dari <math>\sqrt{1+\sqrt{1+x}} = \sqrt[3]{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{1+\sqrt{1+x}} &= \sqrt[3]{x} \\ \sqrt[3]{x} &= n \\ x &= n^3 \\ \sqrt{1+\sqrt{1+n^3}} &= n \\ 1+\sqrt{1+n^3} &= n^2 \\ \sqrt{1+n^3} &= n^2-1 \\ 1+n^3 &= n^4-2n^2+1 \\ n^4-n^3-2n^2 &= 0 \\ n^2(n^2-n-2) &= 0 \\ n^2(n-2)(n+1) &= 0 \\ n=0, n=2 \text{ atau } n=-1 \\ n &= 0 \\ x &= 0^3 \\ &= 0 \\ n &= 2 \\ x &= 2^3 \\ &= 8 \\ n &= -1 \\ x &= (-1)^3 \\ &= -1 \\ \text{yang paling mungkin untuk nilai x adalah } 8 \\ \end{align} </math> </div></div> <ol start=9> <li>Berapa nilai x dari <math>\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} &= \frac{\sqrt{1+x}}{\sqrt{x}} \\ \sqrt{x}(\sqrt{1+x}+\sqrt{x}) &= (\sqrt{1+x}-\sqrt{x})\sqrt{1+x} \\ \sqrt{x(1+x)}+x &= 1+x-\sqrt{x(1+x)} \\ 2\sqrt{x(1+x)} &= 1 \\ \sqrt{x(1+x)} &= \frac{1}{2} \\ x(1+x) &= \frac{1}{4} \\ x^2+x &= \frac{1}{4} \\ 4x^2+4x &= 1 \\ 4x^2+4x-1 &= 0 \\ x &= \frac{-4 \pm \sqrt{4^2-4(4)(-1)}}{2(4)} \\ &= \frac{-4 \pm \sqrt{32}}{8} \\ &= \frac{-4 \pm 4\sqrt{2}}{8} \\ &= \frac{-1 \pm \sqrt{2}}{2} \\ \text{karena akar x harus minimal nol jadi } x = \frac{-1+\sqrt{2}}{2} \\ \end{align} </math> </div></div> <ol start=10> <li>Berapa nilai x dari <math>\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x-\sqrt{x+1}}{x+\sqrt{x+1}} &= \frac{11}{19} \\ \text{misalkan } \sqrt{x+1}=y \text{ dan } x=y^2-1 \\ \frac{y^2-1-y}{y^2-1+y} &= \frac{11}{19} \\ 19(y^2-y-1) &= 11(y^2+y-1) \\ 19y^2-19y-19 &= 11y^2+11y-11 \\ 8y^2-30y-8 &= 0 \\ 4y^2-15y-4 &= 0 \\ (4y+1)(y-4) &= 0 \\ y=-\frac{1}{4} \text{ (TM) atau } & y=4 \\ x &= 4^2-1 \\ &= 15 \\ \end{align} </math> </div></div> <ol start=11> <li>Berapa nilai x dari <math>\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} &= 98 \\ \text{misalkan } \sqrt{x^2-1}=y \\ \frac{x+y}{x-y}+\frac{x-y}{x+y} &= 98 \\ \frac{(x+y)^2+(x-y)^2}{(x-y)(x+y)} &= 98 \\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} &= 98 \\ \frac{2(x^2+y^2)}{x^2-y^2} &= 98 \\ \frac{x^2+y^2}{x^2-y^2} &= 49 \\ x^2+y^2 &= 49(x^2-y^2) \\ x^2+y^2 &= 49x^2-49y^2 \\ 48x^2 &= 50y^2 \\ 24x^2 &= 25y^2 \\ 24x^2 &= 25(\sqrt{x^2-1})^2 \\ 24x^2 &= 25(x^2-1) \\ 24x^2 &= 25x^2-25 \\ x^2 &= 25 \\ x &= \pm 5 \\ \end{align} </math> </div></div> <ol start=12> <li>Berapa nilai x dari <math>\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \sqrt{x}=y \text{ dan } x=y^2 \\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} &= \frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\sqrt{\frac{y^2}{y^2+y}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\frac{y}{\sqrt{y^2+y}} \\ y^2+y-\sqrt{(y^2+y)(y^2-y)} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^4-y^2} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^2(y^2-1)} &= \frac{5}{4}y \\ y(y+1)-y\sqrt{y^2-1} &= \frac{5}{4}y \\ y+1-\sqrt{y^2-1} &= \frac{5}{4} \\ -\sqrt{y^2-1} &= \frac{1}{4}-y \\ y^2-1 &= (\frac{1}{4}-y)^2 \\ y^2-1 &= \frac{1}{16}-\frac{1}{2}y+y^2 \\ -1 &= \frac{1}{16}-\frac{1}{2}y \\ \frac{1}{2}y &= \frac{1}{16}+1 \\ \frac{1}{2}y &= \frac{17}{16} \\ y &= \frac{17}{8} \\ x &= (\frac{17}{8})^2 \\ &= \frac{289}{64} \\ \end{align} </math> </div></div> <ol start=13> <li>Berapa nilai x dari <math>\sqrt[4]{62+x}+\sqrt[4]{275-x}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ misalkan } \sqrt[4]{62+x}=a, 62+x=a^4, \sqrt[4]{275-x}=b \text{ dan } 275-x=b^4 \\ a+b &= 7 \\ (a+b)^2 &= 49 \\ a^2+b^2+2ab &= 49 \\ a^2+b^2 &= 49-2ab \\ a^4+b^4 &= 62+x+275-x \\ (a^2+b^2)^2-2(ab)^2 &= 337 \\ (49-2ab)^2-2(ab)^2 &= 337 \\ 2401-196ab+4(ab)^2-2(ab)^2 &= 337 \\ 2(ab)^2-196ab+2064 &= 0 \\ (ab)^2-98ab+1032 &= 0 \\ (ab-12)(ab-86) &= 0 \\ ab = 12 \text{ atau } & ab = 86 \text{ (TM) karena hasil kali maksimum yaitu 12 } \\ ab =12 \text{ dan } a+b=7 \\ a+b &= 7 \\ b &= 7-a \\ ab &= 12 \\ a(7-a) &= 12 \\ -a^2+7a &= 12 \\ a^2-7a+12 &= 0 \\ (a-3)(a-4) &= 0 \\ a=3 \text{ atau } & a=4 \\ a=3, b=4 \\ 62+x &= a^4 \\ 62+x &= (3)^4 \\ 62+x &= 81 \\ x &= 19 \\ a=4, b=3 \\ 62+x &= a^4 \\ 62+x &= (4)^4 \\ 62+x &= 256 \\ x &= 194 \\ \end{align} </math> </div></div> <ol start=14> <li>Berapa nilai x dari <math>\sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2} &= 7 \\ (\sqrt[3]{8+x})^2-\sqrt[3]{8+x} \sqrt[3]{27-x}+(\sqrt[3]{27-x})^2 &= 7 \\ \text{misalkan } \sqrt[3]{8+x}=a, 8+x=a^3, \sqrt[3]{27-x}=b \text{ dan } 27-x=b^3 \\ a^2-ab+b^2 &= 7 \\ a^3+b^3 &= 8+x+27-x \\ &= 35 \\ a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ 35 &= (a+b)(7) \\ a+b &= 5 \\ b &= 5-a \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ 5^3 &= 35+3ab(5) \\ 125 &= 35+15ab \\ 80 &= 15ab \\ ab &= 6 \\ a(5-a) &= 6 \\ 5a-a^2 &= 6 \\ a^2-5a+6 &= 6 \\ (a-2)(a-3) &= 6 \\ a=2 &\text{ atau } a=3 \\ a=2, b=3 \text{ dan } a=3,b=2 \\ 8+x &= a^3 \\ &= 2^3 \\ &= 8 \\ x &= 0 \\ 8+x &= a^3 \\ &= 3^3 \\ &= 27 \\ x &= 19 \\ \end{align} </math> </div></div> <ol start=15> <li>Berapa nilai x dari <math>3^x+5^x-9^x+15^x-25^x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3^x+5^x-9^x+15^x-25^x &= 1 \\ 3^x+5^x-(3^2)^x+(3 \cdot 5)^x-(5^2)^x &= 1 \\ 3^x+5^x-(3^x)^2+(3^x \cdot 5^x)-(5^x)^2 &= 1 \\ \text{misalkan } 3^x=a \text{ dan } 5^x=b \\ a+b-a^2+ab-b^2 &= 1 \\ a^2-ab+b^2-a-b+1 &= 0 \\ 2a^2-2ab+2b^2-2a-2b+2 &= 0 \\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1 &= 0 \\ (a-b)^2+(a-1)^2+(b-1)^2 &= 0 \\ a-b=0; a-1=0; b-1 &= 0 \\ a=b &= 1 \\ 3^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> <ol start=16> <li>Berapa nilai x dari <math>^6log x^2+^{6x}log \frac{6}{x}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ \text{misalkan } 6x=a \text{ maka } x=\frac{a}{6} \\ ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ ^6log (\frac{a}{6})^2+^{6 \frac{a}{6}}log \frac{6}{\frac{a}{6}} &= 1 \\ ^6log \frac{a^2}{6^2}+^alog \frac{6^2}{a} &= 1 \\ ^6log a^2-^6log 6^2+^alog 6^2-^alog a &= 1 \\ 2 ^6log a-2 ^6log 6+2 ^alog 6-^alog a &= 1 \\ 2 ^6log a-2+2 \frac{1}{^6log a}-1 &= 1 \\ 2 ^6log a+2 \frac{1}{^6log a}-4 &= 0 \\ 2 ^6log^2 a-4 ^6log a+2 &= 0 \\ ^6log^2 a-2 ^6log a+1 &= 0 \\ (^6log a-1)^2 &= 0 \\ ^6log a &= 1 \\ a &= 6 \\ x &= \frac{a}{6} \\ &= \frac{6}{6} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=17> <li>Berapa nilai x dari (x+500)³+x=20?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+500)^3+x &= 20 \\ \text{misalkan } a=x+500 \text{ maka } x=a-500 \\ a^3+a-500 &= 20 \\ a^3+a &= 520 \\ a(a^2+1) &= 8 \cdot 65 \\ a(a^2+1) &= 8(64+1) \\ a(a^2+1) &= 8(8^2+1) \\ a &= 8 \\ x &= 8-500 \\ &= -492 \\ \end{align} </math> </div></div> <ol start=18> <li>Berapa nilai x dari <math>\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2} &= 0 \\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}} &= \frac{1}{2} \\ \frac{x^n+4^n}{x^n+16^n} &= (\frac{1}{2})^n \\ \frac{x^n+4^n}{x^n+16^n} &= \frac{1}{2^n} \\ 2^n(x^n+4^n) &= x^n+16^n \\ 2^n(x^n+2^{2n}) &= x^n+2^{4n} \\ 2^n \cdot x^n+2^{3n} &= x^n+2^{4n} \\ 2^n \cdot x^n-x^n &= 2^{4n}-2^{3n} \\ x^n(2^n-1) &= 2^{3n}(2^n-1) \\ x^n &= 2^{3n} \\ x^n &= (2^3)^n \\ x^n &= 8^n \\ x &= 8 \\ \end{align} </math> </div></div> <ol start=19> <li>Berapa hasil dari <math>\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ x &= \frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5 \cdot 6}+\sqrt{5 \cdot 5}+\sqrt{6 \cdot 4}+\sqrt{5 \cdot 4}}{\sqrt{5 \cdot 4}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5} \cdot \sqrt{6}+\sqrt{5} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{4}}{2 \cdot \sqrt{5}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{6} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{5}+\sqrt{5} \cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}} \\ &= \frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ &= \frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ \frac{1}{x} &= \frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}} \\ &= \frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ &= \frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1} \\ &= \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5} \\ &= \sqrt{6}-\sqrt{4} \\ &= \sqrt{6}-2 \\ x &= \frac{1}{\sqrt{6}-2} \\ &= \frac{\sqrt{6}+2}{6-4} \\ &= \frac{\sqrt{6}+2}{2} \\ &= 1+\frac{\sqrt{6}}{2} \\ \end{align} </math> </div></div> <ol start=20> <li>Berapa hasil dari <math>(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5 \\ \text{misalkan } x=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ x &= \frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ &= \frac{\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}} \\ &= \frac{\sqrt{3}+1}{4} \\ 4x &= \sqrt{3}+1 \\ 4x-1 &= \sqrt{3} \\ (4x-1)^2 &= 3 \\ 16x^2-8x+1 &= 3 \\ 16x^2 &= 8x+2 \\ 8x^2 &= 4x+1 \\ x^2 &= \frac{4x+1}{8} \\ *cara 1 \\ x^3 &= x \cdot x^2 \\ &= x(\frac{4x+1}{8}) \\ &= \frac{4x^2+x}{8} \\ &= \frac{4x^2}{8}+\frac{x}{8} \\ &= \frac{4(\frac{4x+1}{8})}{8}+\frac{x}{8} \\ &= \frac{16x+4}{64}+\frac{x}{8} \\ &= \frac{4x+1}{16}+\frac{x}{8} \\ &= \frac{4x+1+2x}{16} \\ &= \frac{6x+1}{16} \\ x^5 &= x^2 \cdot x^3 \\ &= (\frac{4x+1}{8})(\frac{6x+1}{16}) \\ &= \frac{24x^2+10x+1}{128} \\ &= \frac{24x^2}{128}+\frac{10x+1}{128} \\ &= \frac{24(\frac{4x+1}{8})}{128}+\frac{10x+1}{128} \\ &= \frac{96x+24}{1024}+\frac{10x+1}{128} \\ &= \frac{96x+24+80x+8}{1024} \\ &= \frac{176x+32}{1024} \\ &= \frac{176x}{1024}+\frac{32}{1024} \\ &= \frac{176}{1024}(\frac{\sqrt{3}+1}{4})+\frac{32}{1024} \\ &= \frac{44(\sqrt{3}+1)}{1024}+\frac{32}{1024} \\ &= \frac{44\sqrt{3}+44}{1024}+\frac{32}{1024} \\ &= \frac{76+44\sqrt{3}}{1024} \\ &= \frac{19+11\sqrt{3}}{256} \\ *cara 2 \\ x^4 &= (x^2)^2 \\ &= (\frac{4x+1}{8})^2 \\ &= \frac{16x^2+8x+1}{64} \\ &= \frac{16x^2}{64}+\frac{8x}{64}+\frac{1}{64} \\ &= \frac{x^2}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{\frac{4x+1}{8}}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{4x}{32}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{8}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{4}+\frac{3}{64} \\ x^5 &= x \cdot x^4 \\ &= (\frac{\sqrt{3}+1}{4})(\frac{x}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\frac{\sqrt{3}+1}{4}}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\sqrt{3}+1}{16}+\frac{3}{64}) \\ &= \frac{(\sqrt{3}+1)^2}{64}+(\frac{\sqrt{3}+1}{4})\frac{3}{64} \\ &= \frac{3+2\sqrt{3}+1}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{4+2\sqrt{3}}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{16+8\sqrt{3}}{256}+\frac{3\sqrt{3}+3}{256} \\ &= \frac{19+11\sqrt{3}}{256} \\ \end{align} </math> </div></div> <ol start=21> <li>Berapa hasil dari <math>\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4} &= \frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots) \\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots &= \frac{1}{1-\frac{1}{4}} \\ &= \frac{4}{3} \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{4}{3}) \\ &= \frac{1}{4}+\frac{16}{3} \\ &= \frac{67}{12} \\ x &= \frac{67}{9} \\ \end{align} </math> </div></div> <ol start=22> <li>Berapa nilai y-x jika <math>\frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} = \frac{x}{y}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} &= \frac{x}{y} \\ \frac{\frac{106 \times 107}{2}}{\frac{106}{2}(4+109)} &= \frac{x}{y} \\ \frac{53 \times 107}{53 \times 113} &= \frac{x}{y} \\ y-x &= 113-107 = 6 \\ \end{align} </math> </div></div> <ol start=23> <li>Berapa angka satuan dari hasil 17²⁰²⁴?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan angka satuannya} \\ 17^1 &= 7 \\ 17^2 &= 9 \\ 17^3 &= 3 \\ 17^4 &= 1 \\ 17^5 &= 7 \\ 17^6 &= 9 \\ 17^7 &= 3 \\ 17^8 &= 1 \\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1} \end{align} </math> </div></div> <ol start=24> <li>Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ 1! + 2! + 3! + 4! + \dots + 2024! &= 1 + (1x2) + (1x2x3) + (1x2x3x4) + \dots + 2024! \\ &= 1 + 2 + 6 + 24 + 120 + 720 + \dots + 2024! \\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka } 1+2+6+24 = 33 \text{ jadi angka satuannya adalah } 3 \end{align} </math> </div></div> <ol start=25> <li>Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ \frac{1! + 2! + 3! + 4! + \dots + 2024!}{12} &= \frac{1 + 1x2 + 1x2x3 + 1x2x3x4 + \dots + 2024!}{12} \\ &= \frac{1 + 2 + 6 + 24 + \dots + 2024!}{12} \\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi } 1+2+6 = 9 \end{align} </math> </div></div> <ol start=26> <li>Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{penjumlahan} \\ 1+1+1+1+ \dots \text{ (sebanyak 88 buah) }+x+y &= A \\ 88+x+y &= A \\ \text{perkalian} \\ 1 \times 1 \times 1 \times \dots \text{ (sebanyak 88 buah) }\times x \times y &= A \\ x \times y &= A \\ 88+x+y &= xy \\ xy-y &= 88+x \\ y(x-1) &= 88+x \\ y &= \frac{88+x}{x-1} \\ \text{uji selidiki untuk x=2} \\ y &= \frac{88+2}{2-1} \\ &= 90 \\ \text{buktikan} \\ 88+x+y &= xy \\ 88+2+90 &= 2(90) \\ 180 &= 180 \\ \text{terbukti} \\ \text{nilai A adalah } 180 \\ \end{align} </math> </div></div> <ol start=27> <li>Berapakah nilai x, y dan z dari <math>x+y-z=1, x^2+y^2-z^2=-5 \text{ dan } x^3+y^3-z^3=-53</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+y-z &= 1 \\ x+y &= z+1 \\ x^2+2xy+y^2 &= z^2+2z+1 \\ x^2+y^2-z^2 &= 2z+1-2xy \\ -5 &= 2z+1-2xy \\ 2xy &= 2z+6 \\ xy &= z+3 \\ x^2+y^2-z^2 &= -5 \\ x^2+y^2 &= z^2-5 \\ x^3+y^3-z^3 &= -53 \\ (x+y)(x^2-xy+y^2)-z^3+53 &= 0 \\ (x+y)(x^2+y^2-xy)-z^3+53 &= 0 \\ (z+1)(z^2-5-(z+3))-z^3+53 &= 0 \\ (z+1)(z^2-z-8)-z^3+53 &= 0 \\ z^3-z^2-8z+z^2-z-8-z^3+53 &= 0 \\ -9z+45 &= 0 \\ -9z &= -45 \\ z &= 5 \\ x+y &= 5+1 \\ x+y &= 6 \\ x &= 6-y \\ xy &= 5+3 \\ xy &= 8 \\ (6-y)y &= 8 \\ 6y-y^2 &= 8 \\ y^2-6y+8 &= 0 \\ (y-4)(y-2) &= 0 \\ y=4 \text{ atau } y=2 \\ \text{jika } y=4 \\ x+y &= z+1 \\ x+4 &= 5+1 \\ x &= 2 \\ \text{jika } y=2 \\ x+y &= z+1 \\ x+2 &= 5+1 \\ x &= 4 \\ \end{align} </math> </div></div> <ol start=28> <li>Berapakah nilai titik koordinat (x,y) dari <math>\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}</math> dan <math>\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x+y}+\sqrt{x-y} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+y}-\sqrt{x-y} &= \sqrt{\frac{52y}{3x}} \\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y}) &= \sqrt{\frac{432x}{13y}} \cdot \sqrt{\frac{52y}{3x}} \\ x+y-x+y &= \sqrt{\frac{432x \cdot 52y}{13y \cdot 3x}} \\ 2y &= \sqrt{144 \cdot 4} \\ 2y &= \sqrt{576} \\ 2y &= 24 \\ y &= 12 \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13(12)}} \\ x+12+x-12+2 \cdot \sqrt{x+12} \cdot \sqrt{x-12} &= \frac{36x}{13} \\ 2x+2 \sqrt{x^2-144} &= \frac{36x}{13} \\ 2(x+\sqrt{x^2-144}) &= \frac{36x}{13} \\ x+\sqrt{x^2-144} &= \frac{18x}{13} \\ \sqrt{x^2-144} &= \frac{5x}{13} \\ x^2-144 &= \frac{25x^2}{169} \\ \frac{144x^2}{169}-144 &= 0 \\ \frac{x^2}{169}-1 &= 0 \\ x^2-169 &= 0 \\ (x-13)(x+13) &= 0 \\ x_1=13 &\text{ atau } x_2=-13 \text{ (TM) karena } x>y \\ \end{align} </math> jadi titik koordinat (13,12) </div></div> <ol start=29> <li>Berapakah nilai dari <math>x^2-7x</math> jika <math>(x-2)^2+\frac{1}{(x-2)^2} = 11</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x-2)^2+\frac{1}{(x-2)^2} &= 11 \\ (x-2)^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2)^2} &= 11-2 \\ (x-2-\frac{1}{x-2})^2 &= 9 \\ x-2-\frac{1}{x-2} &= 3 \\ (x-2)^2-1 &= 3(x-2) \\ x^2-4x+4-1 &= 3x-6 \\ x^2-7x &= -9 \\ \end{align} </math> </div></div> <ol start=30> <li>Berapakah nilai dari <math>\frac{(x+y)^2(x+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)}</math> jika xy+yz+xz=1?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy+yz+xz &= 1 \\ x^2+xy+yz+xz &= x^2+1 \\ x(x+y)+z(x+y) &= x^2+1 \\ (x+y)(x+z) &= x^2+1 \\ \text{dengan pola yang sama } \\ (y+x)(y+z) &= y^2+1 \\ (x+z)(y+z) &= z^2+1 \\ \frac{(x+y)^2(y+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)} &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)} \\ &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)^2(y+z)^2(x+z)^2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=31> <li>Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} w+5 &= w+x+y+z+5 \\ x+4 &= w+x+y+z+5 \\ y+3 &= w+x+y+z+5 \\ z+2 &= w+x+y+z+5 \\ \text{jumlahkan keempat persamaan } \\ w+x+y+z+14 &= 4(w+x+y+z+5) \\ w+x+y+z+14 &= 4(w+x+y+z)+20 \\ 3(w+x+y+z) &= -6 \\ w+x+y+z &= -2 \\ \end{align} </math> </div></div> <ol start=32> <li>Berapakah nilai dari <math>\frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2}</math> jika <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3</math> dan x+y+z=xyz?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2} &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2} \\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) \\ 3^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz}) \\ 9 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz}) \\ &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2 \\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} &= 7 \\ \end{align} </math> </div></div> <ol start=33> <li>Berapakah nilai dari <math>\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}</math> jika <math>x^2+y^2+z^2 = -2(ab+bc+ac)</math>?>/li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+z^2 &= -2(xy+yz+xz) \\ x^2+y^2+z^2+2(xy+yz+xz) &= 0 \\ (x+y+z)^2 &= 0 \\ x+y+z &= 0 \\ x+y &= -z \\ x+z &= -y \\ y+z &= -x \\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z} &= \frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x} \\ &= -2-(-5)-(-7) \\ &= 10 \\ \end{align} </math> </div></div> <ol start=34> <li>Berapakah nilai dari <math>\frac{20xyz}{xy+yz+xz}</math> jika <math>16^x = 256^y = 625^z = 40</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 16^x = 256^y = 625^z &= 40 \\ 2^{4x} = 4^{4y} = 5^{4z} &= 40 \\ 2^{4x} &= 40 \\ 2 &= 40^{\frac{1}{4x}} \\ 4^{4y} &= 40 \\ 4 &= 40^{\frac{1}{4y}} \\ 5^{4z} &= 40 \\ 5 &= 40^{\frac{1}{4z}} \\ 2 \cdot 4 \cdot 5 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z}} \\ 1 &= \frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z} \\ 4 &= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{20xyz}{xy+yz+xz} &= 20 \cdot \frac{xyz}{xy+yz+xz} \\ &= 20 \cdot (\frac{xy+yz+xz}{xyz})^{-1} \\ &= 20 \cdot (\frac{1}{z} + \frac{1}{x} + \frac{1}{y})^{-1} \\ &= 20 \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{-1} \\ &= 20 \cdot (4)^{-1} \\ &= 20 \cdot \frac{1}{4} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=35> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+3x^2+1}</math> jika <math>6x^2+25x+6=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 6x^2+25x+6 &= 0 \\ 6x+25+\frac{6}{x} &= 0 \\ 6(x+\frac{1}{x}) &= -25 \\ x+\frac{1}{x} &= \frac{-25}{6} \\ (c+\frac{1}{x})^2 &= (\frac{-25}{6})^2 \\ x^2+2+\frac{1}{x^2} &= \frac{625}{36} \\ x^2+\frac{1}{x^2} &= \frac{625}{36}-2 \\ x^2+\frac{1}{x^2} &= \frac{553}{36} \\ \frac{x^2}{x^4+3x^2+1} &= \frac{1}{x^2+3+\frac{1}{x^2}} \\ &= \frac{1}{a^2+\frac{1}{x^2}+3} \\ &= \frac{1}{\frac{553}{36}+3} \\ &= \frac{1}{\frac{661}{36}} \\ &= \frac{36}{661} \\ \end{align} </math> </div></div> <ol start=36> <li>Berapakah nilai dari <math>\frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5} &= \frac{(9+2\sqrt{20})^{1013}}{((2)^3+3(2)^2(\sqrt{5})+3(2)(\sqrt{5})^2+(\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{((2+\sqrt{5})^2)^{1013}}{((2+\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{(2+\sqrt{5})^{2026}}{(2+\sqrt{5})^{2025}}+6-\sqrt{5} \\ &= 2+\sqrt{5}+6-\sqrt{5} \\ &= 8 \\ \end{align} </math> </div></div> <ol start=37> <li>Berapakah nilai dari <math>27x^3+\frac{8}{x^3}</math> jika <math>3x+\frac{2}{x}=6</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3x+\frac{2}{x} &= 6 \\ (3x+\frac{2}{x})^3 &= 6^3 \\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3} &= 216 \\ 27x^3+18(6)+\frac{8}{x^3} &= 216 \\ 27x^3+108+\frac{8}{x^3} &= 216 \\ 27x^3+\frac{8}{x^3} &= 108 \\ \end{align} </math> </div></div> <ol start=38> <li>Berapakah nilai dari <math>x^6+\frac{8}{x^3}</math> jika <math>x^3+\frac{1}{x^3}=8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3+\frac{1}{x^3} &= 8 \\ x^3 &= 8-\frac{1}{x^3} \\ x^6 &= 8x^3-1 \\ x^6+\frac{8}{x^3} &= 8x^3-1+\frac{8}{x^3} \\ &= 8x^3+\frac{8}{x^3}-1 \\ &= 8(x^3+\frac{1}{x^3})-1 \\ &= 8(8)-1 \\ &= 63 \\ \end{align} </math> </div></div> <ol start=39> <li>Berapakah nilai dari <math>4x+\frac{25}{x}</math> jika <math>2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2\sqrt{x}+\frac{5}{\sqrt{x}} &= 4x-\frac{25}{x} \\ 2\sqrt{x}+\frac{5}{\sqrt{x}} &= (2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}}) \\ 1 &= 2\sqrt{x}-\frac{5}{\sqrt{x}} \\ 1^2 &= (2\sqrt{x}-\frac{5}{\sqrt{x}})^2 \\ 1 &= 4x-20+\frac{25}{x} \\ 4x+\frac{25}{x} &= 21 \\ \end{align} </math> </div></div> <ol start=40> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2-x+1}{x^2+x+1} &= \frac{5}{6} \\ \frac{x^2+1-x}{x^2+1+x} &= \frac{5}{6} \\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} &= \frac{5}{6} \\ \text{ misalkan } x+\frac{1}{x} &= y \\ \frac{y-1}{y+1} &= \frac{5}{6} \\ 6(y-1) &= 5(y+1) \\ 6y-6 &= 5y+5 \\ y &= 11 \\ x+\frac{1}{x} &= 11 \\ \end{align} </math> </div></div> <ol start=41> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt{x}+x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x}+x &= 1 \\ x-1 &= -\sqrt{x} \\ (x-1)^2 &= (-\sqrt{x})^2 \\ x^2-2x+1 &= x \\ x^2-3x+1 &= 0 \\ x-3+\frac{1}{x} &= 0 \\ x+\frac{1}{x} &= 3 \\ \end{align} </math> </div></div> <ol start=42> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt[3]{x}-\sqrt[3]{x-36}=3</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x}-\sqrt[3]{x-36} &= 3 \\ (\sqrt[3]{x}-\sqrt[3]{x-36})^3 &= 3^3 \\ x-(x-36)-3 \sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36}) &= 27 \\ 36-3 \sqrt[3]{x(x-36)}3 &= 27 \\ -9 \sqrt[3]{x(x-36)} &= -9 \\ \sqrt[3]{x(x-36)} &= 1 \\ x(x-36) &= 1 \\ x^2-36x-1 &= 0 \\ x-36-\frac{1}{x} &= 0 \\ x-\frac{1}{x} &= 36 \\ \end{align} </math> </div></div> <ol start=43> <li>Berapakah nilai dari <math>x+\frac{16}{x}</math> jika <math>x-3\sqrt{x}=4</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x-3\sqrt{x} &= 4 \\ x-4 &= 3\sqrt{x} \\ x^2-8x+16 &= 9x \\ x^2-17x+16 &= 0 \\ x-17+\frac{16}{x} &= 0 \\ x+\frac{16}{x} &= 17 \\ \end{align} </math> </div></div> <ol start=44> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+4}</math> jika <math>x^2-7x+2=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2-7x+2 &= 0 \\ x^2+2 &= 7x \\ x+\frac{2}{x} &= 7 \\ x^2+4+\frac{4}{x^2} &= 49 \\ x^2+\frac{4}{x^2} &= 45 \\ \frac{x^4+4}{x^2} &= 45 \\ \frac{x^2}{x^4+4} &= \frac{1}{45} \\ \end{align} </math> </div></div> <ol start=45> <li>Berapakah nilai dari <math>x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}</math> jika <math>x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}} &= 25 \\ x^{\frac{1}{2}}+x^{-\frac{1}{2}} &= 23 \\ x+2+x^{-1} &= 529 \\ x+x^{-1} &= 527 \\ x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+x^{-\frac{3}{4}} &= 110 \\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1} &= x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}} \\ &= 527+110 \\ &= 637 \\ \end{align} </math> </div></div> <ol start=46> <li>Berapakah nilai dari <math>\sqrt{8x^6+x^5+x^4+5x^3+1}</math> jika <math>\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5} &= 0 \\ \frac{x^2+x+1}{x^5} &= 0 \\ x^2+x+1 &= 0 \\ x^2+x+1 &= 0 \\ (x-1)(x^2+x+1) &= 0(x-1) \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ \sqrt{8x^6+x^5+x^4+5x^3+1} &= \sqrt{(2x^3)^2+x^3x^2+x^3x+5x^3+1} \\ &= \sqrt{(2(1))^2+(1)x^2+(1)x+5(1)+1} \\ &= \sqrt{(2)^2+x^2+x+5+1} \\ &= \sqrt{4+x^2+x+1+5} \\ &= \sqrt{4+0+5} \\ &= \sqrt{9} \\ &= 3 \\ \end{align} </math> </div></div> <ol start=47> <li>Berapakah nilai dari <math>f(1)+f(2)+f(3)+ \dots + f(99)</math> jika <math>f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{1}{\sqrt{x+1}+\sqrt{x}} \\ &= \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} \\ &= \sqrt{x+1}-\sqrt{x} \\ f(1)+f(2)+f(3)+ \dots + f(98)+f(99) &= \sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+ \cdot + \sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99} \\ &= \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+ \cdot + \sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ &= \sqrt{100}-\sqrt{1} \\ &= 10-1 \\ &= 9 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots + \frac{2024}{2025})</math> jika <math>h(x)=\frac{3}{3+9^x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} h(x) &= \frac{3}{3+9^x} \\ h(1-x) &= \frac{3}{3+9^{1-x}} \\ &= \frac{3}{3+\frac{9}{9^x}} \\ &= \frac{9^x}{3+9^x} \\ h(x)+h(1-x) &= \frac{3}{3+9^x}+\frac{9^x}{3+9^x} \\ &= \frac{3+9^x}{3+9^x} \\ &= 1 \\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots +(1-\frac{2}{2025})+(1-\frac{1}{2025})) \\ & 5(1+1+1+ \dots +1+1) \text{ sebanyak 1012 kali } \\ & 5(1012) \\ & 5060 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>\frac{7^{2025} - 7^{2023} + 432}{7^{2024} + 7^{2023} + 72}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72} &= \frac{7^{2023}7^{2}-7^{2023} + 48 \times 9}{7^{2023}7^1+7^{2023}+8 \times 9} \\ &= \frac{7^{2023}(7^{2}-1)+48 \times 9}{7^{2023}(7^1+1)+8 \times 9} \\ &= \frac{7^{2023}(49-1)+48 \times 9}{7^{2023}(7+1) + 8 \times 9} \\ &= \frac{7^{2023} \times 48+48 \times 9}{7^{2023} \times 8+8 \times 9} \\ &= \frac{48(7^{2023}+9)}{8(7^{2023}+9)} \\ &= \frac{48}{8} \\ &= 6 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>tan (x+\frac{\pi}{4})</math> jika <math>\frac{1}{cos x}-tan x = \frac{4}{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{cos x}-tan x &= \frac{4}{5} \\ sec x-tan x &= \frac{4}{5} \\ sec^2 x-tan^2 x &= 1 \\ (sec x+tan x)(sec x-tan x) &= 1 \\ (sec x+tan x)\frac{4}{5} &= 1 \\ sec x+tan x &= \frac{5}{4} \\ \text{kedua persamaan dengan cara metode eliminasi } \\ 2 tan x &= \frac{5}{4}-\frac{4}{5} \\ 2 tan x &= \frac{9}{20} \\ tan x &= \frac{9}{40} \\ tan (x+\frac{\pi}{4}) &= \frac{tan x+tan \frac{\pi}{4}}{1-tan x \cdot tan \frac{\pi}{4}} \\ &= \frac{\frac{9}{40}+1}{1-\frac{9}{40} \cdot 1} \\ &= \frac{\frac{49}{40}}{\frac{31}{40}} \\ &= \frac{49}{31} \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>sin^3 x+csc^3 x</math> jika <math>sin x-csc x = 8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ Dengan menggunakan rumus: } (a-b)^3 &= a^3-b^3-3ab(a-b) \\ (sin x-csc x)^3 &= sin^3 x-csc^3 x-3sin x csc x(sin x-csc x) \\ 8^3 &= sin^3 x-csc^3 x-3sin x (\frac{1}{sin x})(8) \\ 512 &= sin^3 x-csc^3 x-24 \\ sin^3 x-csc^3 x &= 512+24 \\ sin^3 x-csc^3 x &= 536 \\ \end{align} </math> </div></div> <ol start=52> <li>Berapakah nilai dari <math>(sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2</math> jika <math>\frac{1}{sin x}+\frac{1}{cos x} = 10</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{sin x}+\frac{1}{cos x} &= 10 \\ \frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} &= 100 \\ (sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2 &= sin^2 x+\frac{2sin x}{cos x}+\frac{1}{cos^2 x}+cos^2 x+\frac{2cos x}{sin x}+\frac{1}{sin^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2(sin^2 x+cos^2 x)}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+100 \\ &= 101 \\ \end{align} </math> </div></div> <ol start=53> <li>Berapakah nilai dari (x-1)⁶ jika <math>x=\frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin 80^\circ &= cos 10^\circ \\ sin 80^\circ-cos 10^\circ &= 0 \\ x &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+2 sin 20^\circ cos 20^\circ}{cos 10^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+4 sin 10^\circ cos 10^\circ cos 20^\circ}{cos 10^\circ} \\ &= 4 cos 55^\circ cos 25^\circ+4 sin 10^\circ cos 20^\circ \\ &= 2(2 cos 55^\circ cos 25^\circ+2 sin 10^\circ cos 20^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ+sin (-10)^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ-sin 10^\circ) \\ &= 2(cos 80^\circ-sin 10^\circ+cos 30^\circ+sin 30^\circ) \\ &= 2(cos 80^\circ-sin (90^\circ-80^\circ)+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(cos 80^\circ-cos 80^\circ+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= \sqrt{3}+1 \\ x-1 &= \sqrt{3} \\ (x-1)^6 &= (\sqrt{3})^6 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=54> <li>Berapakah nilai dari x jika <math>x=\frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ} \\ 2x sin 20^\circ-x sin 40 ^\circ &= x sin 20^\circ-x^2 sin 10^\circ \\ x^2 sin 10^\circ+x sin 20^\circ-x sin 40 ^\circ &= 0 \\ x(x sin 10^\circ+sin 20^\circ-sin 40 ^\circ) &= 0 \\ x = 0 &\text{ atau } x sin 10^\circ+sin 20^\circ-sin 40 ^\circ = 0 \\ x sin 10^\circ+sin 20^\circ-sin 40 ^\circ &= 0 \\ x sin 10^\circ &= sin 40 ^\circ-sin 20^\circ \\ x &= \frac{sin 40 ^\circ-sin 20^\circ}{sin 10^\circ} \\ &= \frac{2 cos 30 ^\circ sin 10^\circ}{sin 10^\circ} \\ &= 2 cos 30 ^\circ \\ &= \frac{2 \sqrt{3}}{2} \\ &= \sqrt{3} \\ \end{align} </math> </div></div> <ol start=55> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x^2}{x^2-16y^2} = \frac{625}{49}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2}{x^2-16y^2} &= \frac{625}{49} \\ \frac{x^2-16y^2}{x^2} &= \frac{49}{625} \text{ (terbalik posisinya)} \\ 1-\frac{16y^2}{x^2} &= \frac{49}{625} \\ \frac{16y^2}{x^2} &= 1 - \frac{49}{625} \\ (\frac{4y}{x})^2 &= \frac{576}{625} \\ (\frac{4y}{x})^2 &= (\frac{24}{25})^2 \\ \frac{4y}{x} &= \frac{24}{25} \\ \frac{y}{x} &= \frac{6}{25} \\ \frac{x}{y} &= \frac{25}{6} \\ \end{align} </math> </div></div> <ol start=56> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x}{y}+\frac{x+10y}{y+10x} = 2</math> serta bilangan real untuk x dan y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x}{y}+\frac{x+10y}{y+10x} &= 2 \\ \frac{x}{y}+\frac{\frac{x}{y}+10}{1+10\frac{x}{y}} &= 2 \\ \text{misalkan } \frac{x}{y} = a \\ a+\frac{a+10}{1+10a} &= 2 \\ a(1+10a)+a+10 &= 2(1+10a) \\ 10a^2+a+a+10 &= 2+20a \\ 10a^2-18a+8 &= 0 \\ 5a^2-9a+4 &= 0 \\ (5a-4)(a-1) &= 0 \\ a = \frac{4}{5} &\text{ atau } a = 1 \\ \text{jadi } \frac{x}{y} = {\frac{4}{5}, 1} \\ \end{align} </math> </div></div> <ol start=57> <li>Berapakah nilai dari xy jika <math>x^4+y^4+x^2y^2=15 \text{ dan } x^2+y^2+xy=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+xy &= 5 \\ x^2+y^2 &= 5-xy \\ x^4+y^4+x^2y^2 &= 15 \\ (x^2)^2+(y^2)^2+2x^2y^2-x^2y^2 &= 15 \\ (x^2+y^2)^2-x^2y^2 &= 15 \\ (5-xy)^2-x^2y^2 &= 15 \\ 25-10xy+x^2y^2-x^2y^2 &= 15 \\ 25-10xy &= 15 \\ 10xy &= 10 \\ xy &= 1 \\ \end{align} </math> </div></div> <ol start=58> <li>Berapakah nilai dari x jika <math>4^x = 63(4^3+1)(4^6+1)(4^{12}+1)+1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 4^x &= 63(4^3+1)(4^6+1)(4^{12}+1)+1 \\ 4^x-1 &= 63(4^3+1)(4^6+1)(4^{12}+1) \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{4^3-1} \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{63} \\ &= (4^3+1)(4^6+1)(4^{12}+1)(4^3-1) \\ &= (4^3-1)(4^3+1)(4^6+1)(4^{12}+1) \\ &= (4^6-1)(4^6+1)(4^{12}+1) \\ &= (4^{12}-1)(4^{12}+1) \\ &= 4^{24}-1 \\ 4^x &= 4^{24} \\ x &= 24 \\ \end{align} </math> </div></div> <ol start=59> <li>Berapakah nilai dari <math>\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}</math> jika <math>x=\sqrt{9+4\sqrt{5}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \sqrt{9+4\sqrt{5}} \\ x &= 2+\sqrt{5} \\ x^2 &= 9+4\sqrt{5} \\ x^2-4x &= 9+4\sqrt{5}-4(2+\sqrt{5}) \\ x^2-4x &= 1 \\ x^2 &= 4x+1 \\ x^3 &= x \cdot x^2 \\ &= x(4x+1) \\ &= 4x^2+x \\ &= 4(4x+1)+x \\ &= 16x+4+x \\ &= 17x+4 \\ x^4 &= x \cdot x^3 \\ &= x(17x+4) \\ &= 17x^2+4x \\ &= 17(4x+1)+4x \\ &= 68x+17+4x \\ &= 72x+17 \\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1} &= \frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1} \\ &= \frac{72x+17-85x-20+8x+2+5x+3}{2} \\ &= \frac{2}{2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=60> <li>Berapakah nilai dari <math>\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}</math> jika 2x-1=<math>\sqrt{61}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \frac{x^3+1}{x^5-x^4-x^3+x^2} = p \\ p &= \frac{x^3+1}{x^5-x^4-x^3+x^2} \\ &= \frac{x^3+1}{x^5-x^4-(x^3-x^2)} \\ &= \frac{x^3+1}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)} \\ &= \frac{x^2-x+1}{x^2(x-1)^2} \\ &= \frac{x^2-x+1}{(x(x-1))^2} \\ &= \frac{x(x-1)+1}{(x(x-1))^2} \\ 2x-1 &= \sqrt{61} \\ x &= \frac{\sqrt{61}+1}{2} \\ x-1 &= \frac{\sqrt{61}-1}{2} \\ x(x-1) &= (\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2}) \\ &= \frac{61-1}{4} \\ &= \frac{60}{4} \\ &= 15 \\ p &= \frac{x(x-1)+1}{(x(x-1))^2} \\ &= \frac{15+1}{15^2} \\ &= \frac{16}{15^2} \\ \sqrt{p} &= \sqrt{\frac{16}{15^2}} \\ &= \frac{4}{15} \\ \end{align} </math> </div></div> <ol start=61> <li>Berapakah nilai dari <math>(\frac{x-3}{x})^{25}</math> jika <math>x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\sqrt[5]{8}+\sqrt[5]{2} &= 1+\sqrt[5]{16}+\sqrt[5]{4} \\ x+(\sqrt[5]{2})^3+\sqrt[5]{2} &= 1+(\sqrt[5]{2})^4+(\sqrt[5]{2})^2 \\ x &= (\sqrt[5]{2})^4-(\sqrt[5]{2})^3+(\sqrt[5]{2})^2-\sqrt[5]{2}+1 \\ \text{misalkan } \sqrt[5]{2} = p \\ x &= p^4-p^3+p^2-p+1 \\ x &= \frac{p^5+1}{p+1} \\ (\frac{x-3}{x})^{25} &= (1-\frac{3}{x})^{25} \\ &= (1-\frac{3}{\frac{p^5+1}{p+1}})^{25} \\ &= (1-\frac{3(p+1)}{p^5+1})^{25} \\ &= (1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2})^5+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{2+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-3\sqrt[5]{2}-3)}{3})^{25} \\ &= (-\sqrt[5]{2})^{25} \\ &= (-2)^5 \\ &= -32 \\ \end{align} </math> </div></div> <ol start=62> <li>Berapakah nilai dari <math>x^{50}+x^{49}+x^{48}+x^{47}+x^{46}</math> jika <math>x^2+x+1=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+x+1 &= 0 \\ x^2+x &= -1 \\ \frac{x^3-1}{x-1} &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46} &= x^{48}(x^2+x+1)+x^{45}(x^2+x) \\ &= x^{48}(0)+(x^3)^{15}(-1) \\ &= 0+(1)^{15}(-1) \\ &= -1 \\ \end{align} </math> </div></div> <ol start=63> <li>Berapakah 2²⁴ dari <math>8^7+8^6+8^5+8^4+8^3+8^2+8+1=A</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= A \\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1) &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8 &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= 8A+1 \\ 8^8+A &= 8A+1 \\ 8^8 &= 7A+1 \\ (2^3)^8 &= 7A+1 \\ 2^{24} &= 7A+1 \\ \end{align} </math> </div></div> <ol start=64> <li>Berapakah nilai dari <math>x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1</math> jika <math>x+\frac{1}{x}=\sqrt{3}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{x} &= \sqrt{3} \\ x^2+2+\frac{1}{x^2} &= 3 \\ x^2-1+\frac{1}{x^2} &= 0 \\ x^2(x^2-1+\frac{1}{x^2}) &= x^2(0) \\ x^4-x^2+1 &= 0 \\ (x^2+1)(x^4-x^2+1) &= (x^2+1)0 \\ x^6-x^4+x^2+x^4-x^2+1 &= 0 \\ x^6+1 &= 0 \\ x^6 &= -1 \\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1 &= {x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1 \\ &= (-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2-1+1 \\ &= -1+1-1+1-1+1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=65> <li>Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari <math>\frac{c-b}{a-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= ax^2+bx+c \\ f(2) &= a(2)^2+2b+c = 4 \\ &= 4a+2b+c = 4 \\ f(7) &= a(7)^2+7b+c = 49 \\ &= 49a+7b+c = 49 \\ 49a+7b+c &= 49 \\ 4a+2b+c &= 4 \\ 45a+5b &= 45 \text{ (f(7) dikurangi f(2)) } \\ 9a+b &= 9 \\ b &= -9a+9 \\ 4a+2b+c &= 4 \\ 4a+2(-9a+9)+c &= 4 \\ 4a-18a+18+c &= 4 \\ -14a+18+c &= 4 \\ c &= 14a-14 \\ \frac{c-b}{a-1} &= \frac{14a-14-(-9a+9)}{a-1} \\ &= \frac{14(a-1)+9(a-1)}{a-1} \\ &= \frac{(14+9)(a-1)}{a-1} \\ &= 23 \\ \end{align} </math> </div></div> <ol start=66> <li>Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+y)^3 &= x^3+y^3+3xy(x+y) \\ 11^3 &= 242+3xy(11) \text{ (dibagi 11)} \\ 11^2 &= 22+3xy \\ 121 &= 22+3xy \\ 99 &= 3xy \\ xy &= 33 \\ (x-y)^2 &= x^2+y^2-2xy \\ &= ((x+y)^2-2xy)-2xy \\ &= (x+y)^2-4xy \\ &= 11^2-4(33) \\ &= 121-132 \\ &= -11 \\ \end{align} </math> </div></div> <ol start=67> <li>Berapa f(1)+f(-1) jika <math>f(\frac{ax-b}{bx-a})</math>=x²-5x+6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ jika} f(1) = f(\frac{ax-b}{bx-a}) \\ 1 &= \frac{ax-b}{bx-a} \\ bx-a &= ax-b \\ (b-a)x &= -b+a \\ &= -(b-a) \\ &= -1 \\ f(1) &= x^2-5x+6 \\ &= (-1)^2-5(-1)+6 \\ &= 12 \\ \text{ jika} f(-1) = f(\frac{ax-b}{bx-a}) \\ -1 &= \frac{ax-b}{bx-a} \\ -(bx-a) &= ax-b \\ -bx+a &= ax-b \\ (-b-a)x &= -b-a \\ &= 1 \\ f(-1) &= x^2-5x+6 \\ &= (1)^2-5(1)+6 \\ &= 2 \\ f(1)+f(-1) &= 12+2 \\ &= 14 \\ \end{align} </math> </div></div> <ol start=68> <li>berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)-x &= f(x-1) \\ f(x)-f(x-1) &= x \\ x=1 ; f(1)-f(0) &= 1 \\ x=2 ; f(2)-f(1) &= 2 \\ x=3 ; f(3)-f(2) &= 3 \\ x=4 ; f(4)-f(3) &= 4 \\ \dots \\ x=200 ; f(200)-f(199) &= 200 \\ \text{ jumlahkan tersebut menjadi } \\ f(200)-f(0) &= 1+2+3+4+\dots+200 \\ &= \frac{200 \cdot 201}{2} \\ &= 20.100 \\ f(200)-1 &= 20.100 \\ &= 20.101 \\ \end{align} </math> </div></div> <ol start=69> <li>Misalkan f(x) adalah fungsi rekursif yang berlaku ∀x ∈ R sebagai berikut: : f(x)+f(15-x) = 2024 : f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)+f(15-x) &= 2024 \\ f(15+x) &= f(x)+2020 \\ *cara 1 \\ \text{ganti x dengan 15+x } \\ f(15+x)+f(-x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ *cara 2 \\ \text{ganti x dengan -x } \\ f(-x)+f(15+x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ \end{align} </math> </div></div> <ol start=70> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>2f(\frac{2002}{x}) + f(x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2f(\frac{2002}{x}) + f(x) &= 3x \\ \text{ganti x dengan 2 } \\ 2f(\frac{2002}{2}) + f(2) &= 3(2) \\ 2f(1001) + f(2) &= 6 \\ \text{ganti x dengan 1001 } \\ 2f(\frac{2002}{1001}) + f(1001) &= 3(1001) \\ 2f(2) + f(1001) &= 3003 \\ 2f(2) + f(1001) &= 3003 \\ f(1001) &= 3003 - 2f(2) \\ 2f(1001) + f(2) &= 6 \\ 2(3003 - 2f(2)) + f(2) &= 6 \\ 6006 - 4f(2) + f(2) &= 6 \\ 3f(2) &= 6000 \\ f(2) &= 2000 \\ \end{align} </math> </div></div> <ol start=71> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>f(\frac{1}{x}) + \frac{1}{x}f(-x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{1}{x})+\frac{1}{x}f(-x) &= 3x \\ \text{ganti x dengan 1/3 } \\ f(3)+3f(-\frac{1}{3}) &= 1 \\ \text{ganti x dengan -3 } \\ f(-\frac{1}{3}) - \frac{1}{3}f(3) &= -9 \\ \text{dikalikan 3 } \\ 3f(-\frac{1}{3})-f(3) &= -27 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ 2f(3) &= 28 \\ f(3) &= 14 \\ \end{align} </math> </div></div> <ol start=72> <li>Diketahui polinom <math>f(7^b-1)=7^{3b}-10</math>. tentukan nilai f(5)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(5) &= f(7^b-1) \\ 5 &= 7^b-1 \\ 7^b &= 6 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(6-1) &= 6^3-10 \\ f(5) &= 216-10 \\ &= 206 \\ *cara 2 \\ \text{misalkan } 7^b-1=a \text{ maka } 7^b=a+1 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(a) &= (a+1)^3-10 \\ f(5) &= (5+1)^3-10 \\ &= 6^3-10 \\ &= 216-10 \\ &= 206 \\ \end{align} </math> </div></div> <ol start=73> <li>Diketahui polinom <math>f(6^b-7)=6^{3b}-2 \cdot 6^{2b}-4</math>. tentukan nilai f(-2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(-2) &= f(6^b-7) \\ -2 &= 6^b-7 \\ 6^b &= 5 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(5-7) &= 5^3-2 \cdot 5^2-4 \\ f(-2) &= 125-50-4 \\ &= 71 \\ *cara 2 \\ \text{misalkan } 6^b-7=a \text{ maka } 6^b=a+7 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(a) &= (a+7)^3-2(a+7)^2-4 \\ f(-2) &= (-2+7)^3-2(-2+7)^2-4 \\ &= 5^3-2(5)^2-4 \\ &= 125-50-4 \\ &= 71 \\ \end{align} </math> </div></div> <ol start=74> <li>Jika <math>f(xy)=\frac{f(x)}{y}</math> dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(10) &= 7 \\ f(2 \cdot 5) &= 7 \\ f(xy) &= \frac{f(x)}{y} \\ f(2 \cdot 5) &= \frac{f(2)}{5} \\ 7 &= \frac{f(2)}{5} \\ f(2) &= 35 \\ \end{align} </math> </div></div> <ol start=75> <li>Jika <math>f(xy)=\frac{f(x+y)}{xy}</math> dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(15) &= 16 \\ f(3 \cdot 5) &= 16 \\ f(xy) &= \frac{f(x+y)}{xy} \\ f(3 \cdot 5) &= \frac{f(3+5)}{3 \cdot 5} \\ f(15) &= \frac{f(8)}{15} \\ 16 &= \frac{f(8)}{15} \\ f(8) &= 240 \\ \end{align} </math> </div></div> <ol start=76> <li>Jika <math>f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15</math> maka tentukan nilai f(16)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x+\frac{1}{x}+6) &= x^2+\frac{1}{x^2}+15 \\ &= (x+\frac{1}{x})^2-2+15 \\ &= (x+\frac{1}{x})^2+13 \\ \text{misalkan } x+\frac{1}{x} &= p \\ f(x+\frac{1}{x}+6) &= (x+\frac{1}{x})^2+13 \\ f(p+6) &= p^2+13 \\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 } \\ f(p+6) &= p^2+13 \\ f(10+6) &= 10^2+13 \\ f(16) &= 100+13 \\ &= 113 \\ \end{align} </math> </div></div> <ol start=77> <li>Tentukan nilai x jika <math>f(x)=\frac{4}{4-x}</math> dan <math>f(x \cdot f(x))^{\frac{f(4x)}{f(x)}}=256</math>!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{4}{4-x} \\ f(4x) &= \frac{4}{4-4x} \\ \frac{f(4x)}{f(x)} &= \frac{\frac{4}{4-4x}}{\frac{4}{4-x}} \\ &= \frac{4-x}{4-4x} \\ f(x \cdot f(x)) &= f(x(\frac{4}{4-x})) \\ &= f(\frac{4x}{4-x}) \\ &= \frac{4}{4-(\frac{4x}{4-x})} \\ &= \frac{4}{\frac{16-4x-4x}{4-x}} \\ &= \frac{4}{\frac{16-8x}{4-x}} \\ &= \frac{4(4-x)}{4(4-4x)} \\ &= \frac{4-x}{4-4x} \\ \text{misalkan } \frac{4-x}{4-4x} &= a \\ f(x \cdot f(x))^{\frac{f(4x)}{f(x)}} &= 256 \\ a^a &= 256 \\ a^a &= 4^4 \\ a &= 4 \\ \frac{4-x}{4-4x} &= 4 \\ 4-x &= 16-16x \\ 15x &= 12 \\ x &= \frac{4}{5} \\ \end{align} </math> </div></div> <ol start=78> <li>Fungsi <math>f(x) = \frac{kx}{2x+1} \text{dengan } x \neq -\frac{1}{2}</math>. Dengan f(f(x)) = x maka tentukan nilai k!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{kx}{2x+1} \\ f(f(x)) &= x \\ f(\frac{kx}{2x+1}) &= x \\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1} &= x \\ \frac{\frac{k^2x}{2x+1}}{\frac{2kx+2x+1}{2x+1}} &= x \\ \frac{k^2x}{2kx+2x+1} &= x \\ \frac{k^2}{2kx+2x+1} &= 1 \\ k^2 &= 2kx+2x+1 \\ k^2-2kx &= 2x+1 \\ k^2-2kx+x^2 &= x^2+2x+1 \\ (k-x)^2 &= (x+1)^2 \\ (k-x)^2-(x+1)^2 &= 0 \\ (k-x+x+1)(k-x-(x+1)) &= 0 \\ k=-1 &\text{ atau } k=2x+1 &\text{ (TM) } \\ \end{align} </math> </div></div> <ol start=79> <li>Jika n = 2023²+2024² maka berapa hasil dari <math>\sqrt{2n-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n &= 2023^2+2024^2 \\ &= 2023^2+(2023+1)^2 \\ \text{misalkan 2023 = p} \\ n &= p^2+(p+1)^2 \\ &= p^2+p^2+2p+1 \\ &= 2p^2+2p+1 \\ \sqrt{2n-1} &= \sqrt{2(2p^2+2p+1)-1} \\ &= \sqrt{4p^2+4p+2-1} \\ &= \sqrt{4p^2+4p+1} \\ &= \sqrt{(2p+1)^2} \\ &= 2p+1 \\ &= 2(2023)+1 \\ &= 4046+1 \\ &= 4047 \\ \end{align} </math> </div></div> <ol start=80> <li>Tentukan nilai dari a+b+c merupakan bilangan bulat positif jika ab = 2, bc = 3 dan ac = 6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab \cdot bc \cdot ac &= 2 \cdot 3 \cdot 6 \\ (abc)^2 &= 36 \\ abc &= \pm 6 \\ abc &= 6 \\ \frac{abc}{ab} &= c = \frac{6}{2} = 3 \\ \frac{abc}{bc} &= a = \frac{6}{3} = 2 \\ \frac{abc}{ac} &= b = \frac{6}{6} = 1 \\ a+b+c &= 6 \\ \end{align} </math> </div></div> <ol start=81> <li>Tentukan nilai dari (a-c)^b jika <math>\frac{ab}{a+b} = \frac{1}{3}</math>, <math>\frac{bc}{b+c} = \frac{1}{4}</math> dan <math>\frac{ac}{a+c} = \frac{1}{9}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{ab}{a+b} &= \frac{1}{3} \\ \frac{a+b}{ab} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{b} + \frac{1}{a} &= 3 \\ \frac{bc}{b+c} &= \frac{1}{4} \\ \frac{b+c}{bc} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{b} &= 4 \\ \frac{ac}{a+c} &= \frac{1}{9} \\ \frac{a+c}{ac} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{a} &= 9 \\ \text{Misalkan 1/a = x, 1/b = y dan 1/c = z} \\ x+y &= 3 \\ y+z &= 4 \\ x+z &= 9 \\ x+y &= 3 \\ y+z &= 4 \\ x-z &= -1 \\ x-z &= -1 \\ x+z &= 9 \\ 2x &= 8 \\ x &= 4 \\ x-z &= -1 \\ 4-z &= -1 \\ z &= 5 \\ x+y &= 3 \\ 4+y &= 3 \\ y &= -1 \\ \frac{1}{a} &= 4 \\ a &= \frac{1}{4} \\ \frac{1}{b} &= -1 \\ b &= -1 \\ \frac{1}{c} &= 5 \\ c &= \frac{1}{5} \\ (a-c)^b &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=82> <li>Tentukan nilai dari a, b dan c jika <math>\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}</math> dan a+2b+3c=28?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan k untuk semua ketiga persamaan tersebut } \\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5} &= k \\ a+b &= 2k \\ a+c &= 4k \\ b+c &= 5k \\ 2a+b+c &= 6k \\ 2a+5k &= 6k \\ k &= 2a \\ a &= \frac{k}{2} \\ b &= \frac{3k}{2} \\ c &= \frac{7k}{2} \\ a+2b+3c &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ a &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ b &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ c &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=83> <li>Tentukan nilai dari (b+c)^a jika <math>\frac{a+b+c}{2} = \sqrt{a-2}+\sqrt{b-1}+\sqrt{c}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{a+b+c}{2} &= \sqrt{a-2}+\sqrt{b-1}+\sqrt{c} \\ a+b+c &= 2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}) \\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c} &= 0 \\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1 &= 0 \\ (\sqrt{a-2}-1)^2+(\sqrt{b-1}-1)^2+(\sqrt{c}-1)^2 &= 0 \\ (\sqrt{a-2}-1)^2 &= 0 \\ \sqrt{a-2}-1 &= 0 \\ \sqrt{a-2} &= 1 \\ a-2 &= 1 \\ a &= 3 \\ (\sqrt{b-1}-1)^2 &= 0 \\ \sqrt{b-1}-1 &= 0 \\ \sqrt{b-1} &= 1 \\ b-1 &= 1 \\ b &= 1 \\ (\sqrt{c}-1)^2 &= 0 \\ \sqrt{c}-1 &= 0 \\ \sqrt{c} &= 1 \\ c &= 1 \\ (b+c)^a &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=84> <li>x dan y merupakan bilangan tak nol. Jika xy = <math>\frac{x}{y}</math> = x-y maka berapa nilai x+y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy &= \frac{x}{y} \\ y^2 &= 1 \\ y^2 - 1 &= 0 \\ (y-1)(y+1) &= 0 \\ y = 1 &\text{ atau } y = -1 \\ \frac{x}{y} &= x-y \\ x &= xy-y^2 \\ x-xy &= -y^2 \\ x(1-y) &= -y^2 \\ x &= \frac{-y^2}{1-y} \\ \text{cek y=1 } \\ x &= \frac{-1^2}{1-1} \\ \text{tidak memenuhi syarat } \\ \text{cek y=-1 } \\ x &= \frac{-(-1)^2}{1-(-1)} \\ &= \frac{-1}{2} \\ x+y &= -1-\frac{1}{2} \\ &= -\frac{3}{2} \\ \end{align} </math> </div></div> <ol start=86> <li>Berapa nilai x dari <math>(\frac{a}{b})^3+(\frac{b}{a})^3 = 2\sqrt{x}</math> jika <math>\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b} \\ \frac{a+b}{ab} &= \frac{1}{a+b} \\ (a+b)^2 &= ab \\ a^2+2ab+b^2 &= ab \\ a^2+b^2 &= -ab \\ \text{misalkan } \frac{a}{b}+\frac{b}{a} = n \\ \frac{a}{b}+\frac{b}{a} &= n \\ \frac{a^2+b^2}{ab} &= n \\ a^2+b^2 &= nab \\ n &= -1 \\ \frac{a}{b}+\frac{b}{a} &= n \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3(\frac{a}{b}+\frac{b}{a}) &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3n &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3 &= n^3-3n \\ &= (-1)^3-3(-1) \\ &= 2 \\ 2\sqrt{x} &= 2 \\ \sqrt{x} &= 1 \\ x &= 1 \\ \end{align} </math> </div></div> <ol start=87> <li>Berapa nilai m dari <math>x^2-mx-1=0</math> jika <math>\sqrt[3]{x_1}+\sqrt[3]{x_2}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x_1} &= a \\ x_1 &= a^3 \\ \sqrt[3]{x_2} &= b \\ x_2 &= b^3 \\ \sqrt[3]{x_1}+\sqrt[3]{x_2} &= 1 \\ a+b &= 1 \\ x^2-mx-1 &= 0 \\ x_1+x_2 &= m \\ x_1 \cdot x_2 &= -1 \\ x_1+x_2 &= m \\ a^3+b^3 &= m \\ x_1 \cdot x_2 &= -1 \\ a^3 \cdot b^3 &= -1 \\ (ab)^2 &= (-1)^3 \\ ab &= -1 \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ (1)^3 &= m+3(-1)(1) \\ 1 &= m-3 \\ m &= 4 \\ \end{align} </math> </div></div> <ol start=88> <li>Berapa nilai <math>\frac{x_1}{x_2}</math> dari <math>ax^2-18x-b=0</math> jika <math>ab=45</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab &= 45 \\ b &= \frac{45}{a} \\ ax^2-18x-b &= 0 \\ ax^2-18x-\frac{45}{a} &= 0 \\ a^2x^2-18ax-45 &= 0 \\ (ax-3)(ax-15) &= 0 \\ ax-3 &= 0 \\ x &= \frac{3}{a} \\ ax-15 &= 0 \\ x &= \frac{15}{a} \\ \frac{x_1}{x_2} &= \frac{\frac{3}{a}}{\frac{15}{a}} \\ &= \frac{3}{15} \\ &= \frac{1}{5} \\ \frac{x_1}{x_2} &= \frac{\frac{15}{a}}{\frac{3}{a}} \\ &= \frac{15}{3} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=89> <li>Jika <math>\frac{u_3}{u_1+u_2} = \frac{7}{8}</math> merupakan barisan aritmetika maka berapa dari <math>\frac{u_2+u_3}{u_1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{u_3}{u_1+u_2} &= \frac{7}{8} \\ \frac{a+2b}{a+a+b} &= \frac{7}{8} \\ \frac{a+2b}{2a+b} &= \frac{7}{8} \\ 8(a+2b) &= 7(2a+b) \\ 8a+16b &= 14a+7b \\ 9b &= 6a \\ b &= \frac{2a}{3} \\ \frac{u_2+u_3}{u_1} &= \frac{a+b+a+2b}{a} \\ &= \frac{2a+3b}{a} \\ &= \frac{2a+3(\frac{2a}{3})}{a} \\ &= \frac{2a+2a}{a} \\ &= \frac{4a}{a} \\ &= 4 \\ \end{align} </math> </div></div> <ol start=90> <li>Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7p+q}{2p+q} &= \frac{17p+q}{7p+q} \\ (7p+q)^2 &= (17p+q)(2p+q) \\ 49p^2+14pq+q^2 &= 34p^2+19pq+q^2 \\ 15p^2 &= 5pq \\ 3p &= q \\ \frac{7p+q}{2p+q} &= \frac{7p+3p}{2p+3p} \\ &= \frac{10p}{5p} \\ &= 2 \\ \end{align} </math> </div></div> <ol start=91> <li>Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{rataan geometris } \\ \sqrt{a \cdot b} &= b-24 \\ a \cdot b &= (b-24)^2 \\ \text{rataan aritmatik } \\ \frac{a+b}{2} &= a+15 \\ a+b &= 2(a+15) \\ a+b &= 2a+30 \\ a &= b-30 \\ a \cdot b &= (b-24)^2 \\ (b-30)b &= (b-24)^2 \\ b^2-30b &= b^2-48b+576 \\ 18b &= 576 \\ b &= 32 \\ a &= b-30 \\ &= 32-30 \\ &= 2 \\ a+b &= 32+2 \\ &= 34 \\ \end{align} </math> </div></div> <ol start=91> <li>Segitiga lancip ABC dengan <math>\frac{a^4+b^4+c^4+a^2b^2}{c^2(a^2+b^2)}=2</math>. tentukan nilai sudut C?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{syarat segitiga lancip semua sudut masing-masing kurang dari } 90^\circ \\ c^2 &= a^2+b^2-2ab cos C \\ cos C &= \frac{a^2+b^2-c^2}{2ab} \\ a^4+b^4+c^4+a^2b^2 &= 2c^2(a^2+b^2) \\ a^4+b^4+a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-2c^2(a^2+b^2)+(c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= (ab)^2 \\ a^2+b^2-c^2 &= \pm ab \\ cos C &= \pm \frac{ab}{2ab} \\ &= \pm \frac{1}{2} \\ &= \frac{1}{2} \text{ (karena sudut harus kurang dari } 90^\circ) \\ C &= 60^\circ \\ \end{align} </math> </div></div> <ol start=92> <li>Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan panjang DA dan EA } = x \text{ dan panjang AB } = y \\ \text{luas segitiga CAB } &= \frac{CA \cdot AB}{2} \\ 45 &= \frac{(x+9)(x+5)}{2} \\ 90 &= x^2+14x+45 \\ x^2+14x &= 45 \\ y^2 &= (x+9)^2+(x+5)^2 \\ &= x^2+18x+81+x^2+10x+25 \\ &= 2x^2+28x+106 \\ &= 2(x^2+14x)+106 \\ &= 2(45)+106 \\ &= 196 \\ y &= 14 \\ \end{align} </math> jadi panjang BC adalah 14 cm </div></div> <ol start=93> <li>Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kita cari ukuran GC yaitu } \\ \frac{GC}{AD} &= \frac{CE}{DE} \\ \frac{GC}{15} &= \frac{6}{18} \\ GC &= 5 \\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC } \\ &= \frac{1}{2} \cdot BC \cdot CF - \frac{1}{2} \cdot GC \cdot CE \\ &= \frac{1}{2} \cdot 15 \cdot 18 - \frac{1}{2} \cdot 5 \cdot 6 \\ &= 135 - 15 \\ &= 120 \\ \end{align} </math> jadi luas daerah BFEG adalah 120 cm² </div></div> <ol start=94> <li>Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ AI &= \frac{1}{5} AF \\ &= \frac{1}{5} 10 \\ &= 2 \\ IF &= AF-AI \\ &= 10-2 \\ &= 8 \\ \text{luas trapesium AFGD } &= \frac{(AD+EF) \cdot AF}{2} \\ &= \frac{(4+6)10}{2} \\ &= 50 \\ \text{luas segitiga AID } &= \frac{AI \cdot AF}{2} \\ &= \frac{(2)4}{2} \\ &= 4 \\ \text{luas segitiga IFG } &= \frac{IF \cdot FG}{2} \\ &= \frac{(8)6}{2} \\ &= 24 \\ \text{luas daerah segitiga IGD } &= \text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG } \\ &= 50-4-24 \\ &= 22 \\ \end{align} </math> jadi luas daerah segitiga IGD adalah 22 cm² </div></div> <ol start=96> <li>Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-<math>\pi</math>) cm³ maka berapa luas permukaan kerucut tersebut?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ \text{volume balok} \\ V_b &= x^2(12) \\ \text{volume kerucut} \\ V_b &= \frac{1}{3}\pi x^2(12) \\ &= 4\pi x^2 \\ V_{b-k} &= Vb-Vk \\ 100(3-\pi) &= 12x^2-4\pi x^2 \\ 100(3-\pi) &= 4x^2(3-\pi) \\ x^2 &= 25 \\ x &= 5 \\ s &= \sqrt{12^2+5^2} \\ &= \sqrt{144+25} \\ &= \sqrt{169} \\ &= 13 \\ \text{luas permukaan kerucut } &= \pi r(r+s) \\ &= \pi(5)(5+13) \\ &= 90\pi \\ \end{align} </math> jadi luas daerah permukaan kerucut adalah 90<math>\pi</math> cm² </div></div> <ol start=97> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 3a+1 \\ B &= 3b+2 \\ A(A+1)+3B \\ (3a+1)(3a+1+1)+3(3b+2) \\ (3a+1)(3a+2)+9b+6 \\ 9a^2+9a+2+9b+6 \\ 9a^2+9a+9b+8 \\ 9(a^2+a+b)+8 \\ \text{sisa pembagiannya adalah } 8 \\ \end{align} </math> </div></div> <ol start=98> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 9a+7 \\ B &= 9b+8 \\ A(A-5)+9B \\ (9a+7)(9a+7-5)+9(9b+8) \\ (9a+7)(9a+2)+81b+72 \\ 81a^2+81a+14+81b+72 \\ 81a^2+81a+81b+86 \\ 81a^2+81a+81b+81+5 \\ 81(a^2+a+b+1)+5 \\ \text{sisa pembagiannya adalah } 5 \\ \end{align} </math> </div></div> <ol start=99> <li>Jika <math>\begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix}</math> maka berapa hasil dari A²¹+A²⁵+A⁴⁶?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A^2 &= A \cdot A \\ &= \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \\ A^3 &= A^2 \cdot A \\ &= \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ &= - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= -I \\ A^{21}+A^{25}+A^{46} &= A^{21} \cdot (I+A^4+A^{25}) \\ &= A^{21} \cdot (I+A^3 \cdot A +A^{24} \cdot A) \\ &= (A^3)^7 \cdot (I+A^3 \cdot A +(A^3)^8 \cdot A) \\ &= (-I)^7 \cdot (I-I \cdot A +(-I)^8 \cdot A) \\ &= -I \cdot (I-A+A) \\ &= -I \cdot I \\ &= -I \\ &= -\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ \end{align} </math> </div></div> <ol start=100> <li>Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida?</li></ol> : bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15 : ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9). : ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8) : maka Ida menulis bilangan 8. <ol start=101> <li>Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9?</li></ol> : Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu. : untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja. : untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja. : Jadi banyaknya bilangan mungkin 2. <ol start=102> <li>Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli!</li></ol> ;cara 1 # 8¹ = 1 # 8² = 1 (8²=8¹x8¹ sama dengan 1x1) # 8³ = 1 (8³=8¹x8² sama dengan 1x1) # 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2) # 8⁵ = 1 # 8ⁿ = 1 (semua n untuk bilangan asli) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli ;cara 2 # 8ⁿ = b mod 7 # 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil) # (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya) # 8ⁿ = 1ⁿ mod 7 # 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli <ol start=103> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 5?</li></ol> ;cara 1 # 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 # 7¹ = 7 (sisa 1) # 7² = 49 (sisa 2) # 7³ = 343 (sisa 3) # 7⁴ = 2,401 (sisa 0) # 7⁵ = 16,807 # 7⁶ = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3 ;cara 2 :17¹ = 2 :17² = 4 :17³ = 3 :17⁴ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3 :17⁹⁹ = (17⁴)²⁴ x 17³ Untuk 17⁴ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 3. Jadi 17⁹⁹ dibagi 7 bersisa 3 ;cara 3 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 17⁴ ::17⁴ = 1 mod 5 ::(17⁴)²⁴ = 1²⁴ mod 5 ::17⁹⁶ = 1²⁴ mod 5 ::17⁹⁶ = 1 mod 5 ::17⁹⁶ x 17³ = 1 x 17³ mod 5 ::17⁹⁹ = 17³ mod 5 ::17⁹⁹ = 17 x 17 x 17 mod 5 ::17⁹⁹ = 2 x 2 x 2 mod 5 ::17⁹⁹ = 8 mod 5 ::17⁹⁹ = 3 mod 5 Jadi 17⁹⁹ dibagi 5 bersisa 3 <ol start=104> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 7?</li></ol> ;cara 1 :17¹ = 3 :17² = 2 :17³ = 6 :17⁴ = 4 :17⁵ = 5 :17⁶ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3 :17⁹⁹ = (17⁶)¹⁶ x 17³ Untuk 17⁶ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 6. Jadi 17⁹⁹ dibagi 7 bersisa 6 ;cara 2 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 17⁶ ::17⁶ = 1 mod 7 ::(17⁶)¹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1 mod 7 ::17⁹⁶ x 17³ = 1 x 17³ mod 7 ::17⁹⁹ = 17³ mod 7 ::17⁹⁹ = 17 x 17 x 17 mod 7 ::17⁹⁹ = 3 x 3 x 3 mod 7 ::17⁹⁹ = 27 mod 7 ::17⁹⁹ = 6 mod 7 Jadi 17⁹⁹ dibagi 7 bersisa 6 <ol start=105> <li>Berapa hasil sisa dari 41²⁰²⁴ dibagi 33?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 41^{2024} &= 41^{2024} \text{ mod } 33 \\ &= (33 \times 3 + 2)^{2024} \text{ mod } 33 \\ &= 2^{2024} \text{ mod } 33 \\ &= 2^{2020} 2^4 \text{ mod } 33 \\ &= (2^5)^{404} 2^4 \text{ mod } 33 \\ &= (33 - 1)^{404} 2^4 \text{ mod } 33 \\ &= (-1)^{404} 2^4 \text{ mod } 33 \\ &= 2^4 \text{ mod } 33 \\ &= 16 \text{ mod } 33 \\ \text{Jadi hasil sisa adalah } 16 \\ \end{align} </math> </div></div> <ol start=106> <li>Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 99^{99} &= (3^2 \times 11)^{99} \\ &= 3^{198} \times 11^{99} \\ 243^n &= (3^5)^n \\ &= 3^{5n} \\ \text{agar bisa membagi, maka} \\ 5n &= 198 \\ n &= 39.6 \\ \text{jadi bilangan n terbesar adalah } 39 \\ \end{align} </math> </div></div> <ol start=107> <li>Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸?</lu></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 88^{88} &= (8 \times 11)^{88} \\ &= 8^{88} \times 11^{88} \\ &= 8^{87} \times 8 \times 11^{88} \\ &= (8^3)^{29} \times 8 \times 11^{88} \\ &= 512^{29} \times 8 \times 11^{88} \\ 512^n &= 512^{29} \\ \text{jadi bilangan n terbesar adalah } 29 \\ \end{align} </math> </div></div> <ol start=108> <li>Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6!</li></ol> ; cara 1 : KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. : N dibagi 3 sisa 1 : N dibagi 5 sisa 2 : N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a : KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70 : KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21 : KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97 ; cara 2 : Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35 : kemudian ditambahkan sisa masing-masing sesuai dengan KPK. : KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, <b>97</b> : KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, <b>97</b> : KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, <b>97</b> Jadi bilangan bulat positif adalah 97 :: NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 <ol start=109> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |- | 4 || 0 || Semua isi ember B dibuang |- | 1 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. <ol start=110> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |} nah ada ember A berisi 4 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |- | 1 || 0 || Semua isi ember B dibuang |- | 0 || 1 || Tuangkan 1 l dari ember A ke B sehingga ember A kosong |- | 3 || 1 || Isikan 3 l ke ember A |- | 0 || 4 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |} nah ada ember B berisi 4 liter. # Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kemungkinan 1} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 +3 \cdot ((\frac{4}{3})^x)^2 &= 7 \cdot (\frac{4}{3})^x \\ \text{misalkan } (\frac{4}{3})^x = y \\ 4 +3y^2 &= 7y \\ 3y^2-7y+4 &= 0 \\ (3y-4)(y-1) &= 0 \\ y_1 = \frac{4}{3} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{4}{3} \\ (\frac{4}{3})^x &= \frac{4}{3} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{4}{3})^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-6}></math>? [[Kategori:Soal-Soal Matematika]] krtw97t9f5pqr478x26e0145e8peuh6 117486 117485 2026-07-13T01:01:42Z Akuindo 8654 117486 wikitext text/x-wiki contoh soal <ol start=1> <li>Berapa hasil dari <math>\sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Misalkan 2020 = p} \\ \sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64} &= \sqrt{(2020-5) \cdot (2020-3) \cdot (2020+3) \cdot (2020+5) + 64} \\ &= \sqrt{(p-5) \cdot (p-3) \cdot (p+3) \cdot (p+5) + 64} \\ &= \sqrt{(p-5) \cdot (p+5) \cdot (p-3) \cdot (p+3) + 64} \\ &= \sqrt{(p^2-25) \cdot (p^2-9) + 64} \\ &= \sqrt{p^4-34p^2+ 225 + 64} \\ &= \sqrt{p^4-34p^2+ 289} \\ &= \sqrt{(p^2-17)^2} \\ &= p^2-17 \\ &= 2020^2-17 \\ &= (2000+20)^2-17 \\ &= 4.000.000+80.000+400-17 \\ &= 4.080.383 \\ \end{align} </math> </div></div> <ol start=2> <li>Berapa nilai x dari <math>\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} = \frac{9}{10}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} &= \frac{9}{10} \\ \text{misalkan untuk } \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} = p \\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} &= p \\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}} &= p^2 \\ x^2-x-p &= p^2 \\ x^2-2x+1+x-1 &= p^2+p \\ (x-1)^2+(x-1) &= p^2+p \\ x-1 &= p \\ \text{misalkan untuk } \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}} &= q^3 \\ x^2 q &= q^3 \\ x^2 &= q^2 \\ x &= q \\ \frac{x-1}{x} &= \frac{9}{10} \\ x &= 10 \\ \end{align} </math> </div></div> <ol start=3> <li>Berapa nilai x dari <math>(\frac{x}{x+10})^{x+10}=\frac{1}{1024}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{x+10}{x})^{-(x+10)} &= (1024)^{-1} \\ (\frac{x+10}{x})^{x+10} &= 1024 \\ (\frac{x+10}{x})^{x+10} &= 2^{10} \\ (\frac{x+10}{x})^{\frac{x+10}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (\frac{1}{2})^{-1} \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (1+(-\frac{1}{2}))^{(1+(-\frac{2}{1}))} \\ \frac{10}{x} &= -\frac{1}{2} \\ x &= -20 \\ \end{align} </math> </div></div> <ol start=4> <li>Berapa nilai x dari <math>x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{2}+\sqrt{x+\frac{1}{4}} &= (\sqrt{x+\frac{1}{4}})^2+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+(\frac{1}{2})^2 \\ &= (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 \\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}} &= 4 \\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2} &= 4 \\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 4 \\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 &= 4 \\ \sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 2 \\ \sqrt{x+\frac{1}{4}} &= \frac{3}{2} \\ x+\frac{1}{4} &= \frac{9}{4} \\ x &= 2 \\ \end{align} </math> </div></div> <ol start=5> <li>Berapa nilai x dari <math>\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^3}{\sqrt{8-x^2}}+x^2-8 &= 0 \\ \frac{x^3}{\sqrt{8-x^2}} &= 8-x^2 \\ x^3 &= (8-x^2)^{\frac{3}{2}} \\ x &= (8-x^2)^{\frac{1}{2}} \\ x^2 &= 8-x^2 \\ 2x^2-8 &= 0 \\ x^2-4 &= 0 \\ (x-2)(x+2) &= 0 \\ \text{membuktikan } \\ x=2 \text{ maka hasilnya 0 } \\ x=-2 \text{ maka hasilnya -8 } \\ \text{jadi } x=2 \\ \end{align} </math> </div></div> <ol start=6> <li>Berapa nilai x dari <math>\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} = 5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} &= 5 \\ \frac{x^{50}+x^{60}+x^{70}}{31}} &= 5^5 \\ x^{50}+x^{60}+x^{70} &= 5^5 \cdot 31 \\ x^{50}(1+x^{10}+x^{20}) &= 5^5 \cdot 31 \\ (x^{10}^5)(1+x^{10}+(x^{10}^2) &= 5^5 \cdot 31 \\ \text{ misalkan } x^{10} = a \\ a^5(1+a+a^2) &= 5^5 \cdot 31 \\ a &= 5 \\ x^{10} &= 5 \\ x &= ^5 log 10 \\ \end{align} </math> </div></div> <ol start=7> <li>Berapa nilai x dari <math>\sqrt{3x+5+\sqrt{4x+5}} = x</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{3x+5+\sqrt{4x+5}} &= x \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \text{misalkan } \sqrt{4x+5}=y \text{ dan } 4x+5=y^2 \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \sqrt{y^2+y-x} &= x \\ y^2+y &= x^2+x \\ y=x \\ 4x+5 &= y^2 \\ 4x+5 &= x^2 \\ x^2-4x-5 &= 0 \\ (x-5)(x+1) &= 0 \\ x=5 &\text{ atau } x=-1 \text{ (TM) } \\ \end{align} </math> </div></div> <ol start=8> <li>Berapa nilai x dari <math>\sqrt{1+\sqrt{1+x}} = \sqrt[3]{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{1+\sqrt{1+x}} &= \sqrt[3]{x} \\ \sqrt[3]{x} &= n \\ x &= n^3 \\ \sqrt{1+\sqrt{1+n^3}} &= n \\ 1+\sqrt{1+n^3} &= n^2 \\ \sqrt{1+n^3} &= n^2-1 \\ 1+n^3 &= n^4-2n^2+1 \\ n^4-n^3-2n^2 &= 0 \\ n^2(n^2-n-2) &= 0 \\ n^2(n-2)(n+1) &= 0 \\ n=0, n=2 \text{ atau } n=-1 \\ n &= 0 \\ x &= 0^3 \\ &= 0 \\ n &= 2 \\ x &= 2^3 \\ &= 8 \\ n &= -1 \\ x &= (-1)^3 \\ &= -1 \\ \text{yang paling mungkin untuk nilai x adalah } 8 \\ \end{align} </math> </div></div> <ol start=9> <li>Berapa nilai x dari <math>\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} &= \frac{\sqrt{1+x}}{\sqrt{x}} \\ \sqrt{x}(\sqrt{1+x}+\sqrt{x}) &= (\sqrt{1+x}-\sqrt{x})\sqrt{1+x} \\ \sqrt{x(1+x)}+x &= 1+x-\sqrt{x(1+x)} \\ 2\sqrt{x(1+x)} &= 1 \\ \sqrt{x(1+x)} &= \frac{1}{2} \\ x(1+x) &= \frac{1}{4} \\ x^2+x &= \frac{1}{4} \\ 4x^2+4x &= 1 \\ 4x^2+4x-1 &= 0 \\ x &= \frac{-4 \pm \sqrt{4^2-4(4)(-1)}}{2(4)} \\ &= \frac{-4 \pm \sqrt{32}}{8} \\ &= \frac{-4 \pm 4\sqrt{2}}{8} \\ &= \frac{-1 \pm \sqrt{2}}{2} \\ \text{karena akar x harus minimal nol jadi } x = \frac{-1+\sqrt{2}}{2} \\ \end{align} </math> </div></div> <ol start=10> <li>Berapa nilai x dari <math>\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x-\sqrt{x+1}}{x+\sqrt{x+1}} &= \frac{11}{19} \\ \text{misalkan } \sqrt{x+1}=y \text{ dan } x=y^2-1 \\ \frac{y^2-1-y}{y^2-1+y} &= \frac{11}{19} \\ 19(y^2-y-1) &= 11(y^2+y-1) \\ 19y^2-19y-19 &= 11y^2+11y-11 \\ 8y^2-30y-8 &= 0 \\ 4y^2-15y-4 &= 0 \\ (4y+1)(y-4) &= 0 \\ y=-\frac{1}{4} \text{ (TM) atau } & y=4 \\ x &= 4^2-1 \\ &= 15 \\ \end{align} </math> </div></div> <ol start=11> <li>Berapa nilai x dari <math>\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} &= 98 \\ \text{misalkan } \sqrt{x^2-1}=y \\ \frac{x+y}{x-y}+\frac{x-y}{x+y} &= 98 \\ \frac{(x+y)^2+(x-y)^2}{(x-y)(x+y)} &= 98 \\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} &= 98 \\ \frac{2(x^2+y^2)}{x^2-y^2} &= 98 \\ \frac{x^2+y^2}{x^2-y^2} &= 49 \\ x^2+y^2 &= 49(x^2-y^2) \\ x^2+y^2 &= 49x^2-49y^2 \\ 48x^2 &= 50y^2 \\ 24x^2 &= 25y^2 \\ 24x^2 &= 25(\sqrt{x^2-1})^2 \\ 24x^2 &= 25(x^2-1) \\ 24x^2 &= 25x^2-25 \\ x^2 &= 25 \\ x &= \pm 5 \\ \end{align} </math> </div></div> <ol start=12> <li>Berapa nilai x dari <math>\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \sqrt{x}=y \text{ dan } x=y^2 \\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} &= \frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\sqrt{\frac{y^2}{y^2+y}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\frac{y}{\sqrt{y^2+y}} \\ y^2+y-\sqrt{(y^2+y)(y^2-y)} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^4-y^2} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^2(y^2-1)} &= \frac{5}{4}y \\ y(y+1)-y\sqrt{y^2-1} &= \frac{5}{4}y \\ y+1-\sqrt{y^2-1} &= \frac{5}{4} \\ -\sqrt{y^2-1} &= \frac{1}{4}-y \\ y^2-1 &= (\frac{1}{4}-y)^2 \\ y^2-1 &= \frac{1}{16}-\frac{1}{2}y+y^2 \\ -1 &= \frac{1}{16}-\frac{1}{2}y \\ \frac{1}{2}y &= \frac{1}{16}+1 \\ \frac{1}{2}y &= \frac{17}{16} \\ y &= \frac{17}{8} \\ x &= (\frac{17}{8})^2 \\ &= \frac{289}{64} \\ \end{align} </math> </div></div> <ol start=13> <li>Berapa nilai x dari <math>\sqrt[4]{62+x}+\sqrt[4]{275-x}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ misalkan } \sqrt[4]{62+x}=a, 62+x=a^4, \sqrt[4]{275-x}=b \text{ dan } 275-x=b^4 \\ a+b &= 7 \\ (a+b)^2 &= 49 \\ a^2+b^2+2ab &= 49 \\ a^2+b^2 &= 49-2ab \\ a^4+b^4 &= 62+x+275-x \\ (a^2+b^2)^2-2(ab)^2 &= 337 \\ (49-2ab)^2-2(ab)^2 &= 337 \\ 2401-196ab+4(ab)^2-2(ab)^2 &= 337 \\ 2(ab)^2-196ab+2064 &= 0 \\ (ab)^2-98ab+1032 &= 0 \\ (ab-12)(ab-86) &= 0 \\ ab = 12 \text{ atau } & ab = 86 \text{ (TM) karena hasil kali maksimum yaitu 12 } \\ ab =12 \text{ dan } a+b=7 \\ a+b &= 7 \\ b &= 7-a \\ ab &= 12 \\ a(7-a) &= 12 \\ -a^2+7a &= 12 \\ a^2-7a+12 &= 0 \\ (a-3)(a-4) &= 0 \\ a=3 \text{ atau } & a=4 \\ a=3, b=4 \\ 62+x &= a^4 \\ 62+x &= (3)^4 \\ 62+x &= 81 \\ x &= 19 \\ a=4, b=3 \\ 62+x &= a^4 \\ 62+x &= (4)^4 \\ 62+x &= 256 \\ x &= 194 \\ \end{align} </math> </div></div> <ol start=14> <li>Berapa nilai x dari <math>\sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2} &= 7 \\ (\sqrt[3]{8+x})^2-\sqrt[3]{8+x} \sqrt[3]{27-x}+(\sqrt[3]{27-x})^2 &= 7 \\ \text{misalkan } \sqrt[3]{8+x}=a, 8+x=a^3, \sqrt[3]{27-x}=b \text{ dan } 27-x=b^3 \\ a^2-ab+b^2 &= 7 \\ a^3+b^3 &= 8+x+27-x \\ &= 35 \\ a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ 35 &= (a+b)(7) \\ a+b &= 5 \\ b &= 5-a \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ 5^3 &= 35+3ab(5) \\ 125 &= 35+15ab \\ 80 &= 15ab \\ ab &= 6 \\ a(5-a) &= 6 \\ 5a-a^2 &= 6 \\ a^2-5a+6 &= 6 \\ (a-2)(a-3) &= 6 \\ a=2 &\text{ atau } a=3 \\ a=2, b=3 \text{ dan } a=3,b=2 \\ 8+x &= a^3 \\ &= 2^3 \\ &= 8 \\ x &= 0 \\ 8+x &= a^3 \\ &= 3^3 \\ &= 27 \\ x &= 19 \\ \end{align} </math> </div></div> <ol start=15> <li>Berapa nilai x dari <math>3^x+5^x-9^x+15^x-25^x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3^x+5^x-9^x+15^x-25^x &= 1 \\ 3^x+5^x-(3^2)^x+(3 \cdot 5)^x-(5^2)^x &= 1 \\ 3^x+5^x-(3^x)^2+(3^x \cdot 5^x)-(5^x)^2 &= 1 \\ \text{misalkan } 3^x=a \text{ dan } 5^x=b \\ a+b-a^2+ab-b^2 &= 1 \\ a^2-ab+b^2-a-b+1 &= 0 \\ 2a^2-2ab+2b^2-2a-2b+2 &= 0 \\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1 &= 0 \\ (a-b)^2+(a-1)^2+(b-1)^2 &= 0 \\ a-b=0; a-1=0; b-1 &= 0 \\ a=b &= 1 \\ 3^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> <ol start=16> <li>Berapa nilai x dari <math>^6log x^2+^{6x}log \frac{6}{x}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ \text{misalkan } 6x=a \text{ maka } x=\frac{a}{6} \\ ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ ^6log (\frac{a}{6})^2+^{6 \frac{a}{6}}log \frac{6}{\frac{a}{6}} &= 1 \\ ^6log \frac{a^2}{6^2}+^alog \frac{6^2}{a} &= 1 \\ ^6log a^2-^6log 6^2+^alog 6^2-^alog a &= 1 \\ 2 ^6log a-2 ^6log 6+2 ^alog 6-^alog a &= 1 \\ 2 ^6log a-2+2 \frac{1}{^6log a}-1 &= 1 \\ 2 ^6log a+2 \frac{1}{^6log a}-4 &= 0 \\ 2 ^6log^2 a-4 ^6log a+2 &= 0 \\ ^6log^2 a-2 ^6log a+1 &= 0 \\ (^6log a-1)^2 &= 0 \\ ^6log a &= 1 \\ a &= 6 \\ x &= \frac{a}{6} \\ &= \frac{6}{6} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=17> <li>Berapa nilai x dari (x+500)³+x=20?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+500)^3+x &= 20 \\ \text{misalkan } a=x+500 \text{ maka } x=a-500 \\ a^3+a-500 &= 20 \\ a^3+a &= 520 \\ a(a^2+1) &= 8 \cdot 65 \\ a(a^2+1) &= 8(64+1) \\ a(a^2+1) &= 8(8^2+1) \\ a &= 8 \\ x &= 8-500 \\ &= -492 \\ \end{align} </math> </div></div> <ol start=18> <li>Berapa nilai x dari <math>\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2} &= 0 \\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}} &= \frac{1}{2} \\ \frac{x^n+4^n}{x^n+16^n} &= (\frac{1}{2})^n \\ \frac{x^n+4^n}{x^n+16^n} &= \frac{1}{2^n} \\ 2^n(x^n+4^n) &= x^n+16^n \\ 2^n(x^n+2^{2n}) &= x^n+2^{4n} \\ 2^n \cdot x^n+2^{3n} &= x^n+2^{4n} \\ 2^n \cdot x^n-x^n &= 2^{4n}-2^{3n} \\ x^n(2^n-1) &= 2^{3n}(2^n-1) \\ x^n &= 2^{3n} \\ x^n &= (2^3)^n \\ x^n &= 8^n \\ x &= 8 \\ \end{align} </math> </div></div> <ol start=19> <li>Berapa hasil dari <math>\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ x &= \frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5 \cdot 6}+\sqrt{5 \cdot 5}+\sqrt{6 \cdot 4}+\sqrt{5 \cdot 4}}{\sqrt{5 \cdot 4}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5} \cdot \sqrt{6}+\sqrt{5} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{4}}{2 \cdot \sqrt{5}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{6} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{5}+\sqrt{5} \cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}} \\ &= \frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ &= \frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ \frac{1}{x} &= \frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}} \\ &= \frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ &= \frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1} \\ &= \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5} \\ &= \sqrt{6}-\sqrt{4} \\ &= \sqrt{6}-2 \\ x &= \frac{1}{\sqrt{6}-2} \\ &= \frac{\sqrt{6}+2}{6-4} \\ &= \frac{\sqrt{6}+2}{2} \\ &= 1+\frac{\sqrt{6}}{2} \\ \end{align} </math> </div></div> <ol start=20> <li>Berapa hasil dari <math>(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5 \\ \text{misalkan } x=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ x &= \frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ &= \frac{\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}} \\ &= \frac{\sqrt{3}+1}{4} \\ 4x &= \sqrt{3}+1 \\ 4x-1 &= \sqrt{3} \\ (4x-1)^2 &= 3 \\ 16x^2-8x+1 &= 3 \\ 16x^2 &= 8x+2 \\ 8x^2 &= 4x+1 \\ x^2 &= \frac{4x+1}{8} \\ *cara 1 \\ x^3 &= x \cdot x^2 \\ &= x(\frac{4x+1}{8}) \\ &= \frac{4x^2+x}{8} \\ &= \frac{4x^2}{8}+\frac{x}{8} \\ &= \frac{4(\frac{4x+1}{8})}{8}+\frac{x}{8} \\ &= \frac{16x+4}{64}+\frac{x}{8} \\ &= \frac{4x+1}{16}+\frac{x}{8} \\ &= \frac{4x+1+2x}{16} \\ &= \frac{6x+1}{16} \\ x^5 &= x^2 \cdot x^3 \\ &= (\frac{4x+1}{8})(\frac{6x+1}{16}) \\ &= \frac{24x^2+10x+1}{128} \\ &= \frac{24x^2}{128}+\frac{10x+1}{128} \\ &= \frac{24(\frac{4x+1}{8})}{128}+\frac{10x+1}{128} \\ &= \frac{96x+24}{1024}+\frac{10x+1}{128} \\ &= \frac{96x+24+80x+8}{1024} \\ &= \frac{176x+32}{1024} \\ &= \frac{176x}{1024}+\frac{32}{1024} \\ &= \frac{176}{1024}(\frac{\sqrt{3}+1}{4})+\frac{32}{1024} \\ &= \frac{44(\sqrt{3}+1)}{1024}+\frac{32}{1024} \\ &= \frac{44\sqrt{3}+44}{1024}+\frac{32}{1024} \\ &= \frac{76+44\sqrt{3}}{1024} \\ &= \frac{19+11\sqrt{3}}{256} \\ *cara 2 \\ x^4 &= (x^2)^2 \\ &= (\frac{4x+1}{8})^2 \\ &= \frac{16x^2+8x+1}{64} \\ &= \frac{16x^2}{64}+\frac{8x}{64}+\frac{1}{64} \\ &= \frac{x^2}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{\frac{4x+1}{8}}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{4x}{32}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{8}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{4}+\frac{3}{64} \\ x^5 &= x \cdot x^4 \\ &= (\frac{\sqrt{3}+1}{4})(\frac{x}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\frac{\sqrt{3}+1}{4}}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\sqrt{3}+1}{16}+\frac{3}{64}) \\ &= \frac{(\sqrt{3}+1)^2}{64}+(\frac{\sqrt{3}+1}{4})\frac{3}{64} \\ &= \frac{3+2\sqrt{3}+1}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{4+2\sqrt{3}}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{16+8\sqrt{3}}{256}+\frac{3\sqrt{3}+3}{256} \\ &= \frac{19+11\sqrt{3}}{256} \\ \end{align} </math> </div></div> <ol start=21> <li>Berapa hasil dari <math>\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4} &= \frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots) \\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots &= \frac{1}{1-\frac{1}{4}} \\ &= \frac{4}{3} \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{4}{3}) \\ &= \frac{1}{4}+\frac{16}{3} \\ &= \frac{67}{12} \\ x &= \frac{67}{9} \\ \end{align} </math> </div></div> <ol start=22> <li>Berapa nilai y-x jika <math>\frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} = \frac{x}{y}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} &= \frac{x}{y} \\ \frac{\frac{106 \times 107}{2}}{\frac{106}{2}(4+109)} &= \frac{x}{y} \\ \frac{53 \times 107}{53 \times 113} &= \frac{x}{y} \\ y-x &= 113-107 = 6 \\ \end{align} </math> </div></div> <ol start=23> <li>Berapa angka satuan dari hasil 17²⁰²⁴?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan angka satuannya} \\ 17^1 &= 7 \\ 17^2 &= 9 \\ 17^3 &= 3 \\ 17^4 &= 1 \\ 17^5 &= 7 \\ 17^6 &= 9 \\ 17^7 &= 3 \\ 17^8 &= 1 \\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1} \end{align} </math> </div></div> <ol start=24> <li>Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ 1! + 2! + 3! + 4! + \dots + 2024! &= 1 + (1x2) + (1x2x3) + (1x2x3x4) + \dots + 2024! \\ &= 1 + 2 + 6 + 24 + 120 + 720 + \dots + 2024! \\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka } 1+2+6+24 = 33 \text{ jadi angka satuannya adalah } 3 \end{align} </math> </div></div> <ol start=25> <li>Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ \frac{1! + 2! + 3! + 4! + \dots + 2024!}{12} &= \frac{1 + 1x2 + 1x2x3 + 1x2x3x4 + \dots + 2024!}{12} \\ &= \frac{1 + 2 + 6 + 24 + \dots + 2024!}{12} \\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi } 1+2+6 = 9 \end{align} </math> </div></div> <ol start=26> <li>Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{penjumlahan} \\ 1+1+1+1+ \dots \text{ (sebanyak 88 buah) }+x+y &= A \\ 88+x+y &= A \\ \text{perkalian} \\ 1 \times 1 \times 1 \times \dots \text{ (sebanyak 88 buah) }\times x \times y &= A \\ x \times y &= A \\ 88+x+y &= xy \\ xy-y &= 88+x \\ y(x-1) &= 88+x \\ y &= \frac{88+x}{x-1} \\ \text{uji selidiki untuk x=2} \\ y &= \frac{88+2}{2-1} \\ &= 90 \\ \text{buktikan} \\ 88+x+y &= xy \\ 88+2+90 &= 2(90) \\ 180 &= 180 \\ \text{terbukti} \\ \text{nilai A adalah } 180 \\ \end{align} </math> </div></div> <ol start=27> <li>Berapakah nilai x, y dan z dari <math>x+y-z=1, x^2+y^2-z^2=-5 \text{ dan } x^3+y^3-z^3=-53</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+y-z &= 1 \\ x+y &= z+1 \\ x^2+2xy+y^2 &= z^2+2z+1 \\ x^2+y^2-z^2 &= 2z+1-2xy \\ -5 &= 2z+1-2xy \\ 2xy &= 2z+6 \\ xy &= z+3 \\ x^2+y^2-z^2 &= -5 \\ x^2+y^2 &= z^2-5 \\ x^3+y^3-z^3 &= -53 \\ (x+y)(x^2-xy+y^2)-z^3+53 &= 0 \\ (x+y)(x^2+y^2-xy)-z^3+53 &= 0 \\ (z+1)(z^2-5-(z+3))-z^3+53 &= 0 \\ (z+1)(z^2-z-8)-z^3+53 &= 0 \\ z^3-z^2-8z+z^2-z-8-z^3+53 &= 0 \\ -9z+45 &= 0 \\ -9z &= -45 \\ z &= 5 \\ x+y &= 5+1 \\ x+y &= 6 \\ x &= 6-y \\ xy &= 5+3 \\ xy &= 8 \\ (6-y)y &= 8 \\ 6y-y^2 &= 8 \\ y^2-6y+8 &= 0 \\ (y-4)(y-2) &= 0 \\ y=4 \text{ atau } y=2 \\ \text{jika } y=4 \\ x+y &= z+1 \\ x+4 &= 5+1 \\ x &= 2 \\ \text{jika } y=2 \\ x+y &= z+1 \\ x+2 &= 5+1 \\ x &= 4 \\ \end{align} </math> </div></div> <ol start=28> <li>Berapakah nilai titik koordinat (x,y) dari <math>\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}</math> dan <math>\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x+y}+\sqrt{x-y} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+y}-\sqrt{x-y} &= \sqrt{\frac{52y}{3x}} \\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y}) &= \sqrt{\frac{432x}{13y}} \cdot \sqrt{\frac{52y}{3x}} \\ x+y-x+y &= \sqrt{\frac{432x \cdot 52y}{13y \cdot 3x}} \\ 2y &= \sqrt{144 \cdot 4} \\ 2y &= \sqrt{576} \\ 2y &= 24 \\ y &= 12 \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13(12)}} \\ x+12+x-12+2 \cdot \sqrt{x+12} \cdot \sqrt{x-12} &= \frac{36x}{13} \\ 2x+2 \sqrt{x^2-144} &= \frac{36x}{13} \\ 2(x+\sqrt{x^2-144}) &= \frac{36x}{13} \\ x+\sqrt{x^2-144} &= \frac{18x}{13} \\ \sqrt{x^2-144} &= \frac{5x}{13} \\ x^2-144 &= \frac{25x^2}{169} \\ \frac{144x^2}{169}-144 &= 0 \\ \frac{x^2}{169}-1 &= 0 \\ x^2-169 &= 0 \\ (x-13)(x+13) &= 0 \\ x_1=13 &\text{ atau } x_2=-13 \text{ (TM) karena } x>y \\ \end{align} </math> jadi titik koordinat (13,12) </div></div> <ol start=29> <li>Berapakah nilai dari <math>x^2-7x</math> jika <math>(x-2)^2+\frac{1}{(x-2)^2} = 11</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x-2)^2+\frac{1}{(x-2)^2} &= 11 \\ (x-2)^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2)^2} &= 11-2 \\ (x-2-\frac{1}{x-2})^2 &= 9 \\ x-2-\frac{1}{x-2} &= 3 \\ (x-2)^2-1 &= 3(x-2) \\ x^2-4x+4-1 &= 3x-6 \\ x^2-7x &= -9 \\ \end{align} </math> </div></div> <ol start=30> <li>Berapakah nilai dari <math>\frac{(x+y)^2(x+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)}</math> jika xy+yz+xz=1?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy+yz+xz &= 1 \\ x^2+xy+yz+xz &= x^2+1 \\ x(x+y)+z(x+y) &= x^2+1 \\ (x+y)(x+z) &= x^2+1 \\ \text{dengan pola yang sama } \\ (y+x)(y+z) &= y^2+1 \\ (x+z)(y+z) &= z^2+1 \\ \frac{(x+y)^2(y+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)} &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)} \\ &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)^2(y+z)^2(x+z)^2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=31> <li>Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} w+5 &= w+x+y+z+5 \\ x+4 &= w+x+y+z+5 \\ y+3 &= w+x+y+z+5 \\ z+2 &= w+x+y+z+5 \\ \text{jumlahkan keempat persamaan } \\ w+x+y+z+14 &= 4(w+x+y+z+5) \\ w+x+y+z+14 &= 4(w+x+y+z)+20 \\ 3(w+x+y+z) &= -6 \\ w+x+y+z &= -2 \\ \end{align} </math> </div></div> <ol start=32> <li>Berapakah nilai dari <math>\frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2}</math> jika <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3</math> dan x+y+z=xyz?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2} &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2} \\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) \\ 3^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz}) \\ 9 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz}) \\ &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2 \\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} &= 7 \\ \end{align} </math> </div></div> <ol start=33> <li>Berapakah nilai dari <math>\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}</math> jika <math>x^2+y^2+z^2 = -2(ab+bc+ac)</math>?>/li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+z^2 &= -2(xy+yz+xz) \\ x^2+y^2+z^2+2(xy+yz+xz) &= 0 \\ (x+y+z)^2 &= 0 \\ x+y+z &= 0 \\ x+y &= -z \\ x+z &= -y \\ y+z &= -x \\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z} &= \frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x} \\ &= -2-(-5)-(-7) \\ &= 10 \\ \end{align} </math> </div></div> <ol start=34> <li>Berapakah nilai dari <math>\frac{20xyz}{xy+yz+xz}</math> jika <math>16^x = 256^y = 625^z = 40</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 16^x = 256^y = 625^z &= 40 \\ 2^{4x} = 4^{4y} = 5^{4z} &= 40 \\ 2^{4x} &= 40 \\ 2 &= 40^{\frac{1}{4x}} \\ 4^{4y} &= 40 \\ 4 &= 40^{\frac{1}{4y}} \\ 5^{4z} &= 40 \\ 5 &= 40^{\frac{1}{4z}} \\ 2 \cdot 4 \cdot 5 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z}} \\ 1 &= \frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z} \\ 4 &= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{20xyz}{xy+yz+xz} &= 20 \cdot \frac{xyz}{xy+yz+xz} \\ &= 20 \cdot (\frac{xy+yz+xz}{xyz})^{-1} \\ &= 20 \cdot (\frac{1}{z} + \frac{1}{x} + \frac{1}{y})^{-1} \\ &= 20 \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{-1} \\ &= 20 \cdot (4)^{-1} \\ &= 20 \cdot \frac{1}{4} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=35> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+3x^2+1}</math> jika <math>6x^2+25x+6=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 6x^2+25x+6 &= 0 \\ 6x+25+\frac{6}{x} &= 0 \\ 6(x+\frac{1}{x}) &= -25 \\ x+\frac{1}{x} &= \frac{-25}{6} \\ (c+\frac{1}{x})^2 &= (\frac{-25}{6})^2 \\ x^2+2+\frac{1}{x^2} &= \frac{625}{36} \\ x^2+\frac{1}{x^2} &= \frac{625}{36}-2 \\ x^2+\frac{1}{x^2} &= \frac{553}{36} \\ \frac{x^2}{x^4+3x^2+1} &= \frac{1}{x^2+3+\frac{1}{x^2}} \\ &= \frac{1}{a^2+\frac{1}{x^2}+3} \\ &= \frac{1}{\frac{553}{36}+3} \\ &= \frac{1}{\frac{661}{36}} \\ &= \frac{36}{661} \\ \end{align} </math> </div></div> <ol start=36> <li>Berapakah nilai dari <math>\frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5} &= \frac{(9+2\sqrt{20})^{1013}}{((2)^3+3(2)^2(\sqrt{5})+3(2)(\sqrt{5})^2+(\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{((2+\sqrt{5})^2)^{1013}}{((2+\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{(2+\sqrt{5})^{2026}}{(2+\sqrt{5})^{2025}}+6-\sqrt{5} \\ &= 2+\sqrt{5}+6-\sqrt{5} \\ &= 8 \\ \end{align} </math> </div></div> <ol start=37> <li>Berapakah nilai dari <math>27x^3+\frac{8}{x^3}</math> jika <math>3x+\frac{2}{x}=6</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3x+\frac{2}{x} &= 6 \\ (3x+\frac{2}{x})^3 &= 6^3 \\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3} &= 216 \\ 27x^3+18(6)+\frac{8}{x^3} &= 216 \\ 27x^3+108+\frac{8}{x^3} &= 216 \\ 27x^3+\frac{8}{x^3} &= 108 \\ \end{align} </math> </div></div> <ol start=38> <li>Berapakah nilai dari <math>x^6+\frac{8}{x^3}</math> jika <math>x^3+\frac{1}{x^3}=8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3+\frac{1}{x^3} &= 8 \\ x^3 &= 8-\frac{1}{x^3} \\ x^6 &= 8x^3-1 \\ x^6+\frac{8}{x^3} &= 8x^3-1+\frac{8}{x^3} \\ &= 8x^3+\frac{8}{x^3}-1 \\ &= 8(x^3+\frac{1}{x^3})-1 \\ &= 8(8)-1 \\ &= 63 \\ \end{align} </math> </div></div> <ol start=39> <li>Berapakah nilai dari <math>4x+\frac{25}{x}</math> jika <math>2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2\sqrt{x}+\frac{5}{\sqrt{x}} &= 4x-\frac{25}{x} \\ 2\sqrt{x}+\frac{5}{\sqrt{x}} &= (2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}}) \\ 1 &= 2\sqrt{x}-\frac{5}{\sqrt{x}} \\ 1^2 &= (2\sqrt{x}-\frac{5}{\sqrt{x}})^2 \\ 1 &= 4x-20+\frac{25}{x} \\ 4x+\frac{25}{x} &= 21 \\ \end{align} </math> </div></div> <ol start=40> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2-x+1}{x^2+x+1} &= \frac{5}{6} \\ \frac{x^2+1-x}{x^2+1+x} &= \frac{5}{6} \\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} &= \frac{5}{6} \\ \text{ misalkan } x+\frac{1}{x} &= y \\ \frac{y-1}{y+1} &= \frac{5}{6} \\ 6(y-1) &= 5(y+1) \\ 6y-6 &= 5y+5 \\ y &= 11 \\ x+\frac{1}{x} &= 11 \\ \end{align} </math> </div></div> <ol start=41> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt{x}+x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x}+x &= 1 \\ x-1 &= -\sqrt{x} \\ (x-1)^2 &= (-\sqrt{x})^2 \\ x^2-2x+1 &= x \\ x^2-3x+1 &= 0 \\ x-3+\frac{1}{x} &= 0 \\ x+\frac{1}{x} &= 3 \\ \end{align} </math> </div></div> <ol start=42> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt[3]{x}-\sqrt[3]{x-36}=3</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x}-\sqrt[3]{x-36} &= 3 \\ (\sqrt[3]{x}-\sqrt[3]{x-36})^3 &= 3^3 \\ x-(x-36)-3 \sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36}) &= 27 \\ 36-3 \sqrt[3]{x(x-36)}3 &= 27 \\ -9 \sqrt[3]{x(x-36)} &= -9 \\ \sqrt[3]{x(x-36)} &= 1 \\ x(x-36) &= 1 \\ x^2-36x-1 &= 0 \\ x-36-\frac{1}{x} &= 0 \\ x-\frac{1}{x} &= 36 \\ \end{align} </math> </div></div> <ol start=43> <li>Berapakah nilai dari <math>x+\frac{16}{x}</math> jika <math>x-3\sqrt{x}=4</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x-3\sqrt{x} &= 4 \\ x-4 &= 3\sqrt{x} \\ x^2-8x+16 &= 9x \\ x^2-17x+16 &= 0 \\ x-17+\frac{16}{x} &= 0 \\ x+\frac{16}{x} &= 17 \\ \end{align} </math> </div></div> <ol start=44> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+4}</math> jika <math>x^2-7x+2=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2-7x+2 &= 0 \\ x^2+2 &= 7x \\ x+\frac{2}{x} &= 7 \\ x^2+4+\frac{4}{x^2} &= 49 \\ x^2+\frac{4}{x^2} &= 45 \\ \frac{x^4+4}{x^2} &= 45 \\ \frac{x^2}{x^4+4} &= \frac{1}{45} \\ \end{align} </math> </div></div> <ol start=45> <li>Berapakah nilai dari <math>x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}</math> jika <math>x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}} &= 25 \\ x^{\frac{1}{2}}+x^{-\frac{1}{2}} &= 23 \\ x+2+x^{-1} &= 529 \\ x+x^{-1} &= 527 \\ x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+x^{-\frac{3}{4}} &= 110 \\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1} &= x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}} \\ &= 527+110 \\ &= 637 \\ \end{align} </math> </div></div> <ol start=46> <li>Berapakah nilai dari <math>\sqrt{8x^6+x^5+x^4+5x^3+1}</math> jika <math>\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5} &= 0 \\ \frac{x^2+x+1}{x^5} &= 0 \\ x^2+x+1 &= 0 \\ x^2+x+1 &= 0 \\ (x-1)(x^2+x+1) &= 0(x-1) \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ \sqrt{8x^6+x^5+x^4+5x^3+1} &= \sqrt{(2x^3)^2+x^3x^2+x^3x+5x^3+1} \\ &= \sqrt{(2(1))^2+(1)x^2+(1)x+5(1)+1} \\ &= \sqrt{(2)^2+x^2+x+5+1} \\ &= \sqrt{4+x^2+x+1+5} \\ &= \sqrt{4+0+5} \\ &= \sqrt{9} \\ &= 3 \\ \end{align} </math> </div></div> <ol start=47> <li>Berapakah nilai dari <math>f(1)+f(2)+f(3)+ \dots + f(99)</math> jika <math>f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{1}{\sqrt{x+1}+\sqrt{x}} \\ &= \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} \\ &= \sqrt{x+1}-\sqrt{x} \\ f(1)+f(2)+f(3)+ \dots + f(98)+f(99) &= \sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+ \cdot + \sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99} \\ &= \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+ \cdot + \sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ &= \sqrt{100}-\sqrt{1} \\ &= 10-1 \\ &= 9 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots + \frac{2024}{2025})</math> jika <math>h(x)=\frac{3}{3+9^x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} h(x) &= \frac{3}{3+9^x} \\ h(1-x) &= \frac{3}{3+9^{1-x}} \\ &= \frac{3}{3+\frac{9}{9^x}} \\ &= \frac{9^x}{3+9^x} \\ h(x)+h(1-x) &= \frac{3}{3+9^x}+\frac{9^x}{3+9^x} \\ &= \frac{3+9^x}{3+9^x} \\ &= 1 \\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots +(1-\frac{2}{2025})+(1-\frac{1}{2025})) \\ & 5(1+1+1+ \dots +1+1) \text{ sebanyak 1012 kali } \\ & 5(1012) \\ & 5060 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>\frac{7^{2025} - 7^{2023} + 432}{7^{2024} + 7^{2023} + 72}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72} &= \frac{7^{2023}7^{2}-7^{2023} + 48 \times 9}{7^{2023}7^1+7^{2023}+8 \times 9} \\ &= \frac{7^{2023}(7^{2}-1)+48 \times 9}{7^{2023}(7^1+1)+8 \times 9} \\ &= \frac{7^{2023}(49-1)+48 \times 9}{7^{2023}(7+1) + 8 \times 9} \\ &= \frac{7^{2023} \times 48+48 \times 9}{7^{2023} \times 8+8 \times 9} \\ &= \frac{48(7^{2023}+9)}{8(7^{2023}+9)} \\ &= \frac{48}{8} \\ &= 6 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>tan (x+\frac{\pi}{4})</math> jika <math>\frac{1}{cos x}-tan x = \frac{4}{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{cos x}-tan x &= \frac{4}{5} \\ sec x-tan x &= \frac{4}{5} \\ sec^2 x-tan^2 x &= 1 \\ (sec x+tan x)(sec x-tan x) &= 1 \\ (sec x+tan x)\frac{4}{5} &= 1 \\ sec x+tan x &= \frac{5}{4} \\ \text{kedua persamaan dengan cara metode eliminasi } \\ 2 tan x &= \frac{5}{4}-\frac{4}{5} \\ 2 tan x &= \frac{9}{20} \\ tan x &= \frac{9}{40} \\ tan (x+\frac{\pi}{4}) &= \frac{tan x+tan \frac{\pi}{4}}{1-tan x \cdot tan \frac{\pi}{4}} \\ &= \frac{\frac{9}{40}+1}{1-\frac{9}{40} \cdot 1} \\ &= \frac{\frac{49}{40}}{\frac{31}{40}} \\ &= \frac{49}{31} \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>sin^3 x+csc^3 x</math> jika <math>sin x-csc x = 8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ Dengan menggunakan rumus: } (a-b)^3 &= a^3-b^3-3ab(a-b) \\ (sin x-csc x)^3 &= sin^3 x-csc^3 x-3sin x csc x(sin x-csc x) \\ 8^3 &= sin^3 x-csc^3 x-3sin x (\frac{1}{sin x})(8) \\ 512 &= sin^3 x-csc^3 x-24 \\ sin^3 x-csc^3 x &= 512+24 \\ sin^3 x-csc^3 x &= 536 \\ \end{align} </math> </div></div> <ol start=52> <li>Berapakah nilai dari <math>(sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2</math> jika <math>\frac{1}{sin x}+\frac{1}{cos x} = 10</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{sin x}+\frac{1}{cos x} &= 10 \\ \frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} &= 100 \\ (sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2 &= sin^2 x+\frac{2sin x}{cos x}+\frac{1}{cos^2 x}+cos^2 x+\frac{2cos x}{sin x}+\frac{1}{sin^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2(sin^2 x+cos^2 x)}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+100 \\ &= 101 \\ \end{align} </math> </div></div> <ol start=53> <li>Berapakah nilai dari (x-1)⁶ jika <math>x=\frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin 80^\circ &= cos 10^\circ \\ sin 80^\circ-cos 10^\circ &= 0 \\ x &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+2 sin 20^\circ cos 20^\circ}{cos 10^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+4 sin 10^\circ cos 10^\circ cos 20^\circ}{cos 10^\circ} \\ &= 4 cos 55^\circ cos 25^\circ+4 sin 10^\circ cos 20^\circ \\ &= 2(2 cos 55^\circ cos 25^\circ+2 sin 10^\circ cos 20^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ+sin (-10)^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ-sin 10^\circ) \\ &= 2(cos 80^\circ-sin 10^\circ+cos 30^\circ+sin 30^\circ) \\ &= 2(cos 80^\circ-sin (90^\circ-80^\circ)+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(cos 80^\circ-cos 80^\circ+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= \sqrt{3}+1 \\ x-1 &= \sqrt{3} \\ (x-1)^6 &= (\sqrt{3})^6 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=54> <li>Berapakah nilai dari x jika <math>x=\frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ} \\ 2x sin 20^\circ-x sin 40 ^\circ &= x sin 20^\circ-x^2 sin 10^\circ \\ x^2 sin 10^\circ+x sin 20^\circ-x sin 40 ^\circ &= 0 \\ x(x sin 10^\circ+sin 20^\circ-sin 40 ^\circ) &= 0 \\ x = 0 &\text{ atau } x sin 10^\circ+sin 20^\circ-sin 40 ^\circ = 0 \\ x sin 10^\circ+sin 20^\circ-sin 40 ^\circ &= 0 \\ x sin 10^\circ &= sin 40 ^\circ-sin 20^\circ \\ x &= \frac{sin 40 ^\circ-sin 20^\circ}{sin 10^\circ} \\ &= \frac{2 cos 30 ^\circ sin 10^\circ}{sin 10^\circ} \\ &= 2 cos 30 ^\circ \\ &= \frac{2 \sqrt{3}}{2} \\ &= \sqrt{3} \\ \end{align} </math> </div></div> <ol start=55> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x^2}{x^2-16y^2} = \frac{625}{49}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2}{x^2-16y^2} &= \frac{625}{49} \\ \frac{x^2-16y^2}{x^2} &= \frac{49}{625} \text{ (terbalik posisinya)} \\ 1-\frac{16y^2}{x^2} &= \frac{49}{625} \\ \frac{16y^2}{x^2} &= 1 - \frac{49}{625} \\ (\frac{4y}{x})^2 &= \frac{576}{625} \\ (\frac{4y}{x})^2 &= (\frac{24}{25})^2 \\ \frac{4y}{x} &= \frac{24}{25} \\ \frac{y}{x} &= \frac{6}{25} \\ \frac{x}{y} &= \frac{25}{6} \\ \end{align} </math> </div></div> <ol start=56> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x}{y}+\frac{x+10y}{y+10x} = 2</math> serta bilangan real untuk x dan y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x}{y}+\frac{x+10y}{y+10x} &= 2 \\ \frac{x}{y}+\frac{\frac{x}{y}+10}{1+10\frac{x}{y}} &= 2 \\ \text{misalkan } \frac{x}{y} = a \\ a+\frac{a+10}{1+10a} &= 2 \\ a(1+10a)+a+10 &= 2(1+10a) \\ 10a^2+a+a+10 &= 2+20a \\ 10a^2-18a+8 &= 0 \\ 5a^2-9a+4 &= 0 \\ (5a-4)(a-1) &= 0 \\ a = \frac{4}{5} &\text{ atau } a = 1 \\ \text{jadi } \frac{x}{y} = {\frac{4}{5}, 1} \\ \end{align} </math> </div></div> <ol start=57> <li>Berapakah nilai dari xy jika <math>x^4+y^4+x^2y^2=15 \text{ dan } x^2+y^2+xy=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+xy &= 5 \\ x^2+y^2 &= 5-xy \\ x^4+y^4+x^2y^2 &= 15 \\ (x^2)^2+(y^2)^2+2x^2y^2-x^2y^2 &= 15 \\ (x^2+y^2)^2-x^2y^2 &= 15 \\ (5-xy)^2-x^2y^2 &= 15 \\ 25-10xy+x^2y^2-x^2y^2 &= 15 \\ 25-10xy &= 15 \\ 10xy &= 10 \\ xy &= 1 \\ \end{align} </math> </div></div> <ol start=58> <li>Berapakah nilai dari x jika <math>4^x = 63(4^3+1)(4^6+1)(4^{12}+1)+1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 4^x &= 63(4^3+1)(4^6+1)(4^{12}+1)+1 \\ 4^x-1 &= 63(4^3+1)(4^6+1)(4^{12}+1) \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{4^3-1} \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{63} \\ &= (4^3+1)(4^6+1)(4^{12}+1)(4^3-1) \\ &= (4^3-1)(4^3+1)(4^6+1)(4^{12}+1) \\ &= (4^6-1)(4^6+1)(4^{12}+1) \\ &= (4^{12}-1)(4^{12}+1) \\ &= 4^{24}-1 \\ 4^x &= 4^{24} \\ x &= 24 \\ \end{align} </math> </div></div> <ol start=59> <li>Berapakah nilai dari <math>\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}</math> jika <math>x=\sqrt{9+4\sqrt{5}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \sqrt{9+4\sqrt{5}} \\ x &= 2+\sqrt{5} \\ x^2 &= 9+4\sqrt{5} \\ x^2-4x &= 9+4\sqrt{5}-4(2+\sqrt{5}) \\ x^2-4x &= 1 \\ x^2 &= 4x+1 \\ x^3 &= x \cdot x^2 \\ &= x(4x+1) \\ &= 4x^2+x \\ &= 4(4x+1)+x \\ &= 16x+4+x \\ &= 17x+4 \\ x^4 &= x \cdot x^3 \\ &= x(17x+4) \\ &= 17x^2+4x \\ &= 17(4x+1)+4x \\ &= 68x+17+4x \\ &= 72x+17 \\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1} &= \frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1} \\ &= \frac{72x+17-85x-20+8x+2+5x+3}{2} \\ &= \frac{2}{2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=60> <li>Berapakah nilai dari <math>\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}</math> jika 2x-1=<math>\sqrt{61}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \frac{x^3+1}{x^5-x^4-x^3+x^2} = p \\ p &= \frac{x^3+1}{x^5-x^4-x^3+x^2} \\ &= \frac{x^3+1}{x^5-x^4-(x^3-x^2)} \\ &= \frac{x^3+1}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)} \\ &= \frac{x^2-x+1}{x^2(x-1)^2} \\ &= \frac{x^2-x+1}{(x(x-1))^2} \\ &= \frac{x(x-1)+1}{(x(x-1))^2} \\ 2x-1 &= \sqrt{61} \\ x &= \frac{\sqrt{61}+1}{2} \\ x-1 &= \frac{\sqrt{61}-1}{2} \\ x(x-1) &= (\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2}) \\ &= \frac{61-1}{4} \\ &= \frac{60}{4} \\ &= 15 \\ p &= \frac{x(x-1)+1}{(x(x-1))^2} \\ &= \frac{15+1}{15^2} \\ &= \frac{16}{15^2} \\ \sqrt{p} &= \sqrt{\frac{16}{15^2}} \\ &= \frac{4}{15} \\ \end{align} </math> </div></div> <ol start=61> <li>Berapakah nilai dari <math>(\frac{x-3}{x})^{25}</math> jika <math>x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\sqrt[5]{8}+\sqrt[5]{2} &= 1+\sqrt[5]{16}+\sqrt[5]{4} \\ x+(\sqrt[5]{2})^3+\sqrt[5]{2} &= 1+(\sqrt[5]{2})^4+(\sqrt[5]{2})^2 \\ x &= (\sqrt[5]{2})^4-(\sqrt[5]{2})^3+(\sqrt[5]{2})^2-\sqrt[5]{2}+1 \\ \text{misalkan } \sqrt[5]{2} = p \\ x &= p^4-p^3+p^2-p+1 \\ x &= \frac{p^5+1}{p+1} \\ (\frac{x-3}{x})^{25} &= (1-\frac{3}{x})^{25} \\ &= (1-\frac{3}{\frac{p^5+1}{p+1}})^{25} \\ &= (1-\frac{3(p+1)}{p^5+1})^{25} \\ &= (1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2})^5+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{2+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-3\sqrt[5]{2}-3)}{3})^{25} \\ &= (-\sqrt[5]{2})^{25} \\ &= (-2)^5 \\ &= -32 \\ \end{align} </math> </div></div> <ol start=62> <li>Berapakah nilai dari <math>x^{50}+x^{49}+x^{48}+x^{47}+x^{46}</math> jika <math>x^2+x+1=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+x+1 &= 0 \\ x^2+x &= -1 \\ \frac{x^3-1}{x-1} &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46} &= x^{48}(x^2+x+1)+x^{45}(x^2+x) \\ &= x^{48}(0)+(x^3)^{15}(-1) \\ &= 0+(1)^{15}(-1) \\ &= -1 \\ \end{align} </math> </div></div> <ol start=63> <li>Berapakah 2²⁴ dari <math>8^7+8^6+8^5+8^4+8^3+8^2+8+1=A</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= A \\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1) &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8 &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= 8A+1 \\ 8^8+A &= 8A+1 \\ 8^8 &= 7A+1 \\ (2^3)^8 &= 7A+1 \\ 2^{24} &= 7A+1 \\ \end{align} </math> </div></div> <ol start=64> <li>Berapakah nilai dari <math>x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1</math> jika <math>x+\frac{1}{x}=\sqrt{3}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{x} &= \sqrt{3} \\ x^2+2+\frac{1}{x^2} &= 3 \\ x^2-1+\frac{1}{x^2} &= 0 \\ x^2(x^2-1+\frac{1}{x^2}) &= x^2(0) \\ x^4-x^2+1 &= 0 \\ (x^2+1)(x^4-x^2+1) &= (x^2+1)0 \\ x^6-x^4+x^2+x^4-x^2+1 &= 0 \\ x^6+1 &= 0 \\ x^6 &= -1 \\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1 &= {x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1 \\ &= (-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2-1+1 \\ &= -1+1-1+1-1+1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=65> <li>Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari <math>\frac{c-b}{a-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= ax^2+bx+c \\ f(2) &= a(2)^2+2b+c = 4 \\ &= 4a+2b+c = 4 \\ f(7) &= a(7)^2+7b+c = 49 \\ &= 49a+7b+c = 49 \\ 49a+7b+c &= 49 \\ 4a+2b+c &= 4 \\ 45a+5b &= 45 \text{ (f(7) dikurangi f(2)) } \\ 9a+b &= 9 \\ b &= -9a+9 \\ 4a+2b+c &= 4 \\ 4a+2(-9a+9)+c &= 4 \\ 4a-18a+18+c &= 4 \\ -14a+18+c &= 4 \\ c &= 14a-14 \\ \frac{c-b}{a-1} &= \frac{14a-14-(-9a+9)}{a-1} \\ &= \frac{14(a-1)+9(a-1)}{a-1} \\ &= \frac{(14+9)(a-1)}{a-1} \\ &= 23 \\ \end{align} </math> </div></div> <ol start=66> <li>Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+y)^3 &= x^3+y^3+3xy(x+y) \\ 11^3 &= 242+3xy(11) \text{ (dibagi 11)} \\ 11^2 &= 22+3xy \\ 121 &= 22+3xy \\ 99 &= 3xy \\ xy &= 33 \\ (x-y)^2 &= x^2+y^2-2xy \\ &= ((x+y)^2-2xy)-2xy \\ &= (x+y)^2-4xy \\ &= 11^2-4(33) \\ &= 121-132 \\ &= -11 \\ \end{align} </math> </div></div> <ol start=67> <li>Berapa f(1)+f(-1) jika <math>f(\frac{ax-b}{bx-a})</math>=x²-5x+6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ jika} f(1) = f(\frac{ax-b}{bx-a}) \\ 1 &= \frac{ax-b}{bx-a} \\ bx-a &= ax-b \\ (b-a)x &= -b+a \\ &= -(b-a) \\ &= -1 \\ f(1) &= x^2-5x+6 \\ &= (-1)^2-5(-1)+6 \\ &= 12 \\ \text{ jika} f(-1) = f(\frac{ax-b}{bx-a}) \\ -1 &= \frac{ax-b}{bx-a} \\ -(bx-a) &= ax-b \\ -bx+a &= ax-b \\ (-b-a)x &= -b-a \\ &= 1 \\ f(-1) &= x^2-5x+6 \\ &= (1)^2-5(1)+6 \\ &= 2 \\ f(1)+f(-1) &= 12+2 \\ &= 14 \\ \end{align} </math> </div></div> <ol start=68> <li>berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)-x &= f(x-1) \\ f(x)-f(x-1) &= x \\ x=1 ; f(1)-f(0) &= 1 \\ x=2 ; f(2)-f(1) &= 2 \\ x=3 ; f(3)-f(2) &= 3 \\ x=4 ; f(4)-f(3) &= 4 \\ \dots \\ x=200 ; f(200)-f(199) &= 200 \\ \text{ jumlahkan tersebut menjadi } \\ f(200)-f(0) &= 1+2+3+4+\dots+200 \\ &= \frac{200 \cdot 201}{2} \\ &= 20.100 \\ f(200)-1 &= 20.100 \\ &= 20.101 \\ \end{align} </math> </div></div> <ol start=69> <li>Misalkan f(x) adalah fungsi rekursif yang berlaku ∀x ∈ R sebagai berikut: : f(x)+f(15-x) = 2024 : f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)+f(15-x) &= 2024 \\ f(15+x) &= f(x)+2020 \\ *cara 1 \\ \text{ganti x dengan 15+x } \\ f(15+x)+f(-x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ *cara 2 \\ \text{ganti x dengan -x } \\ f(-x)+f(15+x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ \end{align} </math> </div></div> <ol start=70> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>2f(\frac{2002}{x}) + f(x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2f(\frac{2002}{x}) + f(x) &= 3x \\ \text{ganti x dengan 2 } \\ 2f(\frac{2002}{2}) + f(2) &= 3(2) \\ 2f(1001) + f(2) &= 6 \\ \text{ganti x dengan 1001 } \\ 2f(\frac{2002}{1001}) + f(1001) &= 3(1001) \\ 2f(2) + f(1001) &= 3003 \\ 2f(2) + f(1001) &= 3003 \\ f(1001) &= 3003 - 2f(2) \\ 2f(1001) + f(2) &= 6 \\ 2(3003 - 2f(2)) + f(2) &= 6 \\ 6006 - 4f(2) + f(2) &= 6 \\ 3f(2) &= 6000 \\ f(2) &= 2000 \\ \end{align} </math> </div></div> <ol start=71> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>f(\frac{1}{x}) + \frac{1}{x}f(-x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{1}{x})+\frac{1}{x}f(-x) &= 3x \\ \text{ganti x dengan 1/3 } \\ f(3)+3f(-\frac{1}{3}) &= 1 \\ \text{ganti x dengan -3 } \\ f(-\frac{1}{3}) - \frac{1}{3}f(3) &= -9 \\ \text{dikalikan 3 } \\ 3f(-\frac{1}{3})-f(3) &= -27 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ 2f(3) &= 28 \\ f(3) &= 14 \\ \end{align} </math> </div></div> <ol start=72> <li>Diketahui polinom <math>f(7^b-1)=7^{3b}-10</math>. tentukan nilai f(5)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(5) &= f(7^b-1) \\ 5 &= 7^b-1 \\ 7^b &= 6 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(6-1) &= 6^3-10 \\ f(5) &= 216-10 \\ &= 206 \\ *cara 2 \\ \text{misalkan } 7^b-1=a \text{ maka } 7^b=a+1 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(a) &= (a+1)^3-10 \\ f(5) &= (5+1)^3-10 \\ &= 6^3-10 \\ &= 216-10 \\ &= 206 \\ \end{align} </math> </div></div> <ol start=73> <li>Diketahui polinom <math>f(6^b-7)=6^{3b}-2 \cdot 6^{2b}-4</math>. tentukan nilai f(-2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(-2) &= f(6^b-7) \\ -2 &= 6^b-7 \\ 6^b &= 5 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(5-7) &= 5^3-2 \cdot 5^2-4 \\ f(-2) &= 125-50-4 \\ &= 71 \\ *cara 2 \\ \text{misalkan } 6^b-7=a \text{ maka } 6^b=a+7 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(a) &= (a+7)^3-2(a+7)^2-4 \\ f(-2) &= (-2+7)^3-2(-2+7)^2-4 \\ &= 5^3-2(5)^2-4 \\ &= 125-50-4 \\ &= 71 \\ \end{align} </math> </div></div> <ol start=74> <li>Jika <math>f(xy)=\frac{f(x)}{y}</math> dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(10) &= 7 \\ f(2 \cdot 5) &= 7 \\ f(xy) &= \frac{f(x)}{y} \\ f(2 \cdot 5) &= \frac{f(2)}{5} \\ 7 &= \frac{f(2)}{5} \\ f(2) &= 35 \\ \end{align} </math> </div></div> <ol start=75> <li>Jika <math>f(xy)=\frac{f(x+y)}{xy}</math> dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(15) &= 16 \\ f(3 \cdot 5) &= 16 \\ f(xy) &= \frac{f(x+y)}{xy} \\ f(3 \cdot 5) &= \frac{f(3+5)}{3 \cdot 5} \\ f(15) &= \frac{f(8)}{15} \\ 16 &= \frac{f(8)}{15} \\ f(8) &= 240 \\ \end{align} </math> </div></div> <ol start=76> <li>Jika <math>f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15</math> maka tentukan nilai f(16)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x+\frac{1}{x}+6) &= x^2+\frac{1}{x^2}+15 \\ &= (x+\frac{1}{x})^2-2+15 \\ &= (x+\frac{1}{x})^2+13 \\ \text{misalkan } x+\frac{1}{x} &= p \\ f(x+\frac{1}{x}+6) &= (x+\frac{1}{x})^2+13 \\ f(p+6) &= p^2+13 \\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 } \\ f(p+6) &= p^2+13 \\ f(10+6) &= 10^2+13 \\ f(16) &= 100+13 \\ &= 113 \\ \end{align} </math> </div></div> <ol start=77> <li>Tentukan nilai x jika <math>f(x)=\frac{4}{4-x}</math> dan <math>f(x \cdot f(x))^{\frac{f(4x)}{f(x)}}=256</math>!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{4}{4-x} \\ f(4x) &= \frac{4}{4-4x} \\ \frac{f(4x)}{f(x)} &= \frac{\frac{4}{4-4x}}{\frac{4}{4-x}} \\ &= \frac{4-x}{4-4x} \\ f(x \cdot f(x)) &= f(x(\frac{4}{4-x})) \\ &= f(\frac{4x}{4-x}) \\ &= \frac{4}{4-(\frac{4x}{4-x})} \\ &= \frac{4}{\frac{16-4x-4x}{4-x}} \\ &= \frac{4}{\frac{16-8x}{4-x}} \\ &= \frac{4(4-x)}{4(4-4x)} \\ &= \frac{4-x}{4-4x} \\ \text{misalkan } \frac{4-x}{4-4x} &= a \\ f(x \cdot f(x))^{\frac{f(4x)}{f(x)}} &= 256 \\ a^a &= 256 \\ a^a &= 4^4 \\ a &= 4 \\ \frac{4-x}{4-4x} &= 4 \\ 4-x &= 16-16x \\ 15x &= 12 \\ x &= \frac{4}{5} \\ \end{align} </math> </div></div> <ol start=78> <li>Fungsi <math>f(x) = \frac{kx}{2x+1} \text{dengan } x \neq -\frac{1}{2}</math>. Dengan f(f(x)) = x maka tentukan nilai k!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{kx}{2x+1} \\ f(f(x)) &= x \\ f(\frac{kx}{2x+1}) &= x \\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1} &= x \\ \frac{\frac{k^2x}{2x+1}}{\frac{2kx+2x+1}{2x+1}} &= x \\ \frac{k^2x}{2kx+2x+1} &= x \\ \frac{k^2}{2kx+2x+1} &= 1 \\ k^2 &= 2kx+2x+1 \\ k^2-2kx &= 2x+1 \\ k^2-2kx+x^2 &= x^2+2x+1 \\ (k-x)^2 &= (x+1)^2 \\ (k-x)^2-(x+1)^2 &= 0 \\ (k-x+x+1)(k-x-(x+1)) &= 0 \\ k=-1 &\text{ atau } k=2x+1 &\text{ (TM) } \\ \end{align} </math> </div></div> <ol start=79> <li>Jika n = 2023²+2024² maka berapa hasil dari <math>\sqrt{2n-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n &= 2023^2+2024^2 \\ &= 2023^2+(2023+1)^2 \\ \text{misalkan 2023 = p} \\ n &= p^2+(p+1)^2 \\ &= p^2+p^2+2p+1 \\ &= 2p^2+2p+1 \\ \sqrt{2n-1} &= \sqrt{2(2p^2+2p+1)-1} \\ &= \sqrt{4p^2+4p+2-1} \\ &= \sqrt{4p^2+4p+1} \\ &= \sqrt{(2p+1)^2} \\ &= 2p+1 \\ &= 2(2023)+1 \\ &= 4046+1 \\ &= 4047 \\ \end{align} </math> </div></div> <ol start=80> <li>Tentukan nilai dari a+b+c merupakan bilangan bulat positif jika ab = 2, bc = 3 dan ac = 6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab \cdot bc \cdot ac &= 2 \cdot 3 \cdot 6 \\ (abc)^2 &= 36 \\ abc &= \pm 6 \\ abc &= 6 \\ \frac{abc}{ab} &= c = \frac{6}{2} = 3 \\ \frac{abc}{bc} &= a = \frac{6}{3} = 2 \\ \frac{abc}{ac} &= b = \frac{6}{6} = 1 \\ a+b+c &= 6 \\ \end{align} </math> </div></div> <ol start=81> <li>Tentukan nilai dari (a-c)^b jika <math>\frac{ab}{a+b} = \frac{1}{3}</math>, <math>\frac{bc}{b+c} = \frac{1}{4}</math> dan <math>\frac{ac}{a+c} = \frac{1}{9}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{ab}{a+b} &= \frac{1}{3} \\ \frac{a+b}{ab} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{b} + \frac{1}{a} &= 3 \\ \frac{bc}{b+c} &= \frac{1}{4} \\ \frac{b+c}{bc} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{b} &= 4 \\ \frac{ac}{a+c} &= \frac{1}{9} \\ \frac{a+c}{ac} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{a} &= 9 \\ \text{Misalkan 1/a = x, 1/b = y dan 1/c = z} \\ x+y &= 3 \\ y+z &= 4 \\ x+z &= 9 \\ x+y &= 3 \\ y+z &= 4 \\ x-z &= -1 \\ x-z &= -1 \\ x+z &= 9 \\ 2x &= 8 \\ x &= 4 \\ x-z &= -1 \\ 4-z &= -1 \\ z &= 5 \\ x+y &= 3 \\ 4+y &= 3 \\ y &= -1 \\ \frac{1}{a} &= 4 \\ a &= \frac{1}{4} \\ \frac{1}{b} &= -1 \\ b &= -1 \\ \frac{1}{c} &= 5 \\ c &= \frac{1}{5} \\ (a-c)^b &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=82> <li>Tentukan nilai dari a, b dan c jika <math>\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}</math> dan a+2b+3c=28?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan k untuk semua ketiga persamaan tersebut } \\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5} &= k \\ a+b &= 2k \\ a+c &= 4k \\ b+c &= 5k \\ 2a+b+c &= 6k \\ 2a+5k &= 6k \\ k &= 2a \\ a &= \frac{k}{2} \\ b &= \frac{3k}{2} \\ c &= \frac{7k}{2} \\ a+2b+3c &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ a &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ b &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ c &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=83> <li>Tentukan nilai dari (b+c)^a jika <math>\frac{a+b+c}{2} = \sqrt{a-2}+\sqrt{b-1}+\sqrt{c}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{a+b+c}{2} &= \sqrt{a-2}+\sqrt{b-1}+\sqrt{c} \\ a+b+c &= 2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}) \\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c} &= 0 \\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1 &= 0 \\ (\sqrt{a-2}-1)^2+(\sqrt{b-1}-1)^2+(\sqrt{c}-1)^2 &= 0 \\ (\sqrt{a-2}-1)^2 &= 0 \\ \sqrt{a-2}-1 &= 0 \\ \sqrt{a-2} &= 1 \\ a-2 &= 1 \\ a &= 3 \\ (\sqrt{b-1}-1)^2 &= 0 \\ \sqrt{b-1}-1 &= 0 \\ \sqrt{b-1} &= 1 \\ b-1 &= 1 \\ b &= 1 \\ (\sqrt{c}-1)^2 &= 0 \\ \sqrt{c}-1 &= 0 \\ \sqrt{c} &= 1 \\ c &= 1 \\ (b+c)^a &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=84> <li>x dan y merupakan bilangan tak nol. Jika xy = <math>\frac{x}{y}</math> = x-y maka berapa nilai x+y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy &= \frac{x}{y} \\ y^2 &= 1 \\ y^2 - 1 &= 0 \\ (y-1)(y+1) &= 0 \\ y = 1 &\text{ atau } y = -1 \\ \frac{x}{y} &= x-y \\ x &= xy-y^2 \\ x-xy &= -y^2 \\ x(1-y) &= -y^2 \\ x &= \frac{-y^2}{1-y} \\ \text{cek y=1 } \\ x &= \frac{-1^2}{1-1} \\ \text{tidak memenuhi syarat } \\ \text{cek y=-1 } \\ x &= \frac{-(-1)^2}{1-(-1)} \\ &= \frac{-1}{2} \\ x+y &= -1-\frac{1}{2} \\ &= -\frac{3}{2} \\ \end{align} </math> </div></div> <ol start=86> <li>Berapa nilai x dari <math>(\frac{a}{b})^3+(\frac{b}{a})^3 = 2\sqrt{x}</math> jika <math>\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b} \\ \frac{a+b}{ab} &= \frac{1}{a+b} \\ (a+b)^2 &= ab \\ a^2+2ab+b^2 &= ab \\ a^2+b^2 &= -ab \\ \text{misalkan } \frac{a}{b}+\frac{b}{a} = n \\ \frac{a}{b}+\frac{b}{a} &= n \\ \frac{a^2+b^2}{ab} &= n \\ a^2+b^2 &= nab \\ n &= -1 \\ \frac{a}{b}+\frac{b}{a} &= n \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3(\frac{a}{b}+\frac{b}{a}) &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3n &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3 &= n^3-3n \\ &= (-1)^3-3(-1) \\ &= 2 \\ 2\sqrt{x} &= 2 \\ \sqrt{x} &= 1 \\ x &= 1 \\ \end{align} </math> </div></div> <ol start=87> <li>Berapa nilai m dari <math>x^2-mx-1=0</math> jika <math>\sqrt[3]{x_1}+\sqrt[3]{x_2}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x_1} &= a \\ x_1 &= a^3 \\ \sqrt[3]{x_2} &= b \\ x_2 &= b^3 \\ \sqrt[3]{x_1}+\sqrt[3]{x_2} &= 1 \\ a+b &= 1 \\ x^2-mx-1 &= 0 \\ x_1+x_2 &= m \\ x_1 \cdot x_2 &= -1 \\ x_1+x_2 &= m \\ a^3+b^3 &= m \\ x_1 \cdot x_2 &= -1 \\ a^3 \cdot b^3 &= -1 \\ (ab)^2 &= (-1)^3 \\ ab &= -1 \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ (1)^3 &= m+3(-1)(1) \\ 1 &= m-3 \\ m &= 4 \\ \end{align} </math> </div></div> <ol start=88> <li>Berapa nilai <math>\frac{x_1}{x_2}</math> dari <math>ax^2-18x-b=0</math> jika <math>ab=45</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab &= 45 \\ b &= \frac{45}{a} \\ ax^2-18x-b &= 0 \\ ax^2-18x-\frac{45}{a} &= 0 \\ a^2x^2-18ax-45 &= 0 \\ (ax-3)(ax-15) &= 0 \\ ax-3 &= 0 \\ x &= \frac{3}{a} \\ ax-15 &= 0 \\ x &= \frac{15}{a} \\ \frac{x_1}{x_2} &= \frac{\frac{3}{a}}{\frac{15}{a}} \\ &= \frac{3}{15} \\ &= \frac{1}{5} \\ \frac{x_1}{x_2} &= \frac{\frac{15}{a}}{\frac{3}{a}} \\ &= \frac{15}{3} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=89> <li>Jika <math>\frac{u_3}{u_1+u_2} = \frac{7}{8}</math> merupakan barisan aritmetika maka berapa dari <math>\frac{u_2+u_3}{u_1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{u_3}{u_1+u_2} &= \frac{7}{8} \\ \frac{a+2b}{a+a+b} &= \frac{7}{8} \\ \frac{a+2b}{2a+b} &= \frac{7}{8} \\ 8(a+2b) &= 7(2a+b) \\ 8a+16b &= 14a+7b \\ 9b &= 6a \\ b &= \frac{2a}{3} \\ \frac{u_2+u_3}{u_1} &= \frac{a+b+a+2b}{a} \\ &= \frac{2a+3b}{a} \\ &= \frac{2a+3(\frac{2a}{3})}{a} \\ &= \frac{2a+2a}{a} \\ &= \frac{4a}{a} \\ &= 4 \\ \end{align} </math> </div></div> <ol start=90> <li>Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7p+q}{2p+q} &= \frac{17p+q}{7p+q} \\ (7p+q)^2 &= (17p+q)(2p+q) \\ 49p^2+14pq+q^2 &= 34p^2+19pq+q^2 \\ 15p^2 &= 5pq \\ 3p &= q \\ \frac{7p+q}{2p+q} &= \frac{7p+3p}{2p+3p} \\ &= \frac{10p}{5p} \\ &= 2 \\ \end{align} </math> </div></div> <ol start=91> <li>Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{rataan geometris } \\ \sqrt{a \cdot b} &= b-24 \\ a \cdot b &= (b-24)^2 \\ \text{rataan aritmatik } \\ \frac{a+b}{2} &= a+15 \\ a+b &= 2(a+15) \\ a+b &= 2a+30 \\ a &= b-30 \\ a \cdot b &= (b-24)^2 \\ (b-30)b &= (b-24)^2 \\ b^2-30b &= b^2-48b+576 \\ 18b &= 576 \\ b &= 32 \\ a &= b-30 \\ &= 32-30 \\ &= 2 \\ a+b &= 32+2 \\ &= 34 \\ \end{align} </math> </div></div> <ol start=91> <li>Segitiga lancip ABC dengan <math>\frac{a^4+b^4+c^4+a^2b^2}{c^2(a^2+b^2)}=2</math>. tentukan nilai sudut C?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{syarat segitiga lancip semua sudut masing-masing kurang dari } 90^\circ \\ c^2 &= a^2+b^2-2ab cos C \\ cos C &= \frac{a^2+b^2-c^2}{2ab} \\ a^4+b^4+c^4+a^2b^2 &= 2c^2(a^2+b^2) \\ a^4+b^4+a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-2c^2(a^2+b^2)+(c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= (ab)^2 \\ a^2+b^2-c^2 &= \pm ab \\ cos C &= \pm \frac{ab}{2ab} \\ &= \pm \frac{1}{2} \\ &= \frac{1}{2} \text{ (karena sudut harus kurang dari } 90^\circ) \\ C &= 60^\circ \\ \end{align} </math> </div></div> <ol start=92> <li>Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan panjang DA dan EA } = x \text{ dan panjang AB } = y \\ \text{luas segitiga CAB } &= \frac{CA \cdot AB}{2} \\ 45 &= \frac{(x+9)(x+5)}{2} \\ 90 &= x^2+14x+45 \\ x^2+14x &= 45 \\ y^2 &= (x+9)^2+(x+5)^2 \\ &= x^2+18x+81+x^2+10x+25 \\ &= 2x^2+28x+106 \\ &= 2(x^2+14x)+106 \\ &= 2(45)+106 \\ &= 196 \\ y &= 14 \\ \end{align} </math> jadi panjang BC adalah 14 cm </div></div> <ol start=93> <li>Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kita cari ukuran GC yaitu } \\ \frac{GC}{AD} &= \frac{CE}{DE} \\ \frac{GC}{15} &= \frac{6}{18} \\ GC &= 5 \\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC } \\ &= \frac{1}{2} \cdot BC \cdot CF - \frac{1}{2} \cdot GC \cdot CE \\ &= \frac{1}{2} \cdot 15 \cdot 18 - \frac{1}{2} \cdot 5 \cdot 6 \\ &= 135 - 15 \\ &= 120 \\ \end{align} </math> jadi luas daerah BFEG adalah 120 cm² </div></div> <ol start=94> <li>Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ AI &= \frac{1}{5} AF \\ &= \frac{1}{5} 10 \\ &= 2 \\ IF &= AF-AI \\ &= 10-2 \\ &= 8 \\ \text{luas trapesium AFGD } &= \frac{(AD+EF) \cdot AF}{2} \\ &= \frac{(4+6)10}{2} \\ &= 50 \\ \text{luas segitiga AID } &= \frac{AI \cdot AF}{2} \\ &= \frac{(2)4}{2} \\ &= 4 \\ \text{luas segitiga IFG } &= \frac{IF \cdot FG}{2} \\ &= \frac{(8)6}{2} \\ &= 24 \\ \text{luas daerah segitiga IGD } &= \text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG } \\ &= 50-4-24 \\ &= 22 \\ \end{align} </math> jadi luas daerah segitiga IGD adalah 22 cm² </div></div> <ol start=96> <li>Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-<math>\pi</math>) cm³ maka berapa luas permukaan kerucut tersebut?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ \text{volume balok} \\ V_b &= x^2(12) \\ \text{volume kerucut} \\ V_b &= \frac{1}{3}\pi x^2(12) \\ &= 4\pi x^2 \\ V_{b-k} &= Vb-Vk \\ 100(3-\pi) &= 12x^2-4\pi x^2 \\ 100(3-\pi) &= 4x^2(3-\pi) \\ x^2 &= 25 \\ x &= 5 \\ s &= \sqrt{12^2+5^2} \\ &= \sqrt{144+25} \\ &= \sqrt{169} \\ &= 13 \\ \text{luas permukaan kerucut } &= \pi r(r+s) \\ &= \pi(5)(5+13) \\ &= 90\pi \\ \end{align} </math> jadi luas daerah permukaan kerucut adalah 90<math>\pi</math> cm² </div></div> <ol start=97> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 3a+1 \\ B &= 3b+2 \\ A(A+1)+3B \\ (3a+1)(3a+1+1)+3(3b+2) \\ (3a+1)(3a+2)+9b+6 \\ 9a^2+9a+2+9b+6 \\ 9a^2+9a+9b+8 \\ 9(a^2+a+b)+8 \\ \text{sisa pembagiannya adalah } 8 \\ \end{align} </math> </div></div> <ol start=98> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 9a+7 \\ B &= 9b+8 \\ A(A-5)+9B \\ (9a+7)(9a+7-5)+9(9b+8) \\ (9a+7)(9a+2)+81b+72 \\ 81a^2+81a+14+81b+72 \\ 81a^2+81a+81b+86 \\ 81a^2+81a+81b+81+5 \\ 81(a^2+a+b+1)+5 \\ \text{sisa pembagiannya adalah } 5 \\ \end{align} </math> </div></div> <ol start=99> <li>Jika <math>\begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix}</math> maka berapa hasil dari A²¹+A²⁵+A⁴⁶?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A^2 &= A \cdot A \\ &= \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \\ A^3 &= A^2 \cdot A \\ &= \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ &= - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= -I \\ A^{21}+A^{25}+A^{46} &= A^{21} \cdot (I+A^4+A^{25}) \\ &= A^{21} \cdot (I+A^3 \cdot A +A^{24} \cdot A) \\ &= (A^3)^7 \cdot (I+A^3 \cdot A +(A^3)^8 \cdot A) \\ &= (-I)^7 \cdot (I-I \cdot A +(-I)^8 \cdot A) \\ &= -I \cdot (I-A+A) \\ &= -I \cdot I \\ &= -I \\ &= -\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ \end{align} </math> </div></div> <ol start=100> <li>Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida?</li></ol> : bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15 : ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9). : ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8) : maka Ida menulis bilangan 8. <ol start=101> <li>Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9?</li></ol> : Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu. : untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja. : untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja. : Jadi banyaknya bilangan mungkin 2. <ol start=102> <li>Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli!</li></ol> ;cara 1 # 8¹ = 1 # 8² = 1 (8²=8¹x8¹ sama dengan 1x1) # 8³ = 1 (8³=8¹x8² sama dengan 1x1) # 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2) # 8⁵ = 1 # 8ⁿ = 1 (semua n untuk bilangan asli) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli ;cara 2 # 8ⁿ = b mod 7 # 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil) # (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya) # 8ⁿ = 1ⁿ mod 7 # 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli <ol start=103> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 5?</li></ol> ;cara 1 # 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 # 7¹ = 7 (sisa 1) # 7² = 49 (sisa 2) # 7³ = 343 (sisa 3) # 7⁴ = 2,401 (sisa 0) # 7⁵ = 16,807 # 7⁶ = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3 ;cara 2 :17¹ = 2 :17² = 4 :17³ = 3 :17⁴ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3 :17⁹⁹ = (17⁴)²⁴ x 17³ Untuk 17⁴ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 3. Jadi 17⁹⁹ dibagi 7 bersisa 3 ;cara 3 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 17⁴ ::17⁴ = 1 mod 5 ::(17⁴)²⁴ = 1²⁴ mod 5 ::17⁹⁶ = 1²⁴ mod 5 ::17⁹⁶ = 1 mod 5 ::17⁹⁶ x 17³ = 1 x 17³ mod 5 ::17⁹⁹ = 17³ mod 5 ::17⁹⁹ = 17 x 17 x 17 mod 5 ::17⁹⁹ = 2 x 2 x 2 mod 5 ::17⁹⁹ = 8 mod 5 ::17⁹⁹ = 3 mod 5 Jadi 17⁹⁹ dibagi 5 bersisa 3 <ol start=104> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 7?</li></ol> ;cara 1 :17¹ = 3 :17² = 2 :17³ = 6 :17⁴ = 4 :17⁵ = 5 :17⁶ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3 :17⁹⁹ = (17⁶)¹⁶ x 17³ Untuk 17⁶ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 6. Jadi 17⁹⁹ dibagi 7 bersisa 6 ;cara 2 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 17⁶ ::17⁶ = 1 mod 7 ::(17⁶)¹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1 mod 7 ::17⁹⁶ x 17³ = 1 x 17³ mod 7 ::17⁹⁹ = 17³ mod 7 ::17⁹⁹ = 17 x 17 x 17 mod 7 ::17⁹⁹ = 3 x 3 x 3 mod 7 ::17⁹⁹ = 27 mod 7 ::17⁹⁹ = 6 mod 7 Jadi 17⁹⁹ dibagi 7 bersisa 6 <ol start=105> <li>Berapa hasil sisa dari 41²⁰²⁴ dibagi 33?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 41^{2024} &= 41^{2024} \text{ mod } 33 \\ &= (33 \times 3 + 2)^{2024} \text{ mod } 33 \\ &= 2^{2024} \text{ mod } 33 \\ &= 2^{2020} 2^4 \text{ mod } 33 \\ &= (2^5)^{404} 2^4 \text{ mod } 33 \\ &= (33 - 1)^{404} 2^4 \text{ mod } 33 \\ &= (-1)^{404} 2^4 \text{ mod } 33 \\ &= 2^4 \text{ mod } 33 \\ &= 16 \text{ mod } 33 \\ \text{Jadi hasil sisa adalah } 16 \\ \end{align} </math> </div></div> <ol start=106> <li>Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 99^{99} &= (3^2 \times 11)^{99} \\ &= 3^{198} \times 11^{99} \\ 243^n &= (3^5)^n \\ &= 3^{5n} \\ \text{agar bisa membagi, maka} \\ 5n &= 198 \\ n &= 39.6 \\ \text{jadi bilangan n terbesar adalah } 39 \\ \end{align} </math> </div></div> <ol start=107> <li>Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸?</lu></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 88^{88} &= (8 \times 11)^{88} \\ &= 8^{88} \times 11^{88} \\ &= 8^{87} \times 8 \times 11^{88} \\ &= (8^3)^{29} \times 8 \times 11^{88} \\ &= 512^{29} \times 8 \times 11^{88} \\ 512^n &= 512^{29} \\ \text{jadi bilangan n terbesar adalah } 29 \\ \end{align} </math> </div></div> <ol start=108> <li>Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6!</li></ol> ; cara 1 : KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. : N dibagi 3 sisa 1 : N dibagi 5 sisa 2 : N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a : KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70 : KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21 : KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97 ; cara 2 : Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35 : kemudian ditambahkan sisa masing-masing sesuai dengan KPK. : KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, <b>97</b> : KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, <b>97</b> : KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, <b>97</b> Jadi bilangan bulat positif adalah 97 :: NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 <ol start=109> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |- | 4 || 0 || Semua isi ember B dibuang |- | 1 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. <ol start=110> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |} nah ada ember A berisi 4 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |- | 1 || 0 || Semua isi ember B dibuang |- | 0 || 1 || Tuangkan 1 l dari ember A ke B sehingga ember A kosong |- | 3 || 1 || Isikan 3 l ke ember A |- | 0 || 4 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |} nah ada ember B berisi 4 liter. # Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kemungkinan 1} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 +3 \cdot ((\frac{4}{3})^x)^2 &= 7 \cdot (\frac{4}{3})^x \\ \text{misalkan } (\frac{4}{3})^x = y \\ 4 +3y^2 &= 7y \\ 3y^2-7y+4 &= 0 \\ (3y-4)(y-1) &= 0 \\ y_1 = \frac{4}{3} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{4}{3} \\ (\frac{4}{3})^x &= \frac{4}{3} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{4}{3})^x &= 1 \\ x &= 0 \\ \text{kemungkinan 2} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot ((\frac{3}{4})^x)^2 +3 &= 7 \cdot (\frac{3}{4})^x \\ \text{misalkan } (\frac{3}{4})^x = y \\ 4y^2 +3 &= 7y \\ 4y^2-7y+3 &= 0 \\ (4y-3)(y-1) &= 0 \\ y_1 = \frac{3}{4} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{3}{4} \\ (\frac{3}{4})^x &= \frac{3}{4} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{3}{4})^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-6}></math>? [[Kategori:Soal-Soal Matematika]] fjcd0p0unasrxr7pwkfbek0si00j7jp 117487 117486 2026-07-13T01:04:41Z Akuindo 8654 117487 wikitext text/x-wiki contoh soal <ol start=1> <li>Berapa hasil dari <math>\sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Misalkan 2020 = p} \\ \sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64} &= \sqrt{(2020-5) \cdot (2020-3) \cdot (2020+3) \cdot (2020+5) + 64} \\ &= \sqrt{(p-5) \cdot (p-3) \cdot (p+3) \cdot (p+5) + 64} \\ &= \sqrt{(p-5) \cdot (p+5) \cdot (p-3) \cdot (p+3) + 64} \\ &= \sqrt{(p^2-25) \cdot (p^2-9) + 64} \\ &= \sqrt{p^4-34p^2+ 225 + 64} \\ &= \sqrt{p^4-34p^2+ 289} \\ &= \sqrt{(p^2-17)^2} \\ &= p^2-17 \\ &= 2020^2-17 \\ &= (2000+20)^2-17 \\ &= 4.000.000+80.000+400-17 \\ &= 4.080.383 \\ \end{align} </math> </div></div> <ol start=2> <li>Berapa nilai x dari <math>\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} = \frac{9}{10}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} &= \frac{9}{10} \\ \text{misalkan untuk } \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} = p \\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} &= p \\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}} &= p^2 \\ x^2-x-p &= p^2 \\ x^2-2x+1+x-1 &= p^2+p \\ (x-1)^2+(x-1) &= p^2+p \\ x-1 &= p \\ \text{misalkan untuk } \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}} &= q^3 \\ x^2 q &= q^3 \\ x^2 &= q^2 \\ x &= q \\ \frac{x-1}{x} &= \frac{9}{10} \\ x &= 10 \\ \end{align} </math> </div></div> <ol start=3> <li>Berapa nilai x dari <math>(\frac{x}{x+10})^{x+10}=\frac{1}{1024}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{x+10}{x})^{-(x+10)} &= (1024)^{-1} \\ (\frac{x+10}{x})^{x+10} &= 1024 \\ (\frac{x+10}{x})^{x+10} &= 2^{10} \\ (\frac{x+10}{x})^{\frac{x+10}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (\frac{1}{2})^{-1} \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (1+(-\frac{1}{2}))^{(1+(-\frac{2}{1}))} \\ \frac{10}{x} &= -\frac{1}{2} \\ x &= -20 \\ \end{align} </math> </div></div> <ol start=4> <li>Berapa nilai x dari <math>x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{2}+\sqrt{x+\frac{1}{4}} &= (\sqrt{x+\frac{1}{4}})^2+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+(\frac{1}{2})^2 \\ &= (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 \\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}} &= 4 \\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2} &= 4 \\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 4 \\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 &= 4 \\ \sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 2 \\ \sqrt{x+\frac{1}{4}} &= \frac{3}{2} \\ x+\frac{1}{4} &= \frac{9}{4} \\ x &= 2 \\ \end{align} </math> </div></div> <ol start=5> <li>Berapa nilai x dari <math>\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^3}{\sqrt{8-x^2}}+x^2-8 &= 0 \\ \frac{x^3}{\sqrt{8-x^2}} &= 8-x^2 \\ x^3 &= (8-x^2)^{\frac{3}{2}} \\ x &= (8-x^2)^{\frac{1}{2}} \\ x^2 &= 8-x^2 \\ 2x^2-8 &= 0 \\ x^2-4 &= 0 \\ (x-2)(x+2) &= 0 \\ \text{membuktikan } \\ x=2 \text{ maka hasilnya 0 } \\ x=-2 \text{ maka hasilnya -8 } \\ \text{jadi } x=2 \\ \end{align} </math> </div></div> <ol start=6> <li>Berapa nilai x dari <math>\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} = 5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} &= 5 \\ \frac{x^{50}+x^{60}+x^{70}}{31}} &= 5^5 \\ x^{50}+x^{60}+x^{70} &= 5^5 \cdot 31 \\ x^{50}(1+x^{10}+x^{20}) &= 5^5 \cdot 31 \\ (x^{10}^5)(1+x^{10}+(x^{10}^2) &= 5^5 \cdot 31 \\ \text{ misalkan } x^{10} = a \\ a^5(1+a+a^2) &= 5^5 \cdot 31 \\ a &= 5 \\ x^{10} &= 5 \\ x &= ^5 log 10 \\ \end{align} </math> </div></div> <ol start=7> <li>Berapa nilai x dari <math>\sqrt{3x+5+\sqrt{4x+5}} = x</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{3x+5+\sqrt{4x+5}} &= x \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \text{misalkan } \sqrt{4x+5}=y \text{ dan } 4x+5=y^2 \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \sqrt{y^2+y-x} &= x \\ y^2+y &= x^2+x \\ y=x \\ 4x+5 &= y^2 \\ 4x+5 &= x^2 \\ x^2-4x-5 &= 0 \\ (x-5)(x+1) &= 0 \\ x=5 &\text{ atau } x=-1 \text{ (TM) } \\ \end{align} </math> </div></div> <ol start=8> <li>Berapa nilai x dari <math>\sqrt{1+\sqrt{1+x}} = \sqrt[3]{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{1+\sqrt{1+x}} &= \sqrt[3]{x} \\ \sqrt[3]{x} &= n \\ x &= n^3 \\ \sqrt{1+\sqrt{1+n^3}} &= n \\ 1+\sqrt{1+n^3} &= n^2 \\ \sqrt{1+n^3} &= n^2-1 \\ 1+n^3 &= n^4-2n^2+1 \\ n^4-n^3-2n^2 &= 0 \\ n^2(n^2-n-2) &= 0 \\ n^2(n-2)(n+1) &= 0 \\ n=0, n=2 \text{ atau } n=-1 \\ n &= 0 \\ x &= 0^3 \\ &= 0 \\ n &= 2 \\ x &= 2^3 \\ &= 8 \\ n &= -1 \\ x &= (-1)^3 \\ &= -1 \\ \text{yang paling mungkin untuk nilai x adalah } 8 \\ \end{align} </math> </div></div> <ol start=9> <li>Berapa nilai x dari <math>\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} &= \frac{\sqrt{1+x}}{\sqrt{x}} \\ \sqrt{x}(\sqrt{1+x}+\sqrt{x}) &= (\sqrt{1+x}-\sqrt{x})\sqrt{1+x} \\ \sqrt{x(1+x)}+x &= 1+x-\sqrt{x(1+x)} \\ 2\sqrt{x(1+x)} &= 1 \\ \sqrt{x(1+x)} &= \frac{1}{2} \\ x(1+x) &= \frac{1}{4} \\ x^2+x &= \frac{1}{4} \\ 4x^2+4x &= 1 \\ 4x^2+4x-1 &= 0 \\ x &= \frac{-4 \pm \sqrt{4^2-4(4)(-1)}}{2(4)} \\ &= \frac{-4 \pm \sqrt{32}}{8} \\ &= \frac{-4 \pm 4\sqrt{2}}{8} \\ &= \frac{-1 \pm \sqrt{2}}{2} \\ \text{karena akar x harus minimal nol jadi } x = \frac{-1+\sqrt{2}}{2} \\ \end{align} </math> </div></div> <ol start=10> <li>Berapa nilai x dari <math>\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x-\sqrt{x+1}}{x+\sqrt{x+1}} &= \frac{11}{19} \\ \text{misalkan } \sqrt{x+1}=y \text{ dan } x=y^2-1 \\ \frac{y^2-1-y}{y^2-1+y} &= \frac{11}{19} \\ 19(y^2-y-1) &= 11(y^2+y-1) \\ 19y^2-19y-19 &= 11y^2+11y-11 \\ 8y^2-30y-8 &= 0 \\ 4y^2-15y-4 &= 0 \\ (4y+1)(y-4) &= 0 \\ y=-\frac{1}{4} \text{ (TM) atau } & y=4 \\ x &= 4^2-1 \\ &= 15 \\ \end{align} </math> </div></div> <ol start=11> <li>Berapa nilai x dari <math>\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} &= 98 \\ \text{misalkan } \sqrt{x^2-1}=y \\ \frac{x+y}{x-y}+\frac{x-y}{x+y} &= 98 \\ \frac{(x+y)^2+(x-y)^2}{(x-y)(x+y)} &= 98 \\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} &= 98 \\ \frac{2(x^2+y^2)}{x^2-y^2} &= 98 \\ \frac{x^2+y^2}{x^2-y^2} &= 49 \\ x^2+y^2 &= 49(x^2-y^2) \\ x^2+y^2 &= 49x^2-49y^2 \\ 48x^2 &= 50y^2 \\ 24x^2 &= 25y^2 \\ 24x^2 &= 25(\sqrt{x^2-1})^2 \\ 24x^2 &= 25(x^2-1) \\ 24x^2 &= 25x^2-25 \\ x^2 &= 25 \\ x &= \pm 5 \\ \end{align} </math> </div></div> <ol start=12> <li>Berapa nilai x dari <math>\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \sqrt{x}=y \text{ dan } x=y^2 \\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} &= \frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\sqrt{\frac{y^2}{y^2+y}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\frac{y}{\sqrt{y^2+y}} \\ y^2+y-\sqrt{(y^2+y)(y^2-y)} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^4-y^2} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^2(y^2-1)} &= \frac{5}{4}y \\ y(y+1)-y\sqrt{y^2-1} &= \frac{5}{4}y \\ y+1-\sqrt{y^2-1} &= \frac{5}{4} \\ -\sqrt{y^2-1} &= \frac{1}{4}-y \\ y^2-1 &= (\frac{1}{4}-y)^2 \\ y^2-1 &= \frac{1}{16}-\frac{1}{2}y+y^2 \\ -1 &= \frac{1}{16}-\frac{1}{2}y \\ \frac{1}{2}y &= \frac{1}{16}+1 \\ \frac{1}{2}y &= \frac{17}{16} \\ y &= \frac{17}{8} \\ x &= (\frac{17}{8})^2 \\ &= \frac{289}{64} \\ \end{align} </math> </div></div> <ol start=13> <li>Berapa nilai x dari <math>\sqrt[4]{62+x}+\sqrt[4]{275-x}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ misalkan } \sqrt[4]{62+x}=a, 62+x=a^4, \sqrt[4]{275-x}=b \text{ dan } 275-x=b^4 \\ a+b &= 7 \\ (a+b)^2 &= 49 \\ a^2+b^2+2ab &= 49 \\ a^2+b^2 &= 49-2ab \\ a^4+b^4 &= 62+x+275-x \\ (a^2+b^2)^2-2(ab)^2 &= 337 \\ (49-2ab)^2-2(ab)^2 &= 337 \\ 2401-196ab+4(ab)^2-2(ab)^2 &= 337 \\ 2(ab)^2-196ab+2064 &= 0 \\ (ab)^2-98ab+1032 &= 0 \\ (ab-12)(ab-86) &= 0 \\ ab = 12 \text{ atau } & ab = 86 \text{ (TM) karena hasil kali maksimum yaitu 12 } \\ ab =12 \text{ dan } a+b=7 \\ a+b &= 7 \\ b &= 7-a \\ ab &= 12 \\ a(7-a) &= 12 \\ -a^2+7a &= 12 \\ a^2-7a+12 &= 0 \\ (a-3)(a-4) &= 0 \\ a=3 \text{ atau } & a=4 \\ a=3, b=4 \\ 62+x &= a^4 \\ 62+x &= (3)^4 \\ 62+x &= 81 \\ x &= 19 \\ a=4, b=3 \\ 62+x &= a^4 \\ 62+x &= (4)^4 \\ 62+x &= 256 \\ x &= 194 \\ \end{align} </math> </div></div> <ol start=14> <li>Berapa nilai x dari <math>\sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2} &= 7 \\ (\sqrt[3]{8+x})^2-\sqrt[3]{8+x} \sqrt[3]{27-x}+(\sqrt[3]{27-x})^2 &= 7 \\ \text{misalkan } \sqrt[3]{8+x}=a, 8+x=a^3, \sqrt[3]{27-x}=b \text{ dan } 27-x=b^3 \\ a^2-ab+b^2 &= 7 \\ a^3+b^3 &= 8+x+27-x \\ &= 35 \\ a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ 35 &= (a+b)(7) \\ a+b &= 5 \\ b &= 5-a \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ 5^3 &= 35+3ab(5) \\ 125 &= 35+15ab \\ 80 &= 15ab \\ ab &= 6 \\ a(5-a) &= 6 \\ 5a-a^2 &= 6 \\ a^2-5a+6 &= 6 \\ (a-2)(a-3) &= 6 \\ a=2 &\text{ atau } a=3 \\ a=2, b=3 \text{ dan } a=3,b=2 \\ 8+x &= a^3 \\ &= 2^3 \\ &= 8 \\ x &= 0 \\ 8+x &= a^3 \\ &= 3^3 \\ &= 27 \\ x &= 19 \\ \end{align} </math> </div></div> <ol start=15> <li>Berapa nilai x dari <math>3^x+5^x-9^x+15^x-25^x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3^x+5^x-9^x+15^x-25^x &= 1 \\ 3^x+5^x-(3^2)^x+(3 \cdot 5)^x-(5^2)^x &= 1 \\ 3^x+5^x-(3^x)^2+(3^x \cdot 5^x)-(5^x)^2 &= 1 \\ \text{misalkan } 3^x=a \text{ dan } 5^x=b \\ a+b-a^2+ab-b^2 &= 1 \\ a^2-ab+b^2-a-b+1 &= 0 \\ 2a^2-2ab+2b^2-2a-2b+2 &= 0 \\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1 &= 0 \\ (a-b)^2+(a-1)^2+(b-1)^2 &= 0 \\ a-b=0; a-1=0; b-1 &= 0 \\ a=b &= 1 \\ 3^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> <ol start=16> <li>Berapa nilai x dari <math>^6log x^2+^{6x}log \frac{6}{x}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ \text{misalkan } 6x=a \text{ maka } x=\frac{a}{6} \\ ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ ^6log (\frac{a}{6})^2+^{6 \frac{a}{6}}log \frac{6}{\frac{a}{6}} &= 1 \\ ^6log \frac{a^2}{6^2}+^alog \frac{6^2}{a} &= 1 \\ ^6log a^2-^6log 6^2+^alog 6^2-^alog a &= 1 \\ 2 ^6log a-2 ^6log 6+2 ^alog 6-^alog a &= 1 \\ 2 ^6log a-2+2 \frac{1}{^6log a}-1 &= 1 \\ 2 ^6log a+2 \frac{1}{^6log a}-4 &= 0 \\ 2 ^6log^2 a-4 ^6log a+2 &= 0 \\ ^6log^2 a-2 ^6log a+1 &= 0 \\ (^6log a-1)^2 &= 0 \\ ^6log a &= 1 \\ a &= 6 \\ x &= \frac{a}{6} \\ &= \frac{6}{6} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=17> <li>Berapa nilai x dari (x+500)³+x=20?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+500)^3+x &= 20 \\ \text{misalkan } a=x+500 \text{ maka } x=a-500 \\ a^3+a-500 &= 20 \\ a^3+a &= 520 \\ a(a^2+1) &= 8 \cdot 65 \\ a(a^2+1) &= 8(64+1) \\ a(a^2+1) &= 8(8^2+1) \\ a &= 8 \\ x &= 8-500 \\ &= -492 \\ \end{align} </math> </div></div> <ol start=18> <li>Berapa nilai x dari <math>\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2} &= 0 \\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}} &= \frac{1}{2} \\ \frac{x^n+4^n}{x^n+16^n} &= (\frac{1}{2})^n \\ \frac{x^n+4^n}{x^n+16^n} &= \frac{1}{2^n} \\ 2^n(x^n+4^n) &= x^n+16^n \\ 2^n(x^n+2^{2n}) &= x^n+2^{4n} \\ 2^n \cdot x^n+2^{3n} &= x^n+2^{4n} \\ 2^n \cdot x^n-x^n &= 2^{4n}-2^{3n} \\ x^n(2^n-1) &= 2^{3n}(2^n-1) \\ x^n &= 2^{3n} \\ x^n &= (2^3)^n \\ x^n &= 8^n \\ x &= 8 \\ \end{align} </math> </div></div> <ol start=19> <li>Berapa hasil dari <math>\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ x &= \frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5 \cdot 6}+\sqrt{5 \cdot 5}+\sqrt{6 \cdot 4}+\sqrt{5 \cdot 4}}{\sqrt{5 \cdot 4}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5} \cdot \sqrt{6}+\sqrt{5} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{4}}{2 \cdot \sqrt{5}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{6} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{5}+\sqrt{5} \cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}} \\ &= \frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ &= \frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ \frac{1}{x} &= \frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}} \\ &= \frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ &= \frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1} \\ &= \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5} \\ &= \sqrt{6}-\sqrt{4} \\ &= \sqrt{6}-2 \\ x &= \frac{1}{\sqrt{6}-2} \\ &= \frac{\sqrt{6}+2}{6-4} \\ &= \frac{\sqrt{6}+2}{2} \\ &= 1+\frac{\sqrt{6}}{2} \\ \end{align} </math> </div></div> <ol start=20> <li>Berapa hasil dari <math>(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5 \\ \text{misalkan } x=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ x &= \frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ &= \frac{\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}} \\ &= \frac{\sqrt{3}+1}{4} \\ 4x &= \sqrt{3}+1 \\ 4x-1 &= \sqrt{3} \\ (4x-1)^2 &= 3 \\ 16x^2-8x+1 &= 3 \\ 16x^2 &= 8x+2 \\ 8x^2 &= 4x+1 \\ x^2 &= \frac{4x+1}{8} \\ *cara 1 \\ x^3 &= x \cdot x^2 \\ &= x(\frac{4x+1}{8}) \\ &= \frac{4x^2+x}{8} \\ &= \frac{4x^2}{8}+\frac{x}{8} \\ &= \frac{4(\frac{4x+1}{8})}{8}+\frac{x}{8} \\ &= \frac{16x+4}{64}+\frac{x}{8} \\ &= \frac{4x+1}{16}+\frac{x}{8} \\ &= \frac{4x+1+2x}{16} \\ &= \frac{6x+1}{16} \\ x^5 &= x^2 \cdot x^3 \\ &= (\frac{4x+1}{8})(\frac{6x+1}{16}) \\ &= \frac{24x^2+10x+1}{128} \\ &= \frac{24x^2}{128}+\frac{10x+1}{128} \\ &= \frac{24(\frac{4x+1}{8})}{128}+\frac{10x+1}{128} \\ &= \frac{96x+24}{1024}+\frac{10x+1}{128} \\ &= \frac{96x+24+80x+8}{1024} \\ &= \frac{176x+32}{1024} \\ &= \frac{176x}{1024}+\frac{32}{1024} \\ &= \frac{176}{1024}(\frac{\sqrt{3}+1}{4})+\frac{32}{1024} \\ &= \frac{44(\sqrt{3}+1)}{1024}+\frac{32}{1024} \\ &= \frac{44\sqrt{3}+44}{1024}+\frac{32}{1024} \\ &= \frac{76+44\sqrt{3}}{1024} \\ &= \frac{19+11\sqrt{3}}{256} \\ *cara 2 \\ x^4 &= (x^2)^2 \\ &= (\frac{4x+1}{8})^2 \\ &= \frac{16x^2+8x+1}{64} \\ &= \frac{16x^2}{64}+\frac{8x}{64}+\frac{1}{64} \\ &= \frac{x^2}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{\frac{4x+1}{8}}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{4x}{32}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{8}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{4}+\frac{3}{64} \\ x^5 &= x \cdot x^4 \\ &= (\frac{\sqrt{3}+1}{4})(\frac{x}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\frac{\sqrt{3}+1}{4}}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\sqrt{3}+1}{16}+\frac{3}{64}) \\ &= \frac{(\sqrt{3}+1)^2}{64}+(\frac{\sqrt{3}+1}{4})\frac{3}{64} \\ &= \frac{3+2\sqrt{3}+1}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{4+2\sqrt{3}}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{16+8\sqrt{3}}{256}+\frac{3\sqrt{3}+3}{256} \\ &= \frac{19+11\sqrt{3}}{256} \\ \end{align} </math> </div></div> <ol start=21> <li>Berapa hasil dari <math>\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4} &= \frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots) \\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots &= \frac{1}{1-\frac{1}{4}} \\ &= \frac{4}{3} \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{4}{3}) \\ &= \frac{1}{4}+\frac{16}{3} \\ &= \frac{67}{12} \\ x &= \frac{67}{9} \\ \end{align} </math> </div></div> <ol start=22> <li>Berapa nilai y-x jika <math>\frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} = \frac{x}{y}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} &= \frac{x}{y} \\ \frac{\frac{106 \times 107}{2}}{\frac{106}{2}(4+109)} &= \frac{x}{y} \\ \frac{53 \times 107}{53 \times 113} &= \frac{x}{y} \\ y-x &= 113-107 = 6 \\ \end{align} </math> </div></div> <ol start=23> <li>Berapa angka satuan dari hasil 17²⁰²⁴?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan angka satuannya} \\ 17^1 &= 7 \\ 17^2 &= 9 \\ 17^3 &= 3 \\ 17^4 &= 1 \\ 17^5 &= 7 \\ 17^6 &= 9 \\ 17^7 &= 3 \\ 17^8 &= 1 \\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1} \end{align} </math> </div></div> <ol start=24> <li>Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ 1! + 2! + 3! + 4! + \dots + 2024! &= 1 + (1x2) + (1x2x3) + (1x2x3x4) + \dots + 2024! \\ &= 1 + 2 + 6 + 24 + 120 + 720 + \dots + 2024! \\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka } 1+2+6+24 = 33 \text{ jadi angka satuannya adalah } 3 \end{align} </math> </div></div> <ol start=25> <li>Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ \frac{1! + 2! + 3! + 4! + \dots + 2024!}{12} &= \frac{1 + 1x2 + 1x2x3 + 1x2x3x4 + \dots + 2024!}{12} \\ &= \frac{1 + 2 + 6 + 24 + \dots + 2024!}{12} \\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi } 1+2+6 = 9 \end{align} </math> </div></div> <ol start=26> <li>Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{penjumlahan} \\ 1+1+1+1+ \dots \text{ (sebanyak 88 buah) }+x+y &= A \\ 88+x+y &= A \\ \text{perkalian} \\ 1 \times 1 \times 1 \times \dots \text{ (sebanyak 88 buah) }\times x \times y &= A \\ x \times y &= A \\ 88+x+y &= xy \\ xy-y &= 88+x \\ y(x-1) &= 88+x \\ y &= \frac{88+x}{x-1} \\ \text{uji selidiki untuk x=2} \\ y &= \frac{88+2}{2-1} \\ &= 90 \\ \text{buktikan} \\ 88+x+y &= xy \\ 88+2+90 &= 2(90) \\ 180 &= 180 \\ \text{terbukti} \\ \text{nilai A adalah } 180 \\ \end{align} </math> </div></div> <ol start=27> <li>Berapakah nilai x, y dan z dari <math>x+y-z=1, x^2+y^2-z^2=-5 \text{ dan } x^3+y^3-z^3=-53</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+y-z &= 1 \\ x+y &= z+1 \\ x^2+2xy+y^2 &= z^2+2z+1 \\ x^2+y^2-z^2 &= 2z+1-2xy \\ -5 &= 2z+1-2xy \\ 2xy &= 2z+6 \\ xy &= z+3 \\ x^2+y^2-z^2 &= -5 \\ x^2+y^2 &= z^2-5 \\ x^3+y^3-z^3 &= -53 \\ (x+y)(x^2-xy+y^2)-z^3+53 &= 0 \\ (x+y)(x^2+y^2-xy)-z^3+53 &= 0 \\ (z+1)(z^2-5-(z+3))-z^3+53 &= 0 \\ (z+1)(z^2-z-8)-z^3+53 &= 0 \\ z^3-z^2-8z+z^2-z-8-z^3+53 &= 0 \\ -9z+45 &= 0 \\ -9z &= -45 \\ z &= 5 \\ x+y &= 5+1 \\ x+y &= 6 \\ x &= 6-y \\ xy &= 5+3 \\ xy &= 8 \\ (6-y)y &= 8 \\ 6y-y^2 &= 8 \\ y^2-6y+8 &= 0 \\ (y-4)(y-2) &= 0 \\ y=4 \text{ atau } y=2 \\ \text{jika } y=4 \\ x+y &= z+1 \\ x+4 &= 5+1 \\ x &= 2 \\ \text{jika } y=2 \\ x+y &= z+1 \\ x+2 &= 5+1 \\ x &= 4 \\ \end{align} </math> </div></div> <ol start=28> <li>Berapakah nilai titik koordinat (x,y) dari <math>\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}</math> dan <math>\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x+y}+\sqrt{x-y} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+y}-\sqrt{x-y} &= \sqrt{\frac{52y}{3x}} \\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y}) &= \sqrt{\frac{432x}{13y}} \cdot \sqrt{\frac{52y}{3x}} \\ x+y-x+y &= \sqrt{\frac{432x \cdot 52y}{13y \cdot 3x}} \\ 2y &= \sqrt{144 \cdot 4} \\ 2y &= \sqrt{576} \\ 2y &= 24 \\ y &= 12 \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13(12)}} \\ x+12+x-12+2 \cdot \sqrt{x+12} \cdot \sqrt{x-12} &= \frac{36x}{13} \\ 2x+2 \sqrt{x^2-144} &= \frac{36x}{13} \\ 2(x+\sqrt{x^2-144}) &= \frac{36x}{13} \\ x+\sqrt{x^2-144} &= \frac{18x}{13} \\ \sqrt{x^2-144} &= \frac{5x}{13} \\ x^2-144 &= \frac{25x^2}{169} \\ \frac{144x^2}{169}-144 &= 0 \\ \frac{x^2}{169}-1 &= 0 \\ x^2-169 &= 0 \\ (x-13)(x+13) &= 0 \\ x_1=13 &\text{ atau } x_2=-13 \text{ (TM) karena } x>y \\ \end{align} </math> jadi titik koordinat (13,12) </div></div> <ol start=29> <li>Berapakah nilai dari <math>x^2-7x</math> jika <math>(x-2)^2+\frac{1}{(x-2)^2} = 11</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x-2)^2+\frac{1}{(x-2)^2} &= 11 \\ (x-2)^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2)^2} &= 11-2 \\ (x-2-\frac{1}{x-2})^2 &= 9 \\ x-2-\frac{1}{x-2} &= 3 \\ (x-2)^2-1 &= 3(x-2) \\ x^2-4x+4-1 &= 3x-6 \\ x^2-7x &= -9 \\ \end{align} </math> </div></div> <ol start=30> <li>Berapakah nilai dari <math>\frac{(x+y)^2(x+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)}</math> jika xy+yz+xz=1?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy+yz+xz &= 1 \\ x^2+xy+yz+xz &= x^2+1 \\ x(x+y)+z(x+y) &= x^2+1 \\ (x+y)(x+z) &= x^2+1 \\ \text{dengan pola yang sama } \\ (y+x)(y+z) &= y^2+1 \\ (x+z)(y+z) &= z^2+1 \\ \frac{(x+y)^2(y+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)} &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)} \\ &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)^2(y+z)^2(x+z)^2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=31> <li>Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} w+5 &= w+x+y+z+5 \\ x+4 &= w+x+y+z+5 \\ y+3 &= w+x+y+z+5 \\ z+2 &= w+x+y+z+5 \\ \text{jumlahkan keempat persamaan } \\ w+x+y+z+14 &= 4(w+x+y+z+5) \\ w+x+y+z+14 &= 4(w+x+y+z)+20 \\ 3(w+x+y+z) &= -6 \\ w+x+y+z &= -2 \\ \end{align} </math> </div></div> <ol start=32> <li>Berapakah nilai dari <math>\frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2}</math> jika <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3</math> dan x+y+z=xyz?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2} &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2} \\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) \\ 3^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz}) \\ 9 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz}) \\ &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2 \\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} &= 7 \\ \end{align} </math> </div></div> <ol start=33> <li>Berapakah nilai dari <math>\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}</math> jika <math>x^2+y^2+z^2 = -2(ab+bc+ac)</math>?>/li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+z^2 &= -2(xy+yz+xz) \\ x^2+y^2+z^2+2(xy+yz+xz) &= 0 \\ (x+y+z)^2 &= 0 \\ x+y+z &= 0 \\ x+y &= -z \\ x+z &= -y \\ y+z &= -x \\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z} &= \frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x} \\ &= -2-(-5)-(-7) \\ &= 10 \\ \end{align} </math> </div></div> <ol start=34> <li>Berapakah nilai dari <math>\frac{20xyz}{xy+yz+xz}</math> jika <math>16^x = 256^y = 625^z = 40</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 16^x = 256^y = 625^z &= 40 \\ 2^{4x} = 4^{4y} = 5^{4z} &= 40 \\ 2^{4x} &= 40 \\ 2 &= 40^{\frac{1}{4x}} \\ 4^{4y} &= 40 \\ 4 &= 40^{\frac{1}{4y}} \\ 5^{4z} &= 40 \\ 5 &= 40^{\frac{1}{4z}} \\ 2 \cdot 4 \cdot 5 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z}} \\ 1 &= \frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z} \\ 4 &= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{20xyz}{xy+yz+xz} &= 20 \cdot \frac{xyz}{xy+yz+xz} \\ &= 20 \cdot (\frac{xy+yz+xz}{xyz})^{-1} \\ &= 20 \cdot (\frac{1}{z} + \frac{1}{x} + \frac{1}{y})^{-1} \\ &= 20 \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{-1} \\ &= 20 \cdot (4)^{-1} \\ &= 20 \cdot \frac{1}{4} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=35> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+3x^2+1}</math> jika <math>6x^2+25x+6=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 6x^2+25x+6 &= 0 \\ 6x+25+\frac{6}{x} &= 0 \\ 6(x+\frac{1}{x}) &= -25 \\ x+\frac{1}{x} &= \frac{-25}{6} \\ (c+\frac{1}{x})^2 &= (\frac{-25}{6})^2 \\ x^2+2+\frac{1}{x^2} &= \frac{625}{36} \\ x^2+\frac{1}{x^2} &= \frac{625}{36}-2 \\ x^2+\frac{1}{x^2} &= \frac{553}{36} \\ \frac{x^2}{x^4+3x^2+1} &= \frac{1}{x^2+3+\frac{1}{x^2}} \\ &= \frac{1}{a^2+\frac{1}{x^2}+3} \\ &= \frac{1}{\frac{553}{36}+3} \\ &= \frac{1}{\frac{661}{36}} \\ &= \frac{36}{661} \\ \end{align} </math> </div></div> <ol start=36> <li>Berapakah nilai dari <math>\frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5} &= \frac{(9+2\sqrt{20})^{1013}}{((2)^3+3(2)^2(\sqrt{5})+3(2)(\sqrt{5})^2+(\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{((2+\sqrt{5})^2)^{1013}}{((2+\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{(2+\sqrt{5})^{2026}}{(2+\sqrt{5})^{2025}}+6-\sqrt{5} \\ &= 2+\sqrt{5}+6-\sqrt{5} \\ &= 8 \\ \end{align} </math> </div></div> <ol start=37> <li>Berapakah nilai dari <math>27x^3+\frac{8}{x^3}</math> jika <math>3x+\frac{2}{x}=6</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3x+\frac{2}{x} &= 6 \\ (3x+\frac{2}{x})^3 &= 6^3 \\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3} &= 216 \\ 27x^3+18(6)+\frac{8}{x^3} &= 216 \\ 27x^3+108+\frac{8}{x^3} &= 216 \\ 27x^3+\frac{8}{x^3} &= 108 \\ \end{align} </math> </div></div> <ol start=38> <li>Berapakah nilai dari <math>x^6+\frac{8}{x^3}</math> jika <math>x^3+\frac{1}{x^3}=8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3+\frac{1}{x^3} &= 8 \\ x^3 &= 8-\frac{1}{x^3} \\ x^6 &= 8x^3-1 \\ x^6+\frac{8}{x^3} &= 8x^3-1+\frac{8}{x^3} \\ &= 8x^3+\frac{8}{x^3}-1 \\ &= 8(x^3+\frac{1}{x^3})-1 \\ &= 8(8)-1 \\ &= 63 \\ \end{align} </math> </div></div> <ol start=39> <li>Berapakah nilai dari <math>4x+\frac{25}{x}</math> jika <math>2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2\sqrt{x}+\frac{5}{\sqrt{x}} &= 4x-\frac{25}{x} \\ 2\sqrt{x}+\frac{5}{\sqrt{x}} &= (2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}}) \\ 1 &= 2\sqrt{x}-\frac{5}{\sqrt{x}} \\ 1^2 &= (2\sqrt{x}-\frac{5}{\sqrt{x}})^2 \\ 1 &= 4x-20+\frac{25}{x} \\ 4x+\frac{25}{x} &= 21 \\ \end{align} </math> </div></div> <ol start=40> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2-x+1}{x^2+x+1} &= \frac{5}{6} \\ \frac{x^2+1-x}{x^2+1+x} &= \frac{5}{6} \\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} &= \frac{5}{6} \\ \text{ misalkan } x+\frac{1}{x} &= y \\ \frac{y-1}{y+1} &= \frac{5}{6} \\ 6(y-1) &= 5(y+1) \\ 6y-6 &= 5y+5 \\ y &= 11 \\ x+\frac{1}{x} &= 11 \\ \end{align} </math> </div></div> <ol start=41> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt{x}+x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x}+x &= 1 \\ x-1 &= -\sqrt{x} \\ (x-1)^2 &= (-\sqrt{x})^2 \\ x^2-2x+1 &= x \\ x^2-3x+1 &= 0 \\ x-3+\frac{1}{x} &= 0 \\ x+\frac{1}{x} &= 3 \\ \end{align} </math> </div></div> <ol start=42> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt[3]{x}-\sqrt[3]{x-36}=3</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x}-\sqrt[3]{x-36} &= 3 \\ (\sqrt[3]{x}-\sqrt[3]{x-36})^3 &= 3^3 \\ x-(x-36)-3 \sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36}) &= 27 \\ 36-3 \sqrt[3]{x(x-36)}3 &= 27 \\ -9 \sqrt[3]{x(x-36)} &= -9 \\ \sqrt[3]{x(x-36)} &= 1 \\ x(x-36) &= 1 \\ x^2-36x-1 &= 0 \\ x-36-\frac{1}{x} &= 0 \\ x-\frac{1}{x} &= 36 \\ \end{align} </math> </div></div> <ol start=43> <li>Berapakah nilai dari <math>x+\frac{16}{x}</math> jika <math>x-3\sqrt{x}=4</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x-3\sqrt{x} &= 4 \\ x-4 &= 3\sqrt{x} \\ x^2-8x+16 &= 9x \\ x^2-17x+16 &= 0 \\ x-17+\frac{16}{x} &= 0 \\ x+\frac{16}{x} &= 17 \\ \end{align} </math> </div></div> <ol start=44> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+4}</math> jika <math>x^2-7x+2=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2-7x+2 &= 0 \\ x^2+2 &= 7x \\ x+\frac{2}{x} &= 7 \\ x^2+4+\frac{4}{x^2} &= 49 \\ x^2+\frac{4}{x^2} &= 45 \\ \frac{x^4+4}{x^2} &= 45 \\ \frac{x^2}{x^4+4} &= \frac{1}{45} \\ \end{align} </math> </div></div> <ol start=45> <li>Berapakah nilai dari <math>x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}</math> jika <math>x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}} &= 25 \\ x^{\frac{1}{2}}+x^{-\frac{1}{2}} &= 23 \\ x+2+x^{-1} &= 529 \\ x+x^{-1} &= 527 \\ x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+x^{-\frac{3}{4}} &= 110 \\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1} &= x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}} \\ &= 527+110 \\ &= 637 \\ \end{align} </math> </div></div> <ol start=46> <li>Berapakah nilai dari <math>\sqrt{8x^6+x^5+x^4+5x^3+1}</math> jika <math>\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5} &= 0 \\ \frac{x^2+x+1}{x^5} &= 0 \\ x^2+x+1 &= 0 \\ x^2+x+1 &= 0 \\ (x-1)(x^2+x+1) &= 0(x-1) \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ \sqrt{8x^6+x^5+x^4+5x^3+1} &= \sqrt{(2x^3)^2+x^3x^2+x^3x+5x^3+1} \\ &= \sqrt{(2(1))^2+(1)x^2+(1)x+5(1)+1} \\ &= \sqrt{(2)^2+x^2+x+5+1} \\ &= \sqrt{4+x^2+x+1+5} \\ &= \sqrt{4+0+5} \\ &= \sqrt{9} \\ &= 3 \\ \end{align} </math> </div></div> <ol start=47> <li>Berapakah nilai dari <math>f(1)+f(2)+f(3)+ \dots + f(99)</math> jika <math>f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{1}{\sqrt{x+1}+\sqrt{x}} \\ &= \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} \\ &= \sqrt{x+1}-\sqrt{x} \\ f(1)+f(2)+f(3)+ \dots + f(98)+f(99) &= \sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+ \cdot + \sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99} \\ &= \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+ \cdot + \sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ &= \sqrt{100}-\sqrt{1} \\ &= 10-1 \\ &= 9 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots + \frac{2024}{2025})</math> jika <math>h(x)=\frac{3}{3+9^x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} h(x) &= \frac{3}{3+9^x} \\ h(1-x) &= \frac{3}{3+9^{1-x}} \\ &= \frac{3}{3+\frac{9}{9^x}} \\ &= \frac{9^x}{3+9^x} \\ h(x)+h(1-x) &= \frac{3}{3+9^x}+\frac{9^x}{3+9^x} \\ &= \frac{3+9^x}{3+9^x} \\ &= 1 \\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots +(1-\frac{2}{2025})+(1-\frac{1}{2025})) \\ & 5(1+1+1+ \dots +1+1) \text{ sebanyak 1012 kali } \\ & 5(1012) \\ & 5060 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>\frac{7^{2025} - 7^{2023} + 432}{7^{2024} + 7^{2023} + 72}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72} &= \frac{7^{2023}7^{2}-7^{2023} + 48 \times 9}{7^{2023}7^1+7^{2023}+8 \times 9} \\ &= \frac{7^{2023}(7^{2}-1)+48 \times 9}{7^{2023}(7^1+1)+8 \times 9} \\ &= \frac{7^{2023}(49-1)+48 \times 9}{7^{2023}(7+1) + 8 \times 9} \\ &= \frac{7^{2023} \times 48+48 \times 9}{7^{2023} \times 8+8 \times 9} \\ &= \frac{48(7^{2023}+9)}{8(7^{2023}+9)} \\ &= \frac{48}{8} \\ &= 6 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>tan (x+\frac{\pi}{4})</math> jika <math>\frac{1}{cos x}-tan x = \frac{4}{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{cos x}-tan x &= \frac{4}{5} \\ sec x-tan x &= \frac{4}{5} \\ sec^2 x-tan^2 x &= 1 \\ (sec x+tan x)(sec x-tan x) &= 1 \\ (sec x+tan x)\frac{4}{5} &= 1 \\ sec x+tan x &= \frac{5}{4} \\ \text{kedua persamaan dengan cara metode eliminasi } \\ 2 tan x &= \frac{5}{4}-\frac{4}{5} \\ 2 tan x &= \frac{9}{20} \\ tan x &= \frac{9}{40} \\ tan (x+\frac{\pi}{4}) &= \frac{tan x+tan \frac{\pi}{4}}{1-tan x \cdot tan \frac{\pi}{4}} \\ &= \frac{\frac{9}{40}+1}{1-\frac{9}{40} \cdot 1} \\ &= \frac{\frac{49}{40}}{\frac{31}{40}} \\ &= \frac{49}{31} \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>sin^3 x+csc^3 x</math> jika <math>sin x-csc x = 8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ Dengan menggunakan rumus: } (a-b)^3 &= a^3-b^3-3ab(a-b) \\ (sin x-csc x)^3 &= sin^3 x-csc^3 x-3sin x csc x(sin x-csc x) \\ 8^3 &= sin^3 x-csc^3 x-3sin x (\frac{1}{sin x})(8) \\ 512 &= sin^3 x-csc^3 x-24 \\ sin^3 x-csc^3 x &= 512+24 \\ sin^3 x-csc^3 x &= 536 \\ \end{align} </math> </div></div> <ol start=52> <li>Berapakah nilai dari <math>(sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2</math> jika <math>\frac{1}{sin x}+\frac{1}{cos x} = 10</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{sin x}+\frac{1}{cos x} &= 10 \\ \frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} &= 100 \\ (sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2 &= sin^2 x+\frac{2sin x}{cos x}+\frac{1}{cos^2 x}+cos^2 x+\frac{2cos x}{sin x}+\frac{1}{sin^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2(sin^2 x+cos^2 x)}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+100 \\ &= 101 \\ \end{align} </math> </div></div> <ol start=53> <li>Berapakah nilai dari (x-1)⁶ jika <math>x=\frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin 80^\circ &= cos 10^\circ \\ sin 80^\circ-cos 10^\circ &= 0 \\ x &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+2 sin 20^\circ cos 20^\circ}{cos 10^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+4 sin 10^\circ cos 10^\circ cos 20^\circ}{cos 10^\circ} \\ &= 4 cos 55^\circ cos 25^\circ+4 sin 10^\circ cos 20^\circ \\ &= 2(2 cos 55^\circ cos 25^\circ+2 sin 10^\circ cos 20^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ+sin (-10)^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ-sin 10^\circ) \\ &= 2(cos 80^\circ-sin 10^\circ+cos 30^\circ+sin 30^\circ) \\ &= 2(cos 80^\circ-sin (90^\circ-80^\circ)+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(cos 80^\circ-cos 80^\circ+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= \sqrt{3}+1 \\ x-1 &= \sqrt{3} \\ (x-1)^6 &= (\sqrt{3})^6 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=54> <li>Berapakah nilai dari x jika <math>x=\frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ} \\ 2x sin 20^\circ-x sin 40 ^\circ &= x sin 20^\circ-x^2 sin 10^\circ \\ x^2 sin 10^\circ+x sin 20^\circ-x sin 40 ^\circ &= 0 \\ x(x sin 10^\circ+sin 20^\circ-sin 40 ^\circ) &= 0 \\ x = 0 &\text{ atau } x sin 10^\circ+sin 20^\circ-sin 40 ^\circ = 0 \\ x sin 10^\circ+sin 20^\circ-sin 40 ^\circ &= 0 \\ x sin 10^\circ &= sin 40 ^\circ-sin 20^\circ \\ x &= \frac{sin 40 ^\circ-sin 20^\circ}{sin 10^\circ} \\ &= \frac{2 cos 30 ^\circ sin 10^\circ}{sin 10^\circ} \\ &= 2 cos 30 ^\circ \\ &= \frac{2 \sqrt{3}}{2} \\ &= \sqrt{3} \\ \end{align} </math> </div></div> <ol start=55> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x^2}{x^2-16y^2} = \frac{625}{49}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2}{x^2-16y^2} &= \frac{625}{49} \\ \frac{x^2-16y^2}{x^2} &= \frac{49}{625} \text{ (terbalik posisinya)} \\ 1-\frac{16y^2}{x^2} &= \frac{49}{625} \\ \frac{16y^2}{x^2} &= 1 - \frac{49}{625} \\ (\frac{4y}{x})^2 &= \frac{576}{625} \\ (\frac{4y}{x})^2 &= (\frac{24}{25})^2 \\ \frac{4y}{x} &= \frac{24}{25} \\ \frac{y}{x} &= \frac{6}{25} \\ \frac{x}{y} &= \frac{25}{6} \\ \end{align} </math> </div></div> <ol start=56> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x}{y}+\frac{x+10y}{y+10x} = 2</math> serta bilangan real untuk x dan y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x}{y}+\frac{x+10y}{y+10x} &= 2 \\ \frac{x}{y}+\frac{\frac{x}{y}+10}{1+10\frac{x}{y}} &= 2 \\ \text{misalkan } \frac{x}{y} = a \\ a+\frac{a+10}{1+10a} &= 2 \\ a(1+10a)+a+10 &= 2(1+10a) \\ 10a^2+a+a+10 &= 2+20a \\ 10a^2-18a+8 &= 0 \\ 5a^2-9a+4 &= 0 \\ (5a-4)(a-1) &= 0 \\ a = \frac{4}{5} &\text{ atau } a = 1 \\ \text{jadi } \frac{x}{y} = {\frac{4}{5}, 1} \\ \end{align} </math> </div></div> <ol start=57> <li>Berapakah nilai dari xy jika <math>x^4+y^4+x^2y^2=15 \text{ dan } x^2+y^2+xy=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+xy &= 5 \\ x^2+y^2 &= 5-xy \\ x^4+y^4+x^2y^2 &= 15 \\ (x^2)^2+(y^2)^2+2x^2y^2-x^2y^2 &= 15 \\ (x^2+y^2)^2-x^2y^2 &= 15 \\ (5-xy)^2-x^2y^2 &= 15 \\ 25-10xy+x^2y^2-x^2y^2 &= 15 \\ 25-10xy &= 15 \\ 10xy &= 10 \\ xy &= 1 \\ \end{align} </math> </div></div> <ol start=58> <li>Berapakah nilai dari x jika <math>4^x = 63(4^3+1)(4^6+1)(4^{12}+1)+1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 4^x &= 63(4^3+1)(4^6+1)(4^{12}+1)+1 \\ 4^x-1 &= 63(4^3+1)(4^6+1)(4^{12}+1) \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{4^3-1} \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{63} \\ &= (4^3+1)(4^6+1)(4^{12}+1)(4^3-1) \\ &= (4^3-1)(4^3+1)(4^6+1)(4^{12}+1) \\ &= (4^6-1)(4^6+1)(4^{12}+1) \\ &= (4^{12}-1)(4^{12}+1) \\ &= 4^{24}-1 \\ 4^x &= 4^{24} \\ x &= 24 \\ \end{align} </math> </div></div> <ol start=59> <li>Berapakah nilai dari <math>\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}</math> jika <math>x=\sqrt{9+4\sqrt{5}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \sqrt{9+4\sqrt{5}} \\ x &= 2+\sqrt{5} \\ x^2 &= 9+4\sqrt{5} \\ x^2-4x &= 9+4\sqrt{5}-4(2+\sqrt{5}) \\ x^2-4x &= 1 \\ x^2 &= 4x+1 \\ x^3 &= x \cdot x^2 \\ &= x(4x+1) \\ &= 4x^2+x \\ &= 4(4x+1)+x \\ &= 16x+4+x \\ &= 17x+4 \\ x^4 &= x \cdot x^3 \\ &= x(17x+4) \\ &= 17x^2+4x \\ &= 17(4x+1)+4x \\ &= 68x+17+4x \\ &= 72x+17 \\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1} &= \frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1} \\ &= \frac{72x+17-85x-20+8x+2+5x+3}{2} \\ &= \frac{2}{2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=60> <li>Berapakah nilai dari <math>\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}</math> jika 2x-1=<math>\sqrt{61}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \frac{x^3+1}{x^5-x^4-x^3+x^2} = p \\ p &= \frac{x^3+1}{x^5-x^4-x^3+x^2} \\ &= \frac{x^3+1}{x^5-x^4-(x^3-x^2)} \\ &= \frac{x^3+1}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)} \\ &= \frac{x^2-x+1}{x^2(x-1)^2} \\ &= \frac{x^2-x+1}{(x(x-1))^2} \\ &= \frac{x(x-1)+1}{(x(x-1))^2} \\ 2x-1 &= \sqrt{61} \\ x &= \frac{\sqrt{61}+1}{2} \\ x-1 &= \frac{\sqrt{61}-1}{2} \\ x(x-1) &= (\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2}) \\ &= \frac{61-1}{4} \\ &= \frac{60}{4} \\ &= 15 \\ p &= \frac{x(x-1)+1}{(x(x-1))^2} \\ &= \frac{15+1}{15^2} \\ &= \frac{16}{15^2} \\ \sqrt{p} &= \sqrt{\frac{16}{15^2}} \\ &= \frac{4}{15} \\ \end{align} </math> </div></div> <ol start=61> <li>Berapakah nilai dari <math>(\frac{x-3}{x})^{25}</math> jika <math>x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\sqrt[5]{8}+\sqrt[5]{2} &= 1+\sqrt[5]{16}+\sqrt[5]{4} \\ x+(\sqrt[5]{2})^3+\sqrt[5]{2} &= 1+(\sqrt[5]{2})^4+(\sqrt[5]{2})^2 \\ x &= (\sqrt[5]{2})^4-(\sqrt[5]{2})^3+(\sqrt[5]{2})^2-\sqrt[5]{2}+1 \\ \text{misalkan } \sqrt[5]{2} = p \\ x &= p^4-p^3+p^2-p+1 \\ x &= \frac{p^5+1}{p+1} \\ (\frac{x-3}{x})^{25} &= (1-\frac{3}{x})^{25} \\ &= (1-\frac{3}{\frac{p^5+1}{p+1}})^{25} \\ &= (1-\frac{3(p+1)}{p^5+1})^{25} \\ &= (1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2})^5+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{2+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-3\sqrt[5]{2}-3)}{3})^{25} \\ &= (-\sqrt[5]{2})^{25} \\ &= (-2)^5 \\ &= -32 \\ \end{align} </math> </div></div> <ol start=62> <li>Berapakah nilai dari <math>x^{50}+x^{49}+x^{48}+x^{47}+x^{46}</math> jika <math>x^2+x+1=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+x+1 &= 0 \\ x^2+x &= -1 \\ \frac{x^3-1}{x-1} &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46} &= x^{48}(x^2+x+1)+x^{45}(x^2+x) \\ &= x^{48}(0)+(x^3)^{15}(-1) \\ &= 0+(1)^{15}(-1) \\ &= -1 \\ \end{align} </math> </div></div> <ol start=63> <li>Berapakah 2²⁴ dari <math>8^7+8^6+8^5+8^4+8^3+8^2+8+1=A</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= A \\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1) &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8 &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= 8A+1 \\ 8^8+A &= 8A+1 \\ 8^8 &= 7A+1 \\ (2^3)^8 &= 7A+1 \\ 2^{24} &= 7A+1 \\ \end{align} </math> </div></div> <ol start=64> <li>Berapakah nilai dari <math>x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1</math> jika <math>x+\frac{1}{x}=\sqrt{3}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{x} &= \sqrt{3} \\ x^2+2+\frac{1}{x^2} &= 3 \\ x^2-1+\frac{1}{x^2} &= 0 \\ x^2(x^2-1+\frac{1}{x^2}) &= x^2(0) \\ x^4-x^2+1 &= 0 \\ (x^2+1)(x^4-x^2+1) &= (x^2+1)0 \\ x^6-x^4+x^2+x^4-x^2+1 &= 0 \\ x^6+1 &= 0 \\ x^6 &= -1 \\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1 &= {x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1 \\ &= (-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2-1+1 \\ &= -1+1-1+1-1+1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=65> <li>Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari <math>\frac{c-b}{a-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= ax^2+bx+c \\ f(2) &= a(2)^2+2b+c = 4 \\ &= 4a+2b+c = 4 \\ f(7) &= a(7)^2+7b+c = 49 \\ &= 49a+7b+c = 49 \\ 49a+7b+c &= 49 \\ 4a+2b+c &= 4 \\ 45a+5b &= 45 \text{ (f(7) dikurangi f(2)) } \\ 9a+b &= 9 \\ b &= -9a+9 \\ 4a+2b+c &= 4 \\ 4a+2(-9a+9)+c &= 4 \\ 4a-18a+18+c &= 4 \\ -14a+18+c &= 4 \\ c &= 14a-14 \\ \frac{c-b}{a-1} &= \frac{14a-14-(-9a+9)}{a-1} \\ &= \frac{14(a-1)+9(a-1)}{a-1} \\ &= \frac{(14+9)(a-1)}{a-1} \\ &= 23 \\ \end{align} </math> </div></div> <ol start=66> <li>Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+y)^3 &= x^3+y^3+3xy(x+y) \\ 11^3 &= 242+3xy(11) \text{ (dibagi 11)} \\ 11^2 &= 22+3xy \\ 121 &= 22+3xy \\ 99 &= 3xy \\ xy &= 33 \\ (x-y)^2 &= x^2+y^2-2xy \\ &= ((x+y)^2-2xy)-2xy \\ &= (x+y)^2-4xy \\ &= 11^2-4(33) \\ &= 121-132 \\ &= -11 \\ \end{align} </math> </div></div> <ol start=67> <li>Berapa f(1)+f(-1) jika <math>f(\frac{ax-b}{bx-a})</math>=x²-5x+6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ jika} f(1) = f(\frac{ax-b}{bx-a}) \\ 1 &= \frac{ax-b}{bx-a} \\ bx-a &= ax-b \\ (b-a)x &= -b+a \\ &= -(b-a) \\ &= -1 \\ f(1) &= x^2-5x+6 \\ &= (-1)^2-5(-1)+6 \\ &= 12 \\ \text{ jika} f(-1) = f(\frac{ax-b}{bx-a}) \\ -1 &= \frac{ax-b}{bx-a} \\ -(bx-a) &= ax-b \\ -bx+a &= ax-b \\ (-b-a)x &= -b-a \\ &= 1 \\ f(-1) &= x^2-5x+6 \\ &= (1)^2-5(1)+6 \\ &= 2 \\ f(1)+f(-1) &= 12+2 \\ &= 14 \\ \end{align} </math> </div></div> <ol start=68> <li>berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)-x &= f(x-1) \\ f(x)-f(x-1) &= x \\ x=1 ; f(1)-f(0) &= 1 \\ x=2 ; f(2)-f(1) &= 2 \\ x=3 ; f(3)-f(2) &= 3 \\ x=4 ; f(4)-f(3) &= 4 \\ \dots \\ x=200 ; f(200)-f(199) &= 200 \\ \text{ jumlahkan tersebut menjadi } \\ f(200)-f(0) &= 1+2+3+4+\dots+200 \\ &= \frac{200 \cdot 201}{2} \\ &= 20.100 \\ f(200)-1 &= 20.100 \\ &= 20.101 \\ \end{align} </math> </div></div> <ol start=69> <li>Misalkan f(x) adalah fungsi rekursif yang berlaku ∀x ∈ R sebagai berikut: : f(x)+f(15-x) = 2024 : f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)+f(15-x) &= 2024 \\ f(15+x) &= f(x)+2020 \\ *cara 1 \\ \text{ganti x dengan 15+x } \\ f(15+x)+f(-x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ *cara 2 \\ \text{ganti x dengan -x } \\ f(-x)+f(15+x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ \end{align} </math> </div></div> <ol start=70> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>2f(\frac{2002}{x}) + f(x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2f(\frac{2002}{x}) + f(x) &= 3x \\ \text{ganti x dengan 2 } \\ 2f(\frac{2002}{2}) + f(2) &= 3(2) \\ 2f(1001) + f(2) &= 6 \\ \text{ganti x dengan 1001 } \\ 2f(\frac{2002}{1001}) + f(1001) &= 3(1001) \\ 2f(2) + f(1001) &= 3003 \\ 2f(2) + f(1001) &= 3003 \\ f(1001) &= 3003 - 2f(2) \\ 2f(1001) + f(2) &= 6 \\ 2(3003 - 2f(2)) + f(2) &= 6 \\ 6006 - 4f(2) + f(2) &= 6 \\ 3f(2) &= 6000 \\ f(2) &= 2000 \\ \end{align} </math> </div></div> <ol start=71> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>f(\frac{1}{x}) + \frac{1}{x}f(-x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{1}{x})+\frac{1}{x}f(-x) &= 3x \\ \text{ganti x dengan 1/3 } \\ f(3)+3f(-\frac{1}{3}) &= 1 \\ \text{ganti x dengan -3 } \\ f(-\frac{1}{3}) - \frac{1}{3}f(3) &= -9 \\ \text{dikalikan 3 } \\ 3f(-\frac{1}{3})-f(3) &= -27 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ 2f(3) &= 28 \\ f(3) &= 14 \\ \end{align} </math> </div></div> <ol start=72> <li>Diketahui polinom <math>f(7^b-1)=7^{3b}-10</math>. tentukan nilai f(5)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(5) &= f(7^b-1) \\ 5 &= 7^b-1 \\ 7^b &= 6 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(6-1) &= 6^3-10 \\ f(5) &= 216-10 \\ &= 206 \\ *cara 2 \\ \text{misalkan } 7^b-1=a \text{ maka } 7^b=a+1 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(a) &= (a+1)^3-10 \\ f(5) &= (5+1)^3-10 \\ &= 6^3-10 \\ &= 216-10 \\ &= 206 \\ \end{align} </math> </div></div> <ol start=73> <li>Diketahui polinom <math>f(6^b-7)=6^{3b}-2 \cdot 6^{2b}-4</math>. tentukan nilai f(-2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(-2) &= f(6^b-7) \\ -2 &= 6^b-7 \\ 6^b &= 5 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(5-7) &= 5^3-2 \cdot 5^2-4 \\ f(-2) &= 125-50-4 \\ &= 71 \\ *cara 2 \\ \text{misalkan } 6^b-7=a \text{ maka } 6^b=a+7 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(a) &= (a+7)^3-2(a+7)^2-4 \\ f(-2) &= (-2+7)^3-2(-2+7)^2-4 \\ &= 5^3-2(5)^2-4 \\ &= 125-50-4 \\ &= 71 \\ \end{align} </math> </div></div> <ol start=74> <li>Jika <math>f(xy)=\frac{f(x)}{y}</math> dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(10) &= 7 \\ f(2 \cdot 5) &= 7 \\ f(xy) &= \frac{f(x)}{y} \\ f(2 \cdot 5) &= \frac{f(2)}{5} \\ 7 &= \frac{f(2)}{5} \\ f(2) &= 35 \\ \end{align} </math> </div></div> <ol start=75> <li>Jika <math>f(xy)=\frac{f(x+y)}{xy}</math> dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(15) &= 16 \\ f(3 \cdot 5) &= 16 \\ f(xy) &= \frac{f(x+y)}{xy} \\ f(3 \cdot 5) &= \frac{f(3+5)}{3 \cdot 5} \\ f(15) &= \frac{f(8)}{15} \\ 16 &= \frac{f(8)}{15} \\ f(8) &= 240 \\ \end{align} </math> </div></div> <ol start=76> <li>Jika <math>f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15</math> maka tentukan nilai f(16)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x+\frac{1}{x}+6) &= x^2+\frac{1}{x^2}+15 \\ &= (x+\frac{1}{x})^2-2+15 \\ &= (x+\frac{1}{x})^2+13 \\ \text{misalkan } x+\frac{1}{x} &= p \\ f(x+\frac{1}{x}+6) &= (x+\frac{1}{x})^2+13 \\ f(p+6) &= p^2+13 \\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 } \\ f(p+6) &= p^2+13 \\ f(10+6) &= 10^2+13 \\ f(16) &= 100+13 \\ &= 113 \\ \end{align} </math> </div></div> <ol start=77> <li>Tentukan nilai x jika <math>f(x)=\frac{4}{4-x}</math> dan <math>f(x \cdot f(x))^{\frac{f(4x)}{f(x)}}=256</math>!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{4}{4-x} \\ f(4x) &= \frac{4}{4-4x} \\ \frac{f(4x)}{f(x)} &= \frac{\frac{4}{4-4x}}{\frac{4}{4-x}} \\ &= \frac{4-x}{4-4x} \\ f(x \cdot f(x)) &= f(x(\frac{4}{4-x})) \\ &= f(\frac{4x}{4-x}) \\ &= \frac{4}{4-(\frac{4x}{4-x})} \\ &= \frac{4}{\frac{16-4x-4x}{4-x}} \\ &= \frac{4}{\frac{16-8x}{4-x}} \\ &= \frac{4(4-x)}{4(4-4x)} \\ &= \frac{4-x}{4-4x} \\ \text{misalkan } \frac{4-x}{4-4x} &= a \\ f(x \cdot f(x))^{\frac{f(4x)}{f(x)}} &= 256 \\ a^a &= 256 \\ a^a &= 4^4 \\ a &= 4 \\ \frac{4-x}{4-4x} &= 4 \\ 4-x &= 16-16x \\ 15x &= 12 \\ x &= \frac{4}{5} \\ \end{align} </math> </div></div> <ol start=78> <li>Fungsi <math>f(x) = \frac{kx}{2x+1} \text{dengan } x \neq -\frac{1}{2}</math>. Dengan f(f(x)) = x maka tentukan nilai k!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{kx}{2x+1} \\ f(f(x)) &= x \\ f(\frac{kx}{2x+1}) &= x \\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1} &= x \\ \frac{\frac{k^2x}{2x+1}}{\frac{2kx+2x+1}{2x+1}} &= x \\ \frac{k^2x}{2kx+2x+1} &= x \\ \frac{k^2}{2kx+2x+1} &= 1 \\ k^2 &= 2kx+2x+1 \\ k^2-2kx &= 2x+1 \\ k^2-2kx+x^2 &= x^2+2x+1 \\ (k-x)^2 &= (x+1)^2 \\ (k-x)^2-(x+1)^2 &= 0 \\ (k-x+x+1)(k-x-(x+1)) &= 0 \\ k=-1 &\text{ atau } k=2x+1 &\text{ (TM) } \\ \end{align} </math> </div></div> <ol start=79> <li>Jika n = 2023²+2024² maka berapa hasil dari <math>\sqrt{2n-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n &= 2023^2+2024^2 \\ &= 2023^2+(2023+1)^2 \\ \text{misalkan 2023 = p} \\ n &= p^2+(p+1)^2 \\ &= p^2+p^2+2p+1 \\ &= 2p^2+2p+1 \\ \sqrt{2n-1} &= \sqrt{2(2p^2+2p+1)-1} \\ &= \sqrt{4p^2+4p+2-1} \\ &= \sqrt{4p^2+4p+1} \\ &= \sqrt{(2p+1)^2} \\ &= 2p+1 \\ &= 2(2023)+1 \\ &= 4046+1 \\ &= 4047 \\ \end{align} </math> </div></div> <ol start=80> <li>Tentukan nilai dari a+b+c merupakan bilangan bulat positif jika ab = 2, bc = 3 dan ac = 6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab \cdot bc \cdot ac &= 2 \cdot 3 \cdot 6 \\ (abc)^2 &= 36 \\ abc &= \pm 6 \\ abc &= 6 \\ \frac{abc}{ab} &= c = \frac{6}{2} = 3 \\ \frac{abc}{bc} &= a = \frac{6}{3} = 2 \\ \frac{abc}{ac} &= b = \frac{6}{6} = 1 \\ a+b+c &= 6 \\ \end{align} </math> </div></div> <ol start=81> <li>Tentukan nilai dari (a-c)^b jika <math>\frac{ab}{a+b} = \frac{1}{3}</math>, <math>\frac{bc}{b+c} = \frac{1}{4}</math> dan <math>\frac{ac}{a+c} = \frac{1}{9}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{ab}{a+b} &= \frac{1}{3} \\ \frac{a+b}{ab} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{b} + \frac{1}{a} &= 3 \\ \frac{bc}{b+c} &= \frac{1}{4} \\ \frac{b+c}{bc} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{b} &= 4 \\ \frac{ac}{a+c} &= \frac{1}{9} \\ \frac{a+c}{ac} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{a} &= 9 \\ \text{Misalkan 1/a = x, 1/b = y dan 1/c = z} \\ x+y &= 3 \\ y+z &= 4 \\ x+z &= 9 \\ x+y &= 3 \\ y+z &= 4 \\ x-z &= -1 \\ x-z &= -1 \\ x+z &= 9 \\ 2x &= 8 \\ x &= 4 \\ x-z &= -1 \\ 4-z &= -1 \\ z &= 5 \\ x+y &= 3 \\ 4+y &= 3 \\ y &= -1 \\ \frac{1}{a} &= 4 \\ a &= \frac{1}{4} \\ \frac{1}{b} &= -1 \\ b &= -1 \\ \frac{1}{c} &= 5 \\ c &= \frac{1}{5} \\ (a-c)^b &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=82> <li>Tentukan nilai dari a, b dan c jika <math>\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}</math> dan a+2b+3c=28?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan k untuk semua ketiga persamaan tersebut } \\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5} &= k \\ a+b &= 2k \\ a+c &= 4k \\ b+c &= 5k \\ 2a+b+c &= 6k \\ 2a+5k &= 6k \\ k &= 2a \\ a &= \frac{k}{2} \\ b &= \frac{3k}{2} \\ c &= \frac{7k}{2} \\ a+2b+3c &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ a &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ b &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ c &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=83> <li>Tentukan nilai dari (b+c)^a jika <math>\frac{a+b+c}{2} = \sqrt{a-2}+\sqrt{b-1}+\sqrt{c}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{a+b+c}{2} &= \sqrt{a-2}+\sqrt{b-1}+\sqrt{c} \\ a+b+c &= 2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}) \\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c} &= 0 \\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1 &= 0 \\ (\sqrt{a-2}-1)^2+(\sqrt{b-1}-1)^2+(\sqrt{c}-1)^2 &= 0 \\ (\sqrt{a-2}-1)^2 &= 0 \\ \sqrt{a-2}-1 &= 0 \\ \sqrt{a-2} &= 1 \\ a-2 &= 1 \\ a &= 3 \\ (\sqrt{b-1}-1)^2 &= 0 \\ \sqrt{b-1}-1 &= 0 \\ \sqrt{b-1} &= 1 \\ b-1 &= 1 \\ b &= 1 \\ (\sqrt{c}-1)^2 &= 0 \\ \sqrt{c}-1 &= 0 \\ \sqrt{c} &= 1 \\ c &= 1 \\ (b+c)^a &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=84> <li>x dan y merupakan bilangan tak nol. Jika xy = <math>\frac{x}{y}</math> = x-y maka berapa nilai x+y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy &= \frac{x}{y} \\ y^2 &= 1 \\ y^2 - 1 &= 0 \\ (y-1)(y+1) &= 0 \\ y = 1 &\text{ atau } y = -1 \\ \frac{x}{y} &= x-y \\ x &= xy-y^2 \\ x-xy &= -y^2 \\ x(1-y) &= -y^2 \\ x &= \frac{-y^2}{1-y} \\ \text{cek y=1 } \\ x &= \frac{-1^2}{1-1} \\ \text{tidak memenuhi syarat } \\ \text{cek y=-1 } \\ x &= \frac{-(-1)^2}{1-(-1)} \\ &= \frac{-1}{2} \\ x+y &= -1-\frac{1}{2} \\ &= -\frac{3}{2} \\ \end{align} </math> </div></div> <ol start=86> <li>Berapa nilai x dari <math>(\frac{a}{b})^3+(\frac{b}{a})^3 = 2\sqrt{x}</math> jika <math>\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b} \\ \frac{a+b}{ab} &= \frac{1}{a+b} \\ (a+b)^2 &= ab \\ a^2+2ab+b^2 &= ab \\ a^2+b^2 &= -ab \\ \text{misalkan } \frac{a}{b}+\frac{b}{a} = n \\ \frac{a}{b}+\frac{b}{a} &= n \\ \frac{a^2+b^2}{ab} &= n \\ a^2+b^2 &= nab \\ n &= -1 \\ \frac{a}{b}+\frac{b}{a} &= n \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3(\frac{a}{b}+\frac{b}{a}) &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3n &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3 &= n^3-3n \\ &= (-1)^3-3(-1) \\ &= 2 \\ 2\sqrt{x} &= 2 \\ \sqrt{x} &= 1 \\ x &= 1 \\ \end{align} </math> </div></div> <ol start=87> <li>Berapa nilai m dari <math>x^2-mx-1=0</math> jika <math>\sqrt[3]{x_1}+\sqrt[3]{x_2}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x_1} &= a \\ x_1 &= a^3 \\ \sqrt[3]{x_2} &= b \\ x_2 &= b^3 \\ \sqrt[3]{x_1}+\sqrt[3]{x_2} &= 1 \\ a+b &= 1 \\ x^2-mx-1 &= 0 \\ x_1+x_2 &= m \\ x_1 \cdot x_2 &= -1 \\ x_1+x_2 &= m \\ a^3+b^3 &= m \\ x_1 \cdot x_2 &= -1 \\ a^3 \cdot b^3 &= -1 \\ (ab)^2 &= (-1)^3 \\ ab &= -1 \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ (1)^3 &= m+3(-1)(1) \\ 1 &= m-3 \\ m &= 4 \\ \end{align} </math> </div></div> <ol start=88> <li>Berapa nilai <math>\frac{x_1}{x_2}</math> dari <math>ax^2-18x-b=0</math> jika <math>ab=45</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab &= 45 \\ b &= \frac{45}{a} \\ ax^2-18x-b &= 0 \\ ax^2-18x-\frac{45}{a} &= 0 \\ a^2x^2-18ax-45 &= 0 \\ (ax-3)(ax-15) &= 0 \\ ax-3 &= 0 \\ x &= \frac{3}{a} \\ ax-15 &= 0 \\ x &= \frac{15}{a} \\ \frac{x_1}{x_2} &= \frac{\frac{3}{a}}{\frac{15}{a}} \\ &= \frac{3}{15} \\ &= \frac{1}{5} \\ \frac{x_1}{x_2} &= \frac{\frac{15}{a}}{\frac{3}{a}} \\ &= \frac{15}{3} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=89> <li>Jika <math>\frac{u_3}{u_1+u_2} = \frac{7}{8}</math> merupakan barisan aritmetika maka berapa dari <math>\frac{u_2+u_3}{u_1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{u_3}{u_1+u_2} &= \frac{7}{8} \\ \frac{a+2b}{a+a+b} &= \frac{7}{8} \\ \frac{a+2b}{2a+b} &= \frac{7}{8} \\ 8(a+2b) &= 7(2a+b) \\ 8a+16b &= 14a+7b \\ 9b &= 6a \\ b &= \frac{2a}{3} \\ \frac{u_2+u_3}{u_1} &= \frac{a+b+a+2b}{a} \\ &= \frac{2a+3b}{a} \\ &= \frac{2a+3(\frac{2a}{3})}{a} \\ &= \frac{2a+2a}{a} \\ &= \frac{4a}{a} \\ &= 4 \\ \end{align} </math> </div></div> <ol start=90> <li>Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7p+q}{2p+q} &= \frac{17p+q}{7p+q} \\ (7p+q)^2 &= (17p+q)(2p+q) \\ 49p^2+14pq+q^2 &= 34p^2+19pq+q^2 \\ 15p^2 &= 5pq \\ 3p &= q \\ \frac{7p+q}{2p+q} &= \frac{7p+3p}{2p+3p} \\ &= \frac{10p}{5p} \\ &= 2 \\ \end{align} </math> </div></div> <ol start=91> <li>Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{rataan geometris } \\ \sqrt{a \cdot b} &= b-24 \\ a \cdot b &= (b-24)^2 \\ \text{rataan aritmatik } \\ \frac{a+b}{2} &= a+15 \\ a+b &= 2(a+15) \\ a+b &= 2a+30 \\ a &= b-30 \\ a \cdot b &= (b-24)^2 \\ (b-30)b &= (b-24)^2 \\ b^2-30b &= b^2-48b+576 \\ 18b &= 576 \\ b &= 32 \\ a &= b-30 \\ &= 32-30 \\ &= 2 \\ a+b &= 32+2 \\ &= 34 \\ \end{align} </math> </div></div> <ol start=91> <li>Segitiga lancip ABC dengan <math>\frac{a^4+b^4+c^4+a^2b^2}{c^2(a^2+b^2)}=2</math>. tentukan nilai sudut C?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{syarat segitiga lancip semua sudut masing-masing kurang dari } 90^\circ \\ c^2 &= a^2+b^2-2ab cos C \\ cos C &= \frac{a^2+b^2-c^2}{2ab} \\ a^4+b^4+c^4+a^2b^2 &= 2c^2(a^2+b^2) \\ a^4+b^4+a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-2c^2(a^2+b^2)+(c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= (ab)^2 \\ a^2+b^2-c^2 &= \pm ab \\ cos C &= \pm \frac{ab}{2ab} \\ &= \pm \frac{1}{2} \\ &= \frac{1}{2} \text{ (karena sudut harus kurang dari } 90^\circ) \\ C &= 60^\circ \\ \end{align} </math> </div></div> <ol start=92> <li>Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan panjang DA dan EA } = x \text{ dan panjang AB } = y \\ \text{luas segitiga CAB } &= \frac{CA \cdot AB}{2} \\ 45 &= \frac{(x+9)(x+5)}{2} \\ 90 &= x^2+14x+45 \\ x^2+14x &= 45 \\ y^2 &= (x+9)^2+(x+5)^2 \\ &= x^2+18x+81+x^2+10x+25 \\ &= 2x^2+28x+106 \\ &= 2(x^2+14x)+106 \\ &= 2(45)+106 \\ &= 196 \\ y &= 14 \\ \end{align} </math> jadi panjang BC adalah 14 cm </div></div> <ol start=93> <li>Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kita cari ukuran GC yaitu } \\ \frac{GC}{AD} &= \frac{CE}{DE} \\ \frac{GC}{15} &= \frac{6}{18} \\ GC &= 5 \\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC } \\ &= \frac{1}{2} \cdot BC \cdot CF - \frac{1}{2} \cdot GC \cdot CE \\ &= \frac{1}{2} \cdot 15 \cdot 18 - \frac{1}{2} \cdot 5 \cdot 6 \\ &= 135 - 15 \\ &= 120 \\ \end{align} </math> jadi luas daerah BFEG adalah 120 cm² </div></div> <ol start=94> <li>Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ AI &= \frac{1}{5} AF \\ &= \frac{1}{5} 10 \\ &= 2 \\ IF &= AF-AI \\ &= 10-2 \\ &= 8 \\ \text{luas trapesium AFGD } &= \frac{(AD+EF) \cdot AF}{2} \\ &= \frac{(4+6)10}{2} \\ &= 50 \\ \text{luas segitiga AID } &= \frac{AI \cdot AF}{2} \\ &= \frac{(2)4}{2} \\ &= 4 \\ \text{luas segitiga IFG } &= \frac{IF \cdot FG}{2} \\ &= \frac{(8)6}{2} \\ &= 24 \\ \text{luas daerah segitiga IGD } &= \text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG } \\ &= 50-4-24 \\ &= 22 \\ \end{align} </math> jadi luas daerah segitiga IGD adalah 22 cm² </div></div> <ol start=96> <li>Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-<math>\pi</math>) cm³ maka berapa luas permukaan kerucut tersebut?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ \text{volume balok} \\ V_b &= x^2(12) \\ \text{volume kerucut} \\ V_b &= \frac{1}{3}\pi x^2(12) \\ &= 4\pi x^2 \\ V_{b-k} &= Vb-Vk \\ 100(3-\pi) &= 12x^2-4\pi x^2 \\ 100(3-\pi) &= 4x^2(3-\pi) \\ x^2 &= 25 \\ x &= 5 \\ s &= \sqrt{12^2+5^2} \\ &= \sqrt{144+25} \\ &= \sqrt{169} \\ &= 13 \\ \text{luas permukaan kerucut } &= \pi r(r+s) \\ &= \pi(5)(5+13) \\ &= 90\pi \\ \end{align} </math> jadi luas daerah permukaan kerucut adalah 90<math>\pi</math> cm² </div></div> <ol start=97> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 3a+1 \\ B &= 3b+2 \\ A(A+1)+3B \\ (3a+1)(3a+1+1)+3(3b+2) \\ (3a+1)(3a+2)+9b+6 \\ 9a^2+9a+2+9b+6 \\ 9a^2+9a+9b+8 \\ 9(a^2+a+b)+8 \\ \text{sisa pembagiannya adalah } 8 \\ \end{align} </math> </div></div> <ol start=98> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 9a+7 \\ B &= 9b+8 \\ A(A-5)+9B \\ (9a+7)(9a+7-5)+9(9b+8) \\ (9a+7)(9a+2)+81b+72 \\ 81a^2+81a+14+81b+72 \\ 81a^2+81a+81b+86 \\ 81a^2+81a+81b+81+5 \\ 81(a^2+a+b+1)+5 \\ \text{sisa pembagiannya adalah } 5 \\ \end{align} </math> </div></div> <ol start=99> <li>Jika <math>\begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix}</math> maka berapa hasil dari A²¹+A²⁵+A⁴⁶?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A^2 &= A \cdot A \\ &= \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \\ A^3 &= A^2 \cdot A \\ &= \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ &= - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= -I \\ A^{21}+A^{25}+A^{46} &= A^{21} \cdot (I+A^4+A^{25}) \\ &= A^{21} \cdot (I+A^3 \cdot A +A^{24} \cdot A) \\ &= (A^3)^7 \cdot (I+A^3 \cdot A +(A^3)^8 \cdot A) \\ &= (-I)^7 \cdot (I-I \cdot A +(-I)^8 \cdot A) \\ &= -I \cdot (I-A+A) \\ &= -I \cdot I \\ &= -I \\ &= -\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ \end{align} </math> </div></div> <ol start=100> <li>Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida?</li></ol> : bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15 : ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9). : ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8) : maka Ida menulis bilangan 8. <ol start=101> <li>Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9?</li></ol> : Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu. : untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja. : untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja. : Jadi banyaknya bilangan mungkin 2. <ol start=102> <li>Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli!</li></ol> ;cara 1 # 8¹ = 1 # 8² = 1 (8²=8¹x8¹ sama dengan 1x1) # 8³ = 1 (8³=8¹x8² sama dengan 1x1) # 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2) # 8⁵ = 1 # 8ⁿ = 1 (semua n untuk bilangan asli) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli ;cara 2 # 8ⁿ = b mod 7 # 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil) # (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya) # 8ⁿ = 1ⁿ mod 7 # 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli <ol start=103> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 5?</li></ol> ;cara 1 # 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 # 7¹ = 7 (sisa 1) # 7² = 49 (sisa 2) # 7³ = 343 (sisa 3) # 7⁴ = 2,401 (sisa 0) # 7⁵ = 16,807 # 7⁶ = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3 ;cara 2 :17¹ = 2 :17² = 4 :17³ = 3 :17⁴ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3 :17⁹⁹ = (17⁴)²⁴ x 17³ Untuk 17⁴ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 3. Jadi 17⁹⁹ dibagi 7 bersisa 3 ;cara 3 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 17⁴ ::17⁴ = 1 mod 5 ::(17⁴)²⁴ = 1²⁴ mod 5 ::17⁹⁶ = 1²⁴ mod 5 ::17⁹⁶ = 1 mod 5 ::17⁹⁶ x 17³ = 1 x 17³ mod 5 ::17⁹⁹ = 17³ mod 5 ::17⁹⁹ = 17 x 17 x 17 mod 5 ::17⁹⁹ = 2 x 2 x 2 mod 5 ::17⁹⁹ = 8 mod 5 ::17⁹⁹ = 3 mod 5 Jadi 17⁹⁹ dibagi 5 bersisa 3 <ol start=104> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 7?</li></ol> ;cara 1 :17¹ = 3 :17² = 2 :17³ = 6 :17⁴ = 4 :17⁵ = 5 :17⁶ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3 :17⁹⁹ = (17⁶)¹⁶ x 17³ Untuk 17⁶ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 6. Jadi 17⁹⁹ dibagi 7 bersisa 6 ;cara 2 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 17⁶ ::17⁶ = 1 mod 7 ::(17⁶)¹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1 mod 7 ::17⁹⁶ x 17³ = 1 x 17³ mod 7 ::17⁹⁹ = 17³ mod 7 ::17⁹⁹ = 17 x 17 x 17 mod 7 ::17⁹⁹ = 3 x 3 x 3 mod 7 ::17⁹⁹ = 27 mod 7 ::17⁹⁹ = 6 mod 7 Jadi 17⁹⁹ dibagi 7 bersisa 6 <ol start=105> <li>Berapa hasil sisa dari 41²⁰²⁴ dibagi 33?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 41^{2024} &= 41^{2024} \text{ mod } 33 \\ &= (33 \times 3 + 2)^{2024} \text{ mod } 33 \\ &= 2^{2024} \text{ mod } 33 \\ &= 2^{2020} 2^4 \text{ mod } 33 \\ &= (2^5)^{404} 2^4 \text{ mod } 33 \\ &= (33 - 1)^{404} 2^4 \text{ mod } 33 \\ &= (-1)^{404} 2^4 \text{ mod } 33 \\ &= 2^4 \text{ mod } 33 \\ &= 16 \text{ mod } 33 \\ \text{Jadi hasil sisa adalah } 16 \\ \end{align} </math> </div></div> <ol start=106> <li>Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 99^{99} &= (3^2 \times 11)^{99} \\ &= 3^{198} \times 11^{99} \\ 243^n &= (3^5)^n \\ &= 3^{5n} \\ \text{agar bisa membagi, maka} \\ 5n &= 198 \\ n &= 39.6 \\ \text{jadi bilangan n terbesar adalah } 39 \\ \end{align} </math> </div></div> <ol start=107> <li>Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸?</lu></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 88^{88} &= (8 \times 11)^{88} \\ &= 8^{88} \times 11^{88} \\ &= 8^{87} \times 8 \times 11^{88} \\ &= (8^3)^{29} \times 8 \times 11^{88} \\ &= 512^{29} \times 8 \times 11^{88} \\ 512^n &= 512^{29} \\ \text{jadi bilangan n terbesar adalah } 29 \\ \end{align} </math> </div></div> <ol start=108> <li>Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6!</li></ol> ; cara 1 : KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. : N dibagi 3 sisa 1 : N dibagi 5 sisa 2 : N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a : KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70 : KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21 : KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97 ; cara 2 : Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35 : kemudian ditambahkan sisa masing-masing sesuai dengan KPK. : KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, <b>97</b> : KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, <b>97</b> : KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, <b>97</b> Jadi bilangan bulat positif adalah 97 :: NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 <ol start=109> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |- | 4 || 0 || Semua isi ember B dibuang |- | 1 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. <ol start=110> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |} nah ada ember A berisi 4 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |- | 1 || 0 || Semua isi ember B dibuang |- | 0 || 1 || Tuangkan 1 l dari ember A ke B sehingga ember A kosong |- | 3 || 1 || Isikan 3 l ke ember A |- | 0 || 4 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |} nah ada ember B berisi 4 liter. # Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kemungkinan 1} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 +3 \cdot ((\frac{4}{3})^x)^2 &= 7 \cdot (\frac{4}{3})^x \\ \text{misalkan } (\frac{4}{3})^x = y \\ 4 +3y^2 &= 7y \\ 3y^2-7y+4 &= 0 \\ (3y-4)(y-1) &= 0 \\ y_1 = \frac{4}{3} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{4}{3} \\ (\frac{4}{3})^x &= \frac{4}{3} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{4}{3})^x &= 1 \\ x &= 0 \\ \text{kemungkinan 2} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot ((\frac{3}{4})^x)^2 +3 &= 7 \cdot (\frac{3}{4})^x \\ \text{misalkan } (\frac{3}{4})^x = y \\ 4y^2 +3 &= 7y \\ 4y^2-7y+3 &= 0 \\ (4y-3)(y-1) &= 0 \\ y_1 = \frac{3}{4} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{3}{4} \\ (\frac{3}{4})^x &= \frac{3}{4} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{3}{4})^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-6}</math>? [[Kategori:Soal-Soal Matematika]] mu2c791su16xjajblobzzoj5f489c13 117491 117487 2026-07-13T04:54:35Z Akuindo 8654 117491 wikitext text/x-wiki contoh soal <ol start=1> <li>Berapa hasil dari <math>\sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Misalkan 2020 = p} \\ \sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64} &= \sqrt{(2020-5) \cdot (2020-3) \cdot (2020+3) \cdot (2020+5) + 64} \\ &= \sqrt{(p-5) \cdot (p-3) \cdot (p+3) \cdot (p+5) + 64} \\ &= \sqrt{(p-5) \cdot (p+5) \cdot (p-3) \cdot (p+3) + 64} \\ &= \sqrt{(p^2-25) \cdot (p^2-9) + 64} \\ &= \sqrt{p^4-34p^2+ 225 + 64} \\ &= \sqrt{p^4-34p^2+ 289} \\ &= \sqrt{(p^2-17)^2} \\ &= p^2-17 \\ &= 2020^2-17 \\ &= (2000+20)^2-17 \\ &= 4.000.000+80.000+400-17 \\ &= 4.080.383 \\ \end{align} </math> </div></div> <ol start=2> <li>Berapa nilai x dari <math>\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} = \frac{9}{10}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} &= \frac{9}{10} \\ \text{misalkan untuk } \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} = p \\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} &= p \\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}} &= p^2 \\ x^2-x-p &= p^2 \\ x^2-2x+1+x-1 &= p^2+p \\ (x-1)^2+(x-1) &= p^2+p \\ x-1 &= p \\ \text{misalkan untuk } \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}} &= q^3 \\ x^2 q &= q^3 \\ x^2 &= q^2 \\ x &= q \\ \frac{x-1}{x} &= \frac{9}{10} \\ x &= 10 \\ \end{align} </math> </div></div> <ol start=3> <li>Berapa nilai x dari <math>(\frac{x}{x+10})^{x+10}=\frac{1}{1024}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{x+10}{x})^{-(x+10)} &= (1024)^{-1} \\ (\frac{x+10}{x})^{x+10} &= 1024 \\ (\frac{x+10}{x})^{x+10} &= 2^{10} \\ (\frac{x+10}{x})^{\frac{x+10}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (\frac{1}{2})^{-1} \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (1+(-\frac{1}{2}))^{(1+(-\frac{2}{1}))} \\ \frac{10}{x} &= -\frac{1}{2} \\ x &= -20 \\ \end{align} </math> </div></div> <ol start=4> <li>Berapa nilai x dari <math>x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{2}+\sqrt{x+\frac{1}{4}} &= (\sqrt{x+\frac{1}{4}})^2+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+(\frac{1}{2})^2 \\ &= (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 \\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}} &= 4 \\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2} &= 4 \\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 4 \\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 &= 4 \\ \sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 2 \\ \sqrt{x+\frac{1}{4}} &= \frac{3}{2} \\ x+\frac{1}{4} &= \frac{9}{4} \\ x &= 2 \\ \end{align} </math> </div></div> <ol start=5> <li>Berapa nilai x dari <math>\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^3}{\sqrt{8-x^2}}+x^2-8 &= 0 \\ \frac{x^3}{\sqrt{8-x^2}} &= 8-x^2 \\ x^3 &= (8-x^2)^{\frac{3}{2}} \\ x &= (8-x^2)^{\frac{1}{2}} \\ x^2 &= 8-x^2 \\ 2x^2-8 &= 0 \\ x^2-4 &= 0 \\ (x-2)(x+2) &= 0 \\ \text{membuktikan } \\ x=2 \text{ maka hasilnya 0 } \\ x=-2 \text{ maka hasilnya -8 } \\ \text{jadi } x=2 \\ \end{align} </math> </div></div> <ol start=6> <li>Berapa nilai x dari <math>\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} = 5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} &= 5 \\ \frac{x^{50}+x^{60}+x^{70}}{31}} &= 5^5 \\ x^{50}+x^{60}+x^{70} &= 5^5 \cdot 31 \\ x^{50}(1+x^{10}+x^{20}) &= 5^5 \cdot 31 \\ (x^{10}^5)(1+x^{10}+(x^{10}^2) &= 5^5 \cdot 31 \\ \text{ misalkan } x^{10} = a \\ a^5(1+a+a^2) &= 5^5 \cdot 31 \\ a &= 5 \\ x^{10} &= 5 \\ x &= ^5 log 10 \\ \end{align} </math> </div></div> <ol start=7> <li>Berapa nilai x dari <math>\sqrt{3x+5+\sqrt{4x+5}} = x</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{3x+5+\sqrt{4x+5}} &= x \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \text{misalkan } \sqrt{4x+5}=y \text{ dan } 4x+5=y^2 \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \sqrt{y^2+y-x} &= x \\ y^2+y &= x^2+x \\ y=x \\ 4x+5 &= y^2 \\ 4x+5 &= x^2 \\ x^2-4x-5 &= 0 \\ (x-5)(x+1) &= 0 \\ x=5 &\text{ atau } x=-1 \text{ (TM) } \\ \end{align} </math> </div></div> <ol start=8> <li>Berapa nilai x dari <math>\sqrt{1+\sqrt{1+x}} = \sqrt[3]{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{1+\sqrt{1+x}} &= \sqrt[3]{x} \\ \sqrt[3]{x} &= n \\ x &= n^3 \\ \sqrt{1+\sqrt{1+n^3}} &= n \\ 1+\sqrt{1+n^3} &= n^2 \\ \sqrt{1+n^3} &= n^2-1 \\ 1+n^3 &= n^4-2n^2+1 \\ n^4-n^3-2n^2 &= 0 \\ n^2(n^2-n-2) &= 0 \\ n^2(n-2)(n+1) &= 0 \\ n=0, n=2 \text{ atau } n=-1 \\ n &= 0 \\ x &= 0^3 \\ &= 0 \\ n &= 2 \\ x &= 2^3 \\ &= 8 \\ n &= -1 \\ x &= (-1)^3 \\ &= -1 \\ \text{yang paling mungkin untuk nilai x adalah } 8 \\ \end{align} </math> </div></div> <ol start=9> <li>Berapa nilai x dari <math>\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} &= \frac{\sqrt{1+x}}{\sqrt{x}} \\ \sqrt{x}(\sqrt{1+x}+\sqrt{x}) &= (\sqrt{1+x}-\sqrt{x})\sqrt{1+x} \\ \sqrt{x(1+x)}+x &= 1+x-\sqrt{x(1+x)} \\ 2\sqrt{x(1+x)} &= 1 \\ \sqrt{x(1+x)} &= \frac{1}{2} \\ x(1+x) &= \frac{1}{4} \\ x^2+x &= \frac{1}{4} \\ 4x^2+4x &= 1 \\ 4x^2+4x-1 &= 0 \\ x &= \frac{-4 \pm \sqrt{4^2-4(4)(-1)}}{2(4)} \\ &= \frac{-4 \pm \sqrt{32}}{8} \\ &= \frac{-4 \pm 4\sqrt{2}}{8} \\ &= \frac{-1 \pm \sqrt{2}}{2} \\ \text{karena akar x harus minimal nol jadi } x = \frac{-1+\sqrt{2}}{2} \\ \end{align} </math> </div></div> <ol start=10> <li>Berapa nilai x dari <math>\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x-\sqrt{x+1}}{x+\sqrt{x+1}} &= \frac{11}{19} \\ \text{misalkan } \sqrt{x+1}=y \text{ dan } x=y^2-1 \\ \frac{y^2-1-y}{y^2-1+y} &= \frac{11}{19} \\ 19(y^2-y-1) &= 11(y^2+y-1) \\ 19y^2-19y-19 &= 11y^2+11y-11 \\ 8y^2-30y-8 &= 0 \\ 4y^2-15y-4 &= 0 \\ (4y+1)(y-4) &= 0 \\ y=-\frac{1}{4} \text{ (TM) atau } & y=4 \\ x &= 4^2-1 \\ &= 15 \\ \end{align} </math> </div></div> <ol start=11> <li>Berapa nilai x dari <math>\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} &= 98 \\ \text{misalkan } \sqrt{x^2-1}=y \\ \frac{x+y}{x-y}+\frac{x-y}{x+y} &= 98 \\ \frac{(x+y)^2+(x-y)^2}{(x-y)(x+y)} &= 98 \\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} &= 98 \\ \frac{2(x^2+y^2)}{x^2-y^2} &= 98 \\ \frac{x^2+y^2}{x^2-y^2} &= 49 \\ x^2+y^2 &= 49(x^2-y^2) \\ x^2+y^2 &= 49x^2-49y^2 \\ 48x^2 &= 50y^2 \\ 24x^2 &= 25y^2 \\ 24x^2 &= 25(\sqrt{x^2-1})^2 \\ 24x^2 &= 25(x^2-1) \\ 24x^2 &= 25x^2-25 \\ x^2 &= 25 \\ x &= \pm 5 \\ \end{align} </math> </div></div> <ol start=12> <li>Berapa nilai x dari <math>\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \sqrt{x}=y \text{ dan } x=y^2 \\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} &= \frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\sqrt{\frac{y^2}{y^2+y}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\frac{y}{\sqrt{y^2+y}} \\ y^2+y-\sqrt{(y^2+y)(y^2-y)} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^4-y^2} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^2(y^2-1)} &= \frac{5}{4}y \\ y(y+1)-y\sqrt{y^2-1} &= \frac{5}{4}y \\ y+1-\sqrt{y^2-1} &= \frac{5}{4} \\ -\sqrt{y^2-1} &= \frac{1}{4}-y \\ y^2-1 &= (\frac{1}{4}-y)^2 \\ y^2-1 &= \frac{1}{16}-\frac{1}{2}y+y^2 \\ -1 &= \frac{1}{16}-\frac{1}{2}y \\ \frac{1}{2}y &= \frac{1}{16}+1 \\ \frac{1}{2}y &= \frac{17}{16} \\ y &= \frac{17}{8} \\ x &= (\frac{17}{8})^2 \\ &= \frac{289}{64} \\ \end{align} </math> </div></div> <ol start=13> <li>Berapa nilai x dari <math>\sqrt[4]{62+x}+\sqrt[4]{275-x}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ misalkan } \sqrt[4]{62+x}=a, 62+x=a^4, \sqrt[4]{275-x}=b \text{ dan } 275-x=b^4 \\ a+b &= 7 \\ (a+b)^2 &= 49 \\ a^2+b^2+2ab &= 49 \\ a^2+b^2 &= 49-2ab \\ a^4+b^4 &= 62+x+275-x \\ (a^2+b^2)^2-2(ab)^2 &= 337 \\ (49-2ab)^2-2(ab)^2 &= 337 \\ 2401-196ab+4(ab)^2-2(ab)^2 &= 337 \\ 2(ab)^2-196ab+2064 &= 0 \\ (ab)^2-98ab+1032 &= 0 \\ (ab-12)(ab-86) &= 0 \\ ab = 12 \text{ atau } & ab = 86 \text{ (TM) karena hasil kali maksimum yaitu 12 } \\ ab =12 \text{ dan } a+b=7 \\ a+b &= 7 \\ b &= 7-a \\ ab &= 12 \\ a(7-a) &= 12 \\ -a^2+7a &= 12 \\ a^2-7a+12 &= 0 \\ (a-3)(a-4) &= 0 \\ a=3 \text{ atau } & a=4 \\ a=3, b=4 \\ 62+x &= a^4 \\ 62+x &= (3)^4 \\ 62+x &= 81 \\ x &= 19 \\ a=4, b=3 \\ 62+x &= a^4 \\ 62+x &= (4)^4 \\ 62+x &= 256 \\ x &= 194 \\ \end{align} </math> </div></div> <ol start=14> <li>Berapa nilai x dari <math>\sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2} &= 7 \\ (\sqrt[3]{8+x})^2-\sqrt[3]{8+x} \sqrt[3]{27-x}+(\sqrt[3]{27-x})^2 &= 7 \\ \text{misalkan } \sqrt[3]{8+x}=a, 8+x=a^3, \sqrt[3]{27-x}=b \text{ dan } 27-x=b^3 \\ a^2-ab+b^2 &= 7 \\ a^3+b^3 &= 8+x+27-x \\ &= 35 \\ a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ 35 &= (a+b)(7) \\ a+b &= 5 \\ b &= 5-a \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ 5^3 &= 35+3ab(5) \\ 125 &= 35+15ab \\ 80 &= 15ab \\ ab &= 6 \\ a(5-a) &= 6 \\ 5a-a^2 &= 6 \\ a^2-5a+6 &= 6 \\ (a-2)(a-3) &= 6 \\ a=2 &\text{ atau } a=3 \\ a=2, b=3 \text{ dan } a=3,b=2 \\ 8+x &= a^3 \\ &= 2^3 \\ &= 8 \\ x &= 0 \\ 8+x &= a^3 \\ &= 3^3 \\ &= 27 \\ x &= 19 \\ \end{align} </math> </div></div> <ol start=15> <li>Berapa nilai x dari <math>3^x+5^x-9^x+15^x-25^x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3^x+5^x-9^x+15^x-25^x &= 1 \\ 3^x+5^x-(3^2)^x+(3 \cdot 5)^x-(5^2)^x &= 1 \\ 3^x+5^x-(3^x)^2+(3^x \cdot 5^x)-(5^x)^2 &= 1 \\ \text{misalkan } 3^x=a \text{ dan } 5^x=b \\ a+b-a^2+ab-b^2 &= 1 \\ a^2-ab+b^2-a-b+1 &= 0 \\ 2a^2-2ab+2b^2-2a-2b+2 &= 0 \\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1 &= 0 \\ (a-b)^2+(a-1)^2+(b-1)^2 &= 0 \\ a-b=0; a-1=0; b-1 &= 0 \\ a=b &= 1 \\ 3^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> <ol start=16> <li>Berapa nilai x dari <math>^6log x^2+^{6x}log \frac{6}{x}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ \text{misalkan } 6x=a \text{ maka } x=\frac{a}{6} \\ ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ ^6log (\frac{a}{6})^2+^{6 \frac{a}{6}}log \frac{6}{\frac{a}{6}} &= 1 \\ ^6log \frac{a^2}{6^2}+^alog \frac{6^2}{a} &= 1 \\ ^6log a^2-^6log 6^2+^alog 6^2-^alog a &= 1 \\ 2 ^6log a-2 ^6log 6+2 ^alog 6-^alog a &= 1 \\ 2 ^6log a-2+2 \frac{1}{^6log a}-1 &= 1 \\ 2 ^6log a+2 \frac{1}{^6log a}-4 &= 0 \\ 2 ^6log^2 a-4 ^6log a+2 &= 0 \\ ^6log^2 a-2 ^6log a+1 &= 0 \\ (^6log a-1)^2 &= 0 \\ ^6log a &= 1 \\ a &= 6 \\ x &= \frac{a}{6} \\ &= \frac{6}{6} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=17> <li>Berapa nilai x dari (x+500)³+x=20?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+500)^3+x &= 20 \\ \text{misalkan } a=x+500 \text{ maka } x=a-500 \\ a^3+a-500 &= 20 \\ a^3+a &= 520 \\ a(a^2+1) &= 8 \cdot 65 \\ a(a^2+1) &= 8(64+1) \\ a(a^2+1) &= 8(8^2+1) \\ a &= 8 \\ x &= 8-500 \\ &= -492 \\ \end{align} </math> </div></div> <ol start=18> <li>Berapa nilai x dari <math>\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2} &= 0 \\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}} &= \frac{1}{2} \\ \frac{x^n+4^n}{x^n+16^n} &= (\frac{1}{2})^n \\ \frac{x^n+4^n}{x^n+16^n} &= \frac{1}{2^n} \\ 2^n(x^n+4^n) &= x^n+16^n \\ 2^n(x^n+2^{2n}) &= x^n+2^{4n} \\ 2^n \cdot x^n+2^{3n} &= x^n+2^{4n} \\ 2^n \cdot x^n-x^n &= 2^{4n}-2^{3n} \\ x^n(2^n-1) &= 2^{3n}(2^n-1) \\ x^n &= 2^{3n} \\ x^n &= (2^3)^n \\ x^n &= 8^n \\ x &= 8 \\ \end{align} </math> </div></div> <ol start=19> <li>Berapa hasil dari <math>\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ x &= \frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5 \cdot 6}+\sqrt{5 \cdot 5}+\sqrt{6 \cdot 4}+\sqrt{5 \cdot 4}}{\sqrt{5 \cdot 4}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5} \cdot \sqrt{6}+\sqrt{5} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{4}}{2 \cdot \sqrt{5}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{6} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{5}+\sqrt{5} \cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}} \\ &= \frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ &= \frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ \frac{1}{x} &= \frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}} \\ &= \frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ &= \frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1} \\ &= \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5} \\ &= \sqrt{6}-\sqrt{4} \\ &= \sqrt{6}-2 \\ x &= \frac{1}{\sqrt{6}-2} \\ &= \frac{\sqrt{6}+2}{6-4} \\ &= \frac{\sqrt{6}+2}{2} \\ &= 1+\frac{\sqrt{6}}{2} \\ \end{align} </math> </div></div> <ol start=20> <li>Berapa hasil dari <math>(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5 \\ \text{misalkan } x=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ x &= \frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ &= \frac{\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}} \\ &= \frac{\sqrt{3}+1}{4} \\ 4x &= \sqrt{3}+1 \\ 4x-1 &= \sqrt{3} \\ (4x-1)^2 &= 3 \\ 16x^2-8x+1 &= 3 \\ 16x^2 &= 8x+2 \\ 8x^2 &= 4x+1 \\ x^2 &= \frac{4x+1}{8} \\ *cara 1 \\ x^3 &= x \cdot x^2 \\ &= x(\frac{4x+1}{8}) \\ &= \frac{4x^2+x}{8} \\ &= \frac{4x^2}{8}+\frac{x}{8} \\ &= \frac{4(\frac{4x+1}{8})}{8}+\frac{x}{8} \\ &= \frac{16x+4}{64}+\frac{x}{8} \\ &= \frac{4x+1}{16}+\frac{x}{8} \\ &= \frac{4x+1+2x}{16} \\ &= \frac{6x+1}{16} \\ x^5 &= x^2 \cdot x^3 \\ &= (\frac{4x+1}{8})(\frac{6x+1}{16}) \\ &= \frac{24x^2+10x+1}{128} \\ &= \frac{24x^2}{128}+\frac{10x+1}{128} \\ &= \frac{24(\frac{4x+1}{8})}{128}+\frac{10x+1}{128} \\ &= \frac{96x+24}{1024}+\frac{10x+1}{128} \\ &= \frac{96x+24+80x+8}{1024} \\ &= \frac{176x+32}{1024} \\ &= \frac{176x}{1024}+\frac{32}{1024} \\ &= \frac{176}{1024}(\frac{\sqrt{3}+1}{4})+\frac{32}{1024} \\ &= \frac{44(\sqrt{3}+1)}{1024}+\frac{32}{1024} \\ &= \frac{44\sqrt{3}+44}{1024}+\frac{32}{1024} \\ &= \frac{76+44\sqrt{3}}{1024} \\ &= \frac{19+11\sqrt{3}}{256} \\ *cara 2 \\ x^4 &= (x^2)^2 \\ &= (\frac{4x+1}{8})^2 \\ &= \frac{16x^2+8x+1}{64} \\ &= \frac{16x^2}{64}+\frac{8x}{64}+\frac{1}{64} \\ &= \frac{x^2}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{\frac{4x+1}{8}}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{4x}{32}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{8}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{4}+\frac{3}{64} \\ x^5 &= x \cdot x^4 \\ &= (\frac{\sqrt{3}+1}{4})(\frac{x}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\frac{\sqrt{3}+1}{4}}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\sqrt{3}+1}{16}+\frac{3}{64}) \\ &= \frac{(\sqrt{3}+1)^2}{64}+(\frac{\sqrt{3}+1}{4})\frac{3}{64} \\ &= \frac{3+2\sqrt{3}+1}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{4+2\sqrt{3}}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{16+8\sqrt{3}}{256}+\frac{3\sqrt{3}+3}{256} \\ &= \frac{19+11\sqrt{3}}{256} \\ \end{align} </math> </div></div> <ol start=21> <li>Berapa hasil dari <math>\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4} &= \frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots) \\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots &= \frac{1}{1-\frac{1}{4}} \\ &= \frac{4}{3} \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{4}{3}) \\ &= \frac{1}{4}+\frac{16}{3} \\ &= \frac{67}{12} \\ x &= \frac{67}{9} \\ \end{align} </math> </div></div> <ol start=22> <li>Berapa nilai y-x jika <math>\frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} = \frac{x}{y}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} &= \frac{x}{y} \\ \frac{\frac{106 \times 107}{2}}{\frac{106}{2}(4+109)} &= \frac{x}{y} \\ \frac{53 \times 107}{53 \times 113} &= \frac{x}{y} \\ y-x &= 113-107 = 6 \\ \end{align} </math> </div></div> <ol start=23> <li>Berapa angka satuan dari hasil 17²⁰²⁴?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan angka satuannya} \\ 17^1 &= 7 \\ 17^2 &= 9 \\ 17^3 &= 3 \\ 17^4 &= 1 \\ 17^5 &= 7 \\ 17^6 &= 9 \\ 17^7 &= 3 \\ 17^8 &= 1 \\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1} \end{align} </math> </div></div> <ol start=24> <li>Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ 1! + 2! + 3! + 4! + \dots + 2024! &= 1 + (1x2) + (1x2x3) + (1x2x3x4) + \dots + 2024! \\ &= 1 + 2 + 6 + 24 + 120 + 720 + \dots + 2024! \\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka } 1+2+6+24 = 33 \text{ jadi angka satuannya adalah } 3 \end{align} </math> </div></div> <ol start=25> <li>Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ \frac{1! + 2! + 3! + 4! + \dots + 2024!}{12} &= \frac{1 + 1x2 + 1x2x3 + 1x2x3x4 + \dots + 2024!}{12} \\ &= \frac{1 + 2 + 6 + 24 + \dots + 2024!}{12} \\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi } 1+2+6 = 9 \end{align} </math> </div></div> <ol start=26> <li>Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{penjumlahan} \\ 1+1+1+1+ \dots \text{ (sebanyak 88 buah) }+x+y &= A \\ 88+x+y &= A \\ \text{perkalian} \\ 1 \times 1 \times 1 \times \dots \text{ (sebanyak 88 buah) }\times x \times y &= A \\ x \times y &= A \\ 88+x+y &= xy \\ xy-y &= 88+x \\ y(x-1) &= 88+x \\ y &= \frac{88+x}{x-1} \\ \text{uji selidiki untuk x=2} \\ y &= \frac{88+2}{2-1} \\ &= 90 \\ \text{buktikan} \\ 88+x+y &= xy \\ 88+2+90 &= 2(90) \\ 180 &= 180 \\ \text{terbukti} \\ \text{nilai A adalah } 180 \\ \end{align} </math> </div></div> <ol start=27> <li>Berapakah nilai x, y dan z dari <math>x+y-z=1, x^2+y^2-z^2=-5 \text{ dan } x^3+y^3-z^3=-53</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+y-z &= 1 \\ x+y &= z+1 \\ x^2+2xy+y^2 &= z^2+2z+1 \\ x^2+y^2-z^2 &= 2z+1-2xy \\ -5 &= 2z+1-2xy \\ 2xy &= 2z+6 \\ xy &= z+3 \\ x^2+y^2-z^2 &= -5 \\ x^2+y^2 &= z^2-5 \\ x^3+y^3-z^3 &= -53 \\ (x+y)(x^2-xy+y^2)-z^3+53 &= 0 \\ (x+y)(x^2+y^2-xy)-z^3+53 &= 0 \\ (z+1)(z^2-5-(z+3))-z^3+53 &= 0 \\ (z+1)(z^2-z-8)-z^3+53 &= 0 \\ z^3-z^2-8z+z^2-z-8-z^3+53 &= 0 \\ -9z+45 &= 0 \\ -9z &= -45 \\ z &= 5 \\ x+y &= 5+1 \\ x+y &= 6 \\ x &= 6-y \\ xy &= 5+3 \\ xy &= 8 \\ (6-y)y &= 8 \\ 6y-y^2 &= 8 \\ y^2-6y+8 &= 0 \\ (y-4)(y-2) &= 0 \\ y=4 \text{ atau } y=2 \\ \text{jika } y=4 \\ x+y &= z+1 \\ x+4 &= 5+1 \\ x &= 2 \\ \text{jika } y=2 \\ x+y &= z+1 \\ x+2 &= 5+1 \\ x &= 4 \\ \end{align} </math> </div></div> <ol start=28> <li>Berapakah nilai titik koordinat (x,y) dari <math>\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}</math> dan <math>\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x+y}+\sqrt{x-y} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+y}-\sqrt{x-y} &= \sqrt{\frac{52y}{3x}} \\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y}) &= \sqrt{\frac{432x}{13y}} \cdot \sqrt{\frac{52y}{3x}} \\ x+y-x+y &= \sqrt{\frac{432x \cdot 52y}{13y \cdot 3x}} \\ 2y &= \sqrt{144 \cdot 4} \\ 2y &= \sqrt{576} \\ 2y &= 24 \\ y &= 12 \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13(12)}} \\ x+12+x-12+2 \cdot \sqrt{x+12} \cdot \sqrt{x-12} &= \frac{36x}{13} \\ 2x+2 \sqrt{x^2-144} &= \frac{36x}{13} \\ 2(x+\sqrt{x^2-144}) &= \frac{36x}{13} \\ x+\sqrt{x^2-144} &= \frac{18x}{13} \\ \sqrt{x^2-144} &= \frac{5x}{13} \\ x^2-144 &= \frac{25x^2}{169} \\ \frac{144x^2}{169}-144 &= 0 \\ \frac{x^2}{169}-1 &= 0 \\ x^2-169 &= 0 \\ (x-13)(x+13) &= 0 \\ x_1=13 &\text{ atau } x_2=-13 \text{ (TM) karena } x>y \\ \end{align} </math> jadi titik koordinat (13,12) </div></div> <ol start=29> <li>Berapakah nilai dari <math>x^2-7x</math> jika <math>(x-2)^2+\frac{1}{(x-2)^2} = 11</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x-2)^2+\frac{1}{(x-2)^2} &= 11 \\ (x-2)^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2)^2} &= 11-2 \\ (x-2-\frac{1}{x-2})^2 &= 9 \\ x-2-\frac{1}{x-2} &= 3 \\ (x-2)^2-1 &= 3(x-2) \\ x^2-4x+4-1 &= 3x-6 \\ x^2-7x &= -9 \\ \end{align} </math> </div></div> <ol start=30> <li>Berapakah nilai dari <math>\frac{(x+y)^2(x+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)}</math> jika xy+yz+xz=1?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy+yz+xz &= 1 \\ x^2+xy+yz+xz &= x^2+1 \\ x(x+y)+z(x+y) &= x^2+1 \\ (x+y)(x+z) &= x^2+1 \\ \text{dengan pola yang sama } \\ (y+x)(y+z) &= y^2+1 \\ (x+z)(y+z) &= z^2+1 \\ \frac{(x+y)^2(y+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)} &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)} \\ &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)^2(y+z)^2(x+z)^2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=31> <li>Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} w+5 &= w+x+y+z+5 \\ x+4 &= w+x+y+z+5 \\ y+3 &= w+x+y+z+5 \\ z+2 &= w+x+y+z+5 \\ \text{jumlahkan keempat persamaan } \\ w+x+y+z+14 &= 4(w+x+y+z+5) \\ w+x+y+z+14 &= 4(w+x+y+z)+20 \\ 3(w+x+y+z) &= -6 \\ w+x+y+z &= -2 \\ \end{align} </math> </div></div> <ol start=32> <li>Berapakah nilai dari <math>\frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2}</math> jika <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3</math> dan x+y+z=xyz?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2} &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2} \\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) \\ 3^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz}) \\ 9 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz}) \\ &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2 \\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} &= 7 \\ \end{align} </math> </div></div> <ol start=33> <li>Berapakah nilai dari <math>\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}</math> jika <math>x^2+y^2+z^2 = -2(ab+bc+ac)</math>?>/li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+z^2 &= -2(xy+yz+xz) \\ x^2+y^2+z^2+2(xy+yz+xz) &= 0 \\ (x+y+z)^2 &= 0 \\ x+y+z &= 0 \\ x+y &= -z \\ x+z &= -y \\ y+z &= -x \\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z} &= \frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x} \\ &= -2-(-5)-(-7) \\ &= 10 \\ \end{align} </math> </div></div> <ol start=34> <li>Berapakah nilai dari <math>\frac{20xyz}{xy+yz+xz}</math> jika <math>16^x = 256^y = 625^z = 40</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 16^x = 256^y = 625^z &= 40 \\ 2^{4x} = 4^{4y} = 5^{4z} &= 40 \\ 2^{4x} &= 40 \\ 2 &= 40^{\frac{1}{4x}} \\ 4^{4y} &= 40 \\ 4 &= 40^{\frac{1}{4y}} \\ 5^{4z} &= 40 \\ 5 &= 40^{\frac{1}{4z}} \\ 2 \cdot 4 \cdot 5 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z}} \\ 1 &= \frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z} \\ 4 &= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{20xyz}{xy+yz+xz} &= 20 \cdot \frac{xyz}{xy+yz+xz} \\ &= 20 \cdot (\frac{xy+yz+xz}{xyz})^{-1} \\ &= 20 \cdot (\frac{1}{z} + \frac{1}{x} + \frac{1}{y})^{-1} \\ &= 20 \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{-1} \\ &= 20 \cdot (4)^{-1} \\ &= 20 \cdot \frac{1}{4} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=35> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+3x^2+1}</math> jika <math>6x^2+25x+6=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 6x^2+25x+6 &= 0 \\ 6x+25+\frac{6}{x} &= 0 \\ 6(x+\frac{1}{x}) &= -25 \\ x+\frac{1}{x} &= \frac{-25}{6} \\ (c+\frac{1}{x})^2 &= (\frac{-25}{6})^2 \\ x^2+2+\frac{1}{x^2} &= \frac{625}{36} \\ x^2+\frac{1}{x^2} &= \frac{625}{36}-2 \\ x^2+\frac{1}{x^2} &= \frac{553}{36} \\ \frac{x^2}{x^4+3x^2+1} &= \frac{1}{x^2+3+\frac{1}{x^2}} \\ &= \frac{1}{a^2+\frac{1}{x^2}+3} \\ &= \frac{1}{\frac{553}{36}+3} \\ &= \frac{1}{\frac{661}{36}} \\ &= \frac{36}{661} \\ \end{align} </math> </div></div> <ol start=36> <li>Berapakah nilai dari <math>\frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5} &= \frac{(9+2\sqrt{20})^{1013}}{((2)^3+3(2)^2(\sqrt{5})+3(2)(\sqrt{5})^2+(\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{((2+\sqrt{5})^2)^{1013}}{((2+\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{(2+\sqrt{5})^{2026}}{(2+\sqrt{5})^{2025}}+6-\sqrt{5} \\ &= 2+\sqrt{5}+6-\sqrt{5} \\ &= 8 \\ \end{align} </math> </div></div> <ol start=37> <li>Berapakah nilai dari <math>27x^3+\frac{8}{x^3}</math> jika <math>3x+\frac{2}{x}=6</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3x+\frac{2}{x} &= 6 \\ (3x+\frac{2}{x})^3 &= 6^3 \\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3} &= 216 \\ 27x^3+18(6)+\frac{8}{x^3} &= 216 \\ 27x^3+108+\frac{8}{x^3} &= 216 \\ 27x^3+\frac{8}{x^3} &= 108 \\ \end{align} </math> </div></div> <ol start=38> <li>Berapakah nilai dari <math>x^6+\frac{8}{x^3}</math> jika <math>x^3+\frac{1}{x^3}=8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3+\frac{1}{x^3} &= 8 \\ x^3 &= 8-\frac{1}{x^3} \\ x^6 &= 8x^3-1 \\ x^6+\frac{8}{x^3} &= 8x^3-1+\frac{8}{x^3} \\ &= 8x^3+\frac{8}{x^3}-1 \\ &= 8(x^3+\frac{1}{x^3})-1 \\ &= 8(8)-1 \\ &= 63 \\ \end{align} </math> </div></div> <ol start=39> <li>Berapakah nilai dari <math>4x+\frac{25}{x}</math> jika <math>2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2\sqrt{x}+\frac{5}{\sqrt{x}} &= 4x-\frac{25}{x} \\ 2\sqrt{x}+\frac{5}{\sqrt{x}} &= (2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}}) \\ 1 &= 2\sqrt{x}-\frac{5}{\sqrt{x}} \\ 1^2 &= (2\sqrt{x}-\frac{5}{\sqrt{x}})^2 \\ 1 &= 4x-20+\frac{25}{x} \\ 4x+\frac{25}{x} &= 21 \\ \end{align} </math> </div></div> <ol start=40> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2-x+1}{x^2+x+1} &= \frac{5}{6} \\ \frac{x^2+1-x}{x^2+1+x} &= \frac{5}{6} \\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} &= \frac{5}{6} \\ \text{ misalkan } x+\frac{1}{x} &= y \\ \frac{y-1}{y+1} &= \frac{5}{6} \\ 6(y-1) &= 5(y+1) \\ 6y-6 &= 5y+5 \\ y &= 11 \\ x+\frac{1}{x} &= 11 \\ \end{align} </math> </div></div> <ol start=41> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt{x}+x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x}+x &= 1 \\ x-1 &= -\sqrt{x} \\ (x-1)^2 &= (-\sqrt{x})^2 \\ x^2-2x+1 &= x \\ x^2-3x+1 &= 0 \\ x-3+\frac{1}{x} &= 0 \\ x+\frac{1}{x} &= 3 \\ \end{align} </math> </div></div> <ol start=42> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt[3]{x}-\sqrt[3]{x-36}=3</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x}-\sqrt[3]{x-36} &= 3 \\ (\sqrt[3]{x}-\sqrt[3]{x-36})^3 &= 3^3 \\ x-(x-36)-3 \sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36}) &= 27 \\ 36-3 \sqrt[3]{x(x-36)}3 &= 27 \\ -9 \sqrt[3]{x(x-36)} &= -9 \\ \sqrt[3]{x(x-36)} &= 1 \\ x(x-36) &= 1 \\ x^2-36x-1 &= 0 \\ x-36-\frac{1}{x} &= 0 \\ x-\frac{1}{x} &= 36 \\ \end{align} </math> </div></div> <ol start=43> <li>Berapakah nilai dari <math>x+\frac{16}{x}</math> jika <math>x-3\sqrt{x}=4</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x-3\sqrt{x} &= 4 \\ x-4 &= 3\sqrt{x} \\ x^2-8x+16 &= 9x \\ x^2-17x+16 &= 0 \\ x-17+\frac{16}{x} &= 0 \\ x+\frac{16}{x} &= 17 \\ \end{align} </math> </div></div> <ol start=44> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+4}</math> jika <math>x^2-7x+2=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2-7x+2 &= 0 \\ x^2+2 &= 7x \\ x+\frac{2}{x} &= 7 \\ x^2+4+\frac{4}{x^2} &= 49 \\ x^2+\frac{4}{x^2} &= 45 \\ \frac{x^4+4}{x^2} &= 45 \\ \frac{x^2}{x^4+4} &= \frac{1}{45} \\ \end{align} </math> </div></div> <ol start=45> <li>Berapakah nilai dari <math>x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}</math> jika <math>x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}} &= 25 \\ x^{\frac{1}{2}}+x^{-\frac{1}{2}} &= 23 \\ x+2+x^{-1} &= 529 \\ x+x^{-1} &= 527 \\ x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+x^{-\frac{3}{4}} &= 110 \\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1} &= x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}} \\ &= 527+110 \\ &= 637 \\ \end{align} </math> </div></div> <ol start=46> <li>Berapakah nilai dari <math>\sqrt{8x^6+x^5+x^4+5x^3+1}</math> jika <math>\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5} &= 0 \\ \frac{x^2+x+1}{x^5} &= 0 \\ x^2+x+1 &= 0 \\ x^2+x+1 &= 0 \\ (x-1)(x^2+x+1) &= 0(x-1) \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ \sqrt{8x^6+x^5+x^4+5x^3+1} &= \sqrt{(2x^3)^2+x^3x^2+x^3x+5x^3+1} \\ &= \sqrt{(2(1))^2+(1)x^2+(1)x+5(1)+1} \\ &= \sqrt{(2)^2+x^2+x+5+1} \\ &= \sqrt{4+x^2+x+1+5} \\ &= \sqrt{4+0+5} \\ &= \sqrt{9} \\ &= 3 \\ \end{align} </math> </div></div> <ol start=47> <li>Berapakah nilai dari <math>f(1)+f(2)+f(3)+ \dots + f(99)</math> jika <math>f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{1}{\sqrt{x+1}+\sqrt{x}} \\ &= \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} \\ &= \sqrt{x+1}-\sqrt{x} \\ f(1)+f(2)+f(3)+ \dots + f(98)+f(99) &= \sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+ \cdot + \sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99} \\ &= \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+ \cdot + \sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ &= \sqrt{100}-\sqrt{1} \\ &= 10-1 \\ &= 9 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots + \frac{2024}{2025})</math> jika <math>h(x)=\frac{3}{3+9^x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} h(x) &= \frac{3}{3+9^x} \\ h(1-x) &= \frac{3}{3+9^{1-x}} \\ &= \frac{3}{3+\frac{9}{9^x}} \\ &= \frac{9^x}{3+9^x} \\ h(x)+h(1-x) &= \frac{3}{3+9^x}+\frac{9^x}{3+9^x} \\ &= \frac{3+9^x}{3+9^x} \\ &= 1 \\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots +(1-\frac{2}{2025})+(1-\frac{1}{2025})) \\ & 5(1+1+1+ \dots +1+1) \text{ sebanyak 1012 kali } \\ & 5(1012) \\ & 5060 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>\frac{7^{2025} - 7^{2023} + 432}{7^{2024} + 7^{2023} + 72}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72} &= \frac{7^{2023}7^{2}-7^{2023} + 48 \times 9}{7^{2023}7^1+7^{2023}+8 \times 9} \\ &= \frac{7^{2023}(7^{2}-1)+48 \times 9}{7^{2023}(7^1+1)+8 \times 9} \\ &= \frac{7^{2023}(49-1)+48 \times 9}{7^{2023}(7+1) + 8 \times 9} \\ &= \frac{7^{2023} \times 48+48 \times 9}{7^{2023} \times 8+8 \times 9} \\ &= \frac{48(7^{2023}+9)}{8(7^{2023}+9)} \\ &= \frac{48}{8} \\ &= 6 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>tan (x+\frac{\pi}{4})</math> jika <math>\frac{1}{cos x}-tan x = \frac{4}{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{cos x}-tan x &= \frac{4}{5} \\ sec x-tan x &= \frac{4}{5} \\ sec^2 x-tan^2 x &= 1 \\ (sec x+tan x)(sec x-tan x) &= 1 \\ (sec x+tan x)\frac{4}{5} &= 1 \\ sec x+tan x &= \frac{5}{4} \\ \text{kedua persamaan dengan cara metode eliminasi } \\ 2 tan x &= \frac{5}{4}-\frac{4}{5} \\ 2 tan x &= \frac{9}{20} \\ tan x &= \frac{9}{40} \\ tan (x+\frac{\pi}{4}) &= \frac{tan x+tan \frac{\pi}{4}}{1-tan x \cdot tan \frac{\pi}{4}} \\ &= \frac{\frac{9}{40}+1}{1-\frac{9}{40} \cdot 1} \\ &= \frac{\frac{49}{40}}{\frac{31}{40}} \\ &= \frac{49}{31} \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>sin^3 x+csc^3 x</math> jika <math>sin x-csc x = 8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ Dengan menggunakan rumus: } (a-b)^3 &= a^3-b^3-3ab(a-b) \\ (sin x-csc x)^3 &= sin^3 x-csc^3 x-3sin x csc x(sin x-csc x) \\ 8^3 &= sin^3 x-csc^3 x-3sin x (\frac{1}{sin x})(8) \\ 512 &= sin^3 x-csc^3 x-24 \\ sin^3 x-csc^3 x &= 512+24 \\ sin^3 x-csc^3 x &= 536 \\ \end{align} </math> </div></div> <ol start=52> <li>Berapakah nilai dari <math>(sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2</math> jika <math>\frac{1}{sin x}+\frac{1}{cos x} = 10</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{sin x}+\frac{1}{cos x} &= 10 \\ \frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} &= 100 \\ (sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2 &= sin^2 x+\frac{2sin x}{cos x}+\frac{1}{cos^2 x}+cos^2 x+\frac{2cos x}{sin x}+\frac{1}{sin^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2(sin^2 x+cos^2 x)}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+100 \\ &= 101 \\ \end{align} </math> </div></div> <ol start=53> <li>Berapakah nilai dari (x-1)⁶ jika <math>x=\frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin 80^\circ &= cos 10^\circ \\ sin 80^\circ-cos 10^\circ &= 0 \\ x &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+2 sin 20^\circ cos 20^\circ}{cos 10^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+4 sin 10^\circ cos 10^\circ cos 20^\circ}{cos 10^\circ} \\ &= 4 cos 55^\circ cos 25^\circ+4 sin 10^\circ cos 20^\circ \\ &= 2(2 cos 55^\circ cos 25^\circ+2 sin 10^\circ cos 20^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ+sin (-10)^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ-sin 10^\circ) \\ &= 2(cos 80^\circ-sin 10^\circ+cos 30^\circ+sin 30^\circ) \\ &= 2(cos 80^\circ-sin (90^\circ-80^\circ)+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(cos 80^\circ-cos 80^\circ+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= \sqrt{3}+1 \\ x-1 &= \sqrt{3} \\ (x-1)^6 &= (\sqrt{3})^6 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=54> <li>Berapakah nilai dari x jika <math>x=\frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ} \\ 2x sin 20^\circ-x sin 40 ^\circ &= x sin 20^\circ-x^2 sin 10^\circ \\ x^2 sin 10^\circ+x sin 20^\circ-x sin 40 ^\circ &= 0 \\ x(x sin 10^\circ+sin 20^\circ-sin 40 ^\circ) &= 0 \\ x = 0 &\text{ atau } x sin 10^\circ+sin 20^\circ-sin 40 ^\circ = 0 \\ x sin 10^\circ+sin 20^\circ-sin 40 ^\circ &= 0 \\ x sin 10^\circ &= sin 40 ^\circ-sin 20^\circ \\ x &= \frac{sin 40 ^\circ-sin 20^\circ}{sin 10^\circ} \\ &= \frac{2 cos 30 ^\circ sin 10^\circ}{sin 10^\circ} \\ &= 2 cos 30 ^\circ \\ &= \frac{2 \sqrt{3}}{2} \\ &= \sqrt{3} \\ \end{align} </math> </div></div> <ol start=55> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x^2}{x^2-16y^2} = \frac{625}{49}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2}{x^2-16y^2} &= \frac{625}{49} \\ \frac{x^2-16y^2}{x^2} &= \frac{49}{625} \text{ (terbalik posisinya)} \\ 1-\frac{16y^2}{x^2} &= \frac{49}{625} \\ \frac{16y^2}{x^2} &= 1 - \frac{49}{625} \\ (\frac{4y}{x})^2 &= \frac{576}{625} \\ (\frac{4y}{x})^2 &= (\frac{24}{25})^2 \\ \frac{4y}{x} &= \frac{24}{25} \\ \frac{y}{x} &= \frac{6}{25} \\ \frac{x}{y} &= \frac{25}{6} \\ \end{align} </math> </div></div> <ol start=56> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x}{y}+\frac{x+10y}{y+10x} = 2</math> serta bilangan real untuk x dan y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x}{y}+\frac{x+10y}{y+10x} &= 2 \\ \frac{x}{y}+\frac{\frac{x}{y}+10}{1+10\frac{x}{y}} &= 2 \\ \text{misalkan } \frac{x}{y} = a \\ a+\frac{a+10}{1+10a} &= 2 \\ a(1+10a)+a+10 &= 2(1+10a) \\ 10a^2+a+a+10 &= 2+20a \\ 10a^2-18a+8 &= 0 \\ 5a^2-9a+4 &= 0 \\ (5a-4)(a-1) &= 0 \\ a = \frac{4}{5} &\text{ atau } a = 1 \\ \text{jadi } \frac{x}{y} = {\frac{4}{5}, 1} \\ \end{align} </math> </div></div> <ol start=57> <li>Berapakah nilai dari xy jika <math>x^4+y^4+x^2y^2=15 \text{ dan } x^2+y^2+xy=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+xy &= 5 \\ x^2+y^2 &= 5-xy \\ x^4+y^4+x^2y^2 &= 15 \\ (x^2)^2+(y^2)^2+2x^2y^2-x^2y^2 &= 15 \\ (x^2+y^2)^2-x^2y^2 &= 15 \\ (5-xy)^2-x^2y^2 &= 15 \\ 25-10xy+x^2y^2-x^2y^2 &= 15 \\ 25-10xy &= 15 \\ 10xy &= 10 \\ xy &= 1 \\ \end{align} </math> </div></div> <ol start=58> <li>Berapakah nilai dari x jika <math>4^x = 63(4^3+1)(4^6+1)(4^{12}+1)+1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 4^x &= 63(4^3+1)(4^6+1)(4^{12}+1)+1 \\ 4^x-1 &= 63(4^3+1)(4^6+1)(4^{12}+1) \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{4^3-1} \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{63} \\ &= (4^3+1)(4^6+1)(4^{12}+1)(4^3-1) \\ &= (4^3-1)(4^3+1)(4^6+1)(4^{12}+1) \\ &= (4^6-1)(4^6+1)(4^{12}+1) \\ &= (4^{12}-1)(4^{12}+1) \\ &= 4^{24}-1 \\ 4^x &= 4^{24} \\ x &= 24 \\ \end{align} </math> </div></div> <ol start=59> <li>Berapakah nilai dari <math>\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}</math> jika <math>x=\sqrt{9+4\sqrt{5}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \sqrt{9+4\sqrt{5}} \\ x &= 2+\sqrt{5} \\ x^2 &= 9+4\sqrt{5} \\ x^2-4x &= 9+4\sqrt{5}-4(2+\sqrt{5}) \\ x^2-4x &= 1 \\ x^2 &= 4x+1 \\ x^3 &= x \cdot x^2 \\ &= x(4x+1) \\ &= 4x^2+x \\ &= 4(4x+1)+x \\ &= 16x+4+x \\ &= 17x+4 \\ x^4 &= x \cdot x^3 \\ &= x(17x+4) \\ &= 17x^2+4x \\ &= 17(4x+1)+4x \\ &= 68x+17+4x \\ &= 72x+17 \\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1} &= \frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1} \\ &= \frac{72x+17-85x-20+8x+2+5x+3}{2} \\ &= \frac{2}{2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=60> <li>Berapakah nilai dari <math>\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}</math> jika 2x-1=<math>\sqrt{61}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \frac{x^3+1}{x^5-x^4-x^3+x^2} = p \\ p &= \frac{x^3+1}{x^5-x^4-x^3+x^2} \\ &= \frac{x^3+1}{x^5-x^4-(x^3-x^2)} \\ &= \frac{x^3+1}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)} \\ &= \frac{x^2-x+1}{x^2(x-1)^2} \\ &= \frac{x^2-x+1}{(x(x-1))^2} \\ &= \frac{x(x-1)+1}{(x(x-1))^2} \\ 2x-1 &= \sqrt{61} \\ x &= \frac{\sqrt{61}+1}{2} \\ x-1 &= \frac{\sqrt{61}-1}{2} \\ x(x-1) &= (\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2}) \\ &= \frac{61-1}{4} \\ &= \frac{60}{4} \\ &= 15 \\ p &= \frac{x(x-1)+1}{(x(x-1))^2} \\ &= \frac{15+1}{15^2} \\ &= \frac{16}{15^2} \\ \sqrt{p} &= \sqrt{\frac{16}{15^2}} \\ &= \frac{4}{15} \\ \end{align} </math> </div></div> <ol start=61> <li>Berapakah nilai dari <math>(\frac{x-3}{x})^{25}</math> jika <math>x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\sqrt[5]{8}+\sqrt[5]{2} &= 1+\sqrt[5]{16}+\sqrt[5]{4} \\ x+(\sqrt[5]{2})^3+\sqrt[5]{2} &= 1+(\sqrt[5]{2})^4+(\sqrt[5]{2})^2 \\ x &= (\sqrt[5]{2})^4-(\sqrt[5]{2})^3+(\sqrt[5]{2})^2-\sqrt[5]{2}+1 \\ \text{misalkan } \sqrt[5]{2} = p \\ x &= p^4-p^3+p^2-p+1 \\ x &= \frac{p^5+1}{p+1} \\ (\frac{x-3}{x})^{25} &= (1-\frac{3}{x})^{25} \\ &= (1-\frac{3}{\frac{p^5+1}{p+1}})^{25} \\ &= (1-\frac{3(p+1)}{p^5+1})^{25} \\ &= (1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2})^5+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{2+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-3\sqrt[5]{2}-3)}{3})^{25} \\ &= (-\sqrt[5]{2})^{25} \\ &= (-2)^5 \\ &= -32 \\ \end{align} </math> </div></div> <ol start=62> <li>Berapakah nilai dari <math>x^{50}+x^{49}+x^{48}+x^{47}+x^{46}</math> jika <math>x^2+x+1=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+x+1 &= 0 \\ x^2+x &= -1 \\ \frac{x^3-1}{x-1} &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46} &= x^{48}(x^2+x+1)+x^{45}(x^2+x) \\ &= x^{48}(0)+(x^3)^{15}(-1) \\ &= 0+(1)^{15}(-1) \\ &= -1 \\ \end{align} </math> </div></div> <ol start=63> <li>Berapakah 2²⁴ dari <math>8^7+8^6+8^5+8^4+8^3+8^2+8+1=A</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= A \\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1) &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8 &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= 8A+1 \\ 8^8+A &= 8A+1 \\ 8^8 &= 7A+1 \\ (2^3)^8 &= 7A+1 \\ 2^{24} &= 7A+1 \\ \end{align} </math> </div></div> <ol start=64> <li>Berapakah nilai dari <math>x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1</math> jika <math>x+\frac{1}{x}=\sqrt{3}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{x} &= \sqrt{3} \\ x^2+2+\frac{1}{x^2} &= 3 \\ x^2-1+\frac{1}{x^2} &= 0 \\ x^2(x^2-1+\frac{1}{x^2}) &= x^2(0) \\ x^4-x^2+1 &= 0 \\ (x^2+1)(x^4-x^2+1) &= (x^2+1)0 \\ x^6-x^4+x^2+x^4-x^2+1 &= 0 \\ x^6+1 &= 0 \\ x^6 &= -1 \\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1 &= {x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1 \\ &= (-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2-1+1 \\ &= -1+1-1+1-1+1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=65> <li>Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari <math>\frac{c-b}{a-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= ax^2+bx+c \\ f(2) &= a(2)^2+2b+c = 4 \\ &= 4a+2b+c = 4 \\ f(7) &= a(7)^2+7b+c = 49 \\ &= 49a+7b+c = 49 \\ 49a+7b+c &= 49 \\ 4a+2b+c &= 4 \\ 45a+5b &= 45 \text{ (f(7) dikurangi f(2)) } \\ 9a+b &= 9 \\ b &= -9a+9 \\ 4a+2b+c &= 4 \\ 4a+2(-9a+9)+c &= 4 \\ 4a-18a+18+c &= 4 \\ -14a+18+c &= 4 \\ c &= 14a-14 \\ \frac{c-b}{a-1} &= \frac{14a-14-(-9a+9)}{a-1} \\ &= \frac{14(a-1)+9(a-1)}{a-1} \\ &= \frac{(14+9)(a-1)}{a-1} \\ &= 23 \\ \end{align} </math> </div></div> <ol start=66> <li>Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+y)^3 &= x^3+y^3+3xy(x+y) \\ 11^3 &= 242+3xy(11) \text{ (dibagi 11)} \\ 11^2 &= 22+3xy \\ 121 &= 22+3xy \\ 99 &= 3xy \\ xy &= 33 \\ (x-y)^2 &= x^2+y^2-2xy \\ &= ((x+y)^2-2xy)-2xy \\ &= (x+y)^2-4xy \\ &= 11^2-4(33) \\ &= 121-132 \\ &= -11 \\ \end{align} </math> </div></div> <ol start=67> <li>Berapa f(1)+f(-1) jika <math>f(\frac{ax-b}{bx-a})</math>=x²-5x+6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ jika} f(1) = f(\frac{ax-b}{bx-a}) \\ 1 &= \frac{ax-b}{bx-a} \\ bx-a &= ax-b \\ (b-a)x &= -b+a \\ &= -(b-a) \\ &= -1 \\ f(1) &= x^2-5x+6 \\ &= (-1)^2-5(-1)+6 \\ &= 12 \\ \text{ jika} f(-1) = f(\frac{ax-b}{bx-a}) \\ -1 &= \frac{ax-b}{bx-a} \\ -(bx-a) &= ax-b \\ -bx+a &= ax-b \\ (-b-a)x &= -b-a \\ &= 1 \\ f(-1) &= x^2-5x+6 \\ &= (1)^2-5(1)+6 \\ &= 2 \\ f(1)+f(-1) &= 12+2 \\ &= 14 \\ \end{align} </math> </div></div> <ol start=68> <li>berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)-x &= f(x-1) \\ f(x)-f(x-1) &= x \\ x=1 ; f(1)-f(0) &= 1 \\ x=2 ; f(2)-f(1) &= 2 \\ x=3 ; f(3)-f(2) &= 3 \\ x=4 ; f(4)-f(3) &= 4 \\ \dots \\ x=200 ; f(200)-f(199) &= 200 \\ \text{ jumlahkan tersebut menjadi } \\ f(200)-f(0) &= 1+2+3+4+\dots+200 \\ &= \frac{200 \cdot 201}{2} \\ &= 20.100 \\ f(200)-1 &= 20.100 \\ &= 20.101 \\ \end{align} </math> </div></div> <ol start=69> <li>Misalkan f(x) adalah fungsi rekursif yang berlaku ∀x ∈ R sebagai berikut: : f(x)+f(15-x) = 2024 : f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)+f(15-x) &= 2024 \\ f(15+x) &= f(x)+2020 \\ *cara 1 \\ \text{ganti x dengan 15+x } \\ f(15+x)+f(-x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ *cara 2 \\ \text{ganti x dengan -x } \\ f(-x)+f(15+x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ \end{align} </math> </div></div> <ol start=70> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>2f(\frac{2002}{x}) + f(x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2f(\frac{2002}{x}) + f(x) &= 3x \\ \text{ganti x dengan 2 } \\ 2f(\frac{2002}{2}) + f(2) &= 3(2) \\ 2f(1001) + f(2) &= 6 \\ \text{ganti x dengan 1001 } \\ 2f(\frac{2002}{1001}) + f(1001) &= 3(1001) \\ 2f(2) + f(1001) &= 3003 \\ 2f(2) + f(1001) &= 3003 \\ f(1001) &= 3003 - 2f(2) \\ 2f(1001) + f(2) &= 6 \\ 2(3003 - 2f(2)) + f(2) &= 6 \\ 6006 - 4f(2) + f(2) &= 6 \\ 3f(2) &= 6000 \\ f(2) &= 2000 \\ \end{align} </math> </div></div> <ol start=71> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>f(\frac{1}{x}) + \frac{1}{x}f(-x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{1}{x})+\frac{1}{x}f(-x) &= 3x \\ \text{ganti x dengan 1/3 } \\ f(3)+3f(-\frac{1}{3}) &= 1 \\ \text{ganti x dengan -3 } \\ f(-\frac{1}{3}) - \frac{1}{3}f(3) &= -9 \\ \text{dikalikan 3 } \\ 3f(-\frac{1}{3})-f(3) &= -27 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ 2f(3) &= 28 \\ f(3) &= 14 \\ \end{align} </math> </div></div> <ol start=72> <li>Diketahui polinom <math>f(7^b-1)=7^{3b}-10</math>. tentukan nilai f(5)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(5) &= f(7^b-1) \\ 5 &= 7^b-1 \\ 7^b &= 6 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(6-1) &= 6^3-10 \\ f(5) &= 216-10 \\ &= 206 \\ *cara 2 \\ \text{misalkan } 7^b-1=a \text{ maka } 7^b=a+1 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(a) &= (a+1)^3-10 \\ f(5) &= (5+1)^3-10 \\ &= 6^3-10 \\ &= 216-10 \\ &= 206 \\ \end{align} </math> </div></div> <ol start=73> <li>Diketahui polinom <math>f(6^b-7)=6^{3b}-2 \cdot 6^{2b}-4</math>. tentukan nilai f(-2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(-2) &= f(6^b-7) \\ -2 &= 6^b-7 \\ 6^b &= 5 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(5-7) &= 5^3-2 \cdot 5^2-4 \\ f(-2) &= 125-50-4 \\ &= 71 \\ *cara 2 \\ \text{misalkan } 6^b-7=a \text{ maka } 6^b=a+7 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(a) &= (a+7)^3-2(a+7)^2-4 \\ f(-2) &= (-2+7)^3-2(-2+7)^2-4 \\ &= 5^3-2(5)^2-4 \\ &= 125-50-4 \\ &= 71 \\ \end{align} </math> </div></div> <ol start=74> <li>Jika <math>f(xy)=\frac{f(x)}{y}</math> dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(10) &= 7 \\ f(2 \cdot 5) &= 7 \\ f(xy) &= \frac{f(x)}{y} \\ f(2 \cdot 5) &= \frac{f(2)}{5} \\ 7 &= \frac{f(2)}{5} \\ f(2) &= 35 \\ \end{align} </math> </div></div> <ol start=75> <li>Jika <math>f(xy)=\frac{f(x+y)}{xy}</math> dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(15) &= 16 \\ f(3 \cdot 5) &= 16 \\ f(xy) &= \frac{f(x+y)}{xy} \\ f(3 \cdot 5) &= \frac{f(3+5)}{3 \cdot 5} \\ f(15) &= \frac{f(8)}{15} \\ 16 &= \frac{f(8)}{15} \\ f(8) &= 240 \\ \end{align} </math> </div></div> <ol start=76> <li>Jika <math>f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15</math> maka tentukan nilai f(16)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x+\frac{1}{x}+6) &= x^2+\frac{1}{x^2}+15 \\ &= (x+\frac{1}{x})^2-2+15 \\ &= (x+\frac{1}{x})^2+13 \\ \text{misalkan } x+\frac{1}{x} &= p \\ f(x+\frac{1}{x}+6) &= (x+\frac{1}{x})^2+13 \\ f(p+6) &= p^2+13 \\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 } \\ f(p+6) &= p^2+13 \\ f(10+6) &= 10^2+13 \\ f(16) &= 100+13 \\ &= 113 \\ \end{align} </math> </div></div> <ol start=77> <li>Tentukan nilai x jika <math>f(x)=\frac{4}{4-x}</math> dan <math>f(x \cdot f(x))^{\frac{f(4x)}{f(x)}}=256</math>!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{4}{4-x} \\ f(4x) &= \frac{4}{4-4x} \\ \frac{f(4x)}{f(x)} &= \frac{\frac{4}{4-4x}}{\frac{4}{4-x}} \\ &= \frac{4-x}{4-4x} \\ f(x \cdot f(x)) &= f(x(\frac{4}{4-x})) \\ &= f(\frac{4x}{4-x}) \\ &= \frac{4}{4-(\frac{4x}{4-x})} \\ &= \frac{4}{\frac{16-4x-4x}{4-x}} \\ &= \frac{4}{\frac{16-8x}{4-x}} \\ &= \frac{4(4-x)}{4(4-4x)} \\ &= \frac{4-x}{4-4x} \\ \text{misalkan } \frac{4-x}{4-4x} &= a \\ f(x \cdot f(x))^{\frac{f(4x)}{f(x)}} &= 256 \\ a^a &= 256 \\ a^a &= 4^4 \\ a &= 4 \\ \frac{4-x}{4-4x} &= 4 \\ 4-x &= 16-16x \\ 15x &= 12 \\ x &= \frac{4}{5} \\ \end{align} </math> </div></div> <ol start=78> <li>Fungsi <math>f(x) = \frac{kx}{2x+1} \text{dengan } x \neq -\frac{1}{2}</math>. Dengan f(f(x)) = x maka tentukan nilai k!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{kx}{2x+1} \\ f(f(x)) &= x \\ f(\frac{kx}{2x+1}) &= x \\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1} &= x \\ \frac{\frac{k^2x}{2x+1}}{\frac{2kx+2x+1}{2x+1}} &= x \\ \frac{k^2x}{2kx+2x+1} &= x \\ \frac{k^2}{2kx+2x+1} &= 1 \\ k^2 &= 2kx+2x+1 \\ k^2-2kx &= 2x+1 \\ k^2-2kx+x^2 &= x^2+2x+1 \\ (k-x)^2 &= (x+1)^2 \\ (k-x)^2-(x+1)^2 &= 0 \\ (k-x+x+1)(k-x-(x+1)) &= 0 \\ k=-1 &\text{ atau } k=2x+1 &\text{ (TM) } \\ \end{align} </math> </div></div> <ol start=79> <li>Jika n = 2023²+2024² maka berapa hasil dari <math>\sqrt{2n-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n &= 2023^2+2024^2 \\ &= 2023^2+(2023+1)^2 \\ \text{misalkan 2023 = p} \\ n &= p^2+(p+1)^2 \\ &= p^2+p^2+2p+1 \\ &= 2p^2+2p+1 \\ \sqrt{2n-1} &= \sqrt{2(2p^2+2p+1)-1} \\ &= \sqrt{4p^2+4p+2-1} \\ &= \sqrt{4p^2+4p+1} \\ &= \sqrt{(2p+1)^2} \\ &= 2p+1 \\ &= 2(2023)+1 \\ &= 4046+1 \\ &= 4047 \\ \end{align} </math> </div></div> <ol start=80> <li>Tentukan nilai dari a+b+c merupakan bilangan bulat positif jika ab = 2, bc = 3 dan ac = 6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab \cdot bc \cdot ac &= 2 \cdot 3 \cdot 6 \\ (abc)^2 &= 36 \\ abc &= \pm 6 \\ abc &= 6 \\ \frac{abc}{ab} &= c = \frac{6}{2} = 3 \\ \frac{abc}{bc} &= a = \frac{6}{3} = 2 \\ \frac{abc}{ac} &= b = \frac{6}{6} = 1 \\ a+b+c &= 6 \\ \end{align} </math> </div></div> <ol start=81> <li>Tentukan nilai dari (a-c)^b jika <math>\frac{ab}{a+b} = \frac{1}{3}</math>, <math>\frac{bc}{b+c} = \frac{1}{4}</math> dan <math>\frac{ac}{a+c} = \frac{1}{9}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{ab}{a+b} &= \frac{1}{3} \\ \frac{a+b}{ab} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{b} + \frac{1}{a} &= 3 \\ \frac{bc}{b+c} &= \frac{1}{4} \\ \frac{b+c}{bc} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{b} &= 4 \\ \frac{ac}{a+c} &= \frac{1}{9} \\ \frac{a+c}{ac} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{a} &= 9 \\ \text{Misalkan 1/a = x, 1/b = y dan 1/c = z} \\ x+y &= 3 \\ y+z &= 4 \\ x+z &= 9 \\ x+y &= 3 \\ y+z &= 4 \\ x-z &= -1 \\ x-z &= -1 \\ x+z &= 9 \\ 2x &= 8 \\ x &= 4 \\ x-z &= -1 \\ 4-z &= -1 \\ z &= 5 \\ x+y &= 3 \\ 4+y &= 3 \\ y &= -1 \\ \frac{1}{a} &= 4 \\ a &= \frac{1}{4} \\ \frac{1}{b} &= -1 \\ b &= -1 \\ \frac{1}{c} &= 5 \\ c &= \frac{1}{5} \\ (a-c)^b &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=82> <li>Tentukan nilai dari a, b dan c jika <math>\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}</math> dan a+2b+3c=28?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan k untuk semua ketiga persamaan tersebut } \\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5} &= k \\ a+b &= 2k \\ a+c &= 4k \\ b+c &= 5k \\ 2a+b+c &= 6k \\ 2a+5k &= 6k \\ k &= 2a \\ a &= \frac{k}{2} \\ b &= \frac{3k}{2} \\ c &= \frac{7k}{2} \\ a+2b+3c &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ a &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ b &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ c &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=83> <li>Tentukan nilai dari (b+c)^a jika <math>\frac{a+b+c}{2} = \sqrt{a-2}+\sqrt{b-1}+\sqrt{c}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{a+b+c}{2} &= \sqrt{a-2}+\sqrt{b-1}+\sqrt{c} \\ a+b+c &= 2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}) \\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c} &= 0 \\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1 &= 0 \\ (\sqrt{a-2}-1)^2+(\sqrt{b-1}-1)^2+(\sqrt{c}-1)^2 &= 0 \\ (\sqrt{a-2}-1)^2 &= 0 \\ \sqrt{a-2}-1 &= 0 \\ \sqrt{a-2} &= 1 \\ a-2 &= 1 \\ a &= 3 \\ (\sqrt{b-1}-1)^2 &= 0 \\ \sqrt{b-1}-1 &= 0 \\ \sqrt{b-1} &= 1 \\ b-1 &= 1 \\ b &= 1 \\ (\sqrt{c}-1)^2 &= 0 \\ \sqrt{c}-1 &= 0 \\ \sqrt{c} &= 1 \\ c &= 1 \\ (b+c)^a &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=84> <li>x dan y merupakan bilangan tak nol. Jika xy = <math>\frac{x}{y}</math> = x-y maka berapa nilai x+y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy &= \frac{x}{y} \\ y^2 &= 1 \\ y^2 - 1 &= 0 \\ (y-1)(y+1) &= 0 \\ y = 1 &\text{ atau } y = -1 \\ \frac{x}{y} &= x-y \\ x &= xy-y^2 \\ x-xy &= -y^2 \\ x(1-y) &= -y^2 \\ x &= \frac{-y^2}{1-y} \\ \text{cek y=1 } \\ x &= \frac{-1^2}{1-1} \\ \text{tidak memenuhi syarat } \\ \text{cek y=-1 } \\ x &= \frac{-(-1)^2}{1-(-1)} \\ &= \frac{-1}{2} \\ x+y &= -1-\frac{1}{2} \\ &= -\frac{3}{2} \\ \end{align} </math> </div></div> <ol start=86> <li>Berapa nilai x dari <math>(\frac{a}{b})^3+(\frac{b}{a})^3 = 2\sqrt{x}</math> jika <math>\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b} \\ \frac{a+b}{ab} &= \frac{1}{a+b} \\ (a+b)^2 &= ab \\ a^2+2ab+b^2 &= ab \\ a^2+b^2 &= -ab \\ \text{misalkan } \frac{a}{b}+\frac{b}{a} = n \\ \frac{a}{b}+\frac{b}{a} &= n \\ \frac{a^2+b^2}{ab} &= n \\ a^2+b^2 &= nab \\ n &= -1 \\ \frac{a}{b}+\frac{b}{a} &= n \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3(\frac{a}{b}+\frac{b}{a}) &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3n &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3 &= n^3-3n \\ &= (-1)^3-3(-1) \\ &= 2 \\ 2\sqrt{x} &= 2 \\ \sqrt{x} &= 1 \\ x &= 1 \\ \end{align} </math> </div></div> <ol start=87> <li>Berapa nilai m dari <math>x^2-mx-1=0</math> jika <math>\sqrt[3]{x_1}+\sqrt[3]{x_2}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x_1} &= a \\ x_1 &= a^3 \\ \sqrt[3]{x_2} &= b \\ x_2 &= b^3 \\ \sqrt[3]{x_1}+\sqrt[3]{x_2} &= 1 \\ a+b &= 1 \\ x^2-mx-1 &= 0 \\ x_1+x_2 &= m \\ x_1 \cdot x_2 &= -1 \\ x_1+x_2 &= m \\ a^3+b^3 &= m \\ x_1 \cdot x_2 &= -1 \\ a^3 \cdot b^3 &= -1 \\ (ab)^2 &= (-1)^3 \\ ab &= -1 \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ (1)^3 &= m+3(-1)(1) \\ 1 &= m-3 \\ m &= 4 \\ \end{align} </math> </div></div> <ol start=88> <li>Berapa nilai <math>\frac{x_1}{x_2}</math> dari <math>ax^2-18x-b=0</math> jika <math>ab=45</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab &= 45 \\ b &= \frac{45}{a} \\ ax^2-18x-b &= 0 \\ ax^2-18x-\frac{45}{a} &= 0 \\ a^2x^2-18ax-45 &= 0 \\ (ax-3)(ax-15) &= 0 \\ ax-3 &= 0 \\ x &= \frac{3}{a} \\ ax-15 &= 0 \\ x &= \frac{15}{a} \\ \frac{x_1}{x_2} &= \frac{\frac{3}{a}}{\frac{15}{a}} \\ &= \frac{3}{15} \\ &= \frac{1}{5} \\ \frac{x_1}{x_2} &= \frac{\frac{15}{a}}{\frac{3}{a}} \\ &= \frac{15}{3} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=89> <li>Jika <math>\frac{u_3}{u_1+u_2} = \frac{7}{8}</math> merupakan barisan aritmetika maka berapa dari <math>\frac{u_2+u_3}{u_1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{u_3}{u_1+u_2} &= \frac{7}{8} \\ \frac{a+2b}{a+a+b} &= \frac{7}{8} \\ \frac{a+2b}{2a+b} &= \frac{7}{8} \\ 8(a+2b) &= 7(2a+b) \\ 8a+16b &= 14a+7b \\ 9b &= 6a \\ b &= \frac{2a}{3} \\ \frac{u_2+u_3}{u_1} &= \frac{a+b+a+2b}{a} \\ &= \frac{2a+3b}{a} \\ &= \frac{2a+3(\frac{2a}{3})}{a} \\ &= \frac{2a+2a}{a} \\ &= \frac{4a}{a} \\ &= 4 \\ \end{align} </math> </div></div> <ol start=90> <li>Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7p+q}{2p+q} &= \frac{17p+q}{7p+q} \\ (7p+q)^2 &= (17p+q)(2p+q) \\ 49p^2+14pq+q^2 &= 34p^2+19pq+q^2 \\ 15p^2 &= 5pq \\ 3p &= q \\ \frac{7p+q}{2p+q} &= \frac{7p+3p}{2p+3p} \\ &= \frac{10p}{5p} \\ &= 2 \\ \end{align} </math> </div></div> <ol start=91> <li>Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{rataan geometris } \\ \sqrt{a \cdot b} &= b-24 \\ a \cdot b &= (b-24)^2 \\ \text{rataan aritmatik } \\ \frac{a+b}{2} &= a+15 \\ a+b &= 2(a+15) \\ a+b &= 2a+30 \\ a &= b-30 \\ a \cdot b &= (b-24)^2 \\ (b-30)b &= (b-24)^2 \\ b^2-30b &= b^2-48b+576 \\ 18b &= 576 \\ b &= 32 \\ a &= b-30 \\ &= 32-30 \\ &= 2 \\ a+b &= 32+2 \\ &= 34 \\ \end{align} </math> </div></div> <ol start=91> <li>Segitiga lancip ABC dengan <math>\frac{a^4+b^4+c^4+a^2b^2}{c^2(a^2+b^2)}=2</math>. tentukan nilai sudut C?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{syarat segitiga lancip semua sudut masing-masing kurang dari } 90^\circ \\ c^2 &= a^2+b^2-2ab cos C \\ cos C &= \frac{a^2+b^2-c^2}{2ab} \\ a^4+b^4+c^4+a^2b^2 &= 2c^2(a^2+b^2) \\ a^4+b^4+a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-2c^2(a^2+b^2)+(c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= (ab)^2 \\ a^2+b^2-c^2 &= \pm ab \\ cos C &= \pm \frac{ab}{2ab} \\ &= \pm \frac{1}{2} \\ &= \frac{1}{2} \text{ (karena sudut harus kurang dari } 90^\circ) \\ C &= 60^\circ \\ \end{align} </math> </div></div> <ol start=92> <li>Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan panjang DA dan EA } = x \text{ dan panjang AB } = y \\ \text{luas segitiga CAB } &= \frac{CA \cdot AB}{2} \\ 45 &= \frac{(x+9)(x+5)}{2} \\ 90 &= x^2+14x+45 \\ x^2+14x &= 45 \\ y^2 &= (x+9)^2+(x+5)^2 \\ &= x^2+18x+81+x^2+10x+25 \\ &= 2x^2+28x+106 \\ &= 2(x^2+14x)+106 \\ &= 2(45)+106 \\ &= 196 \\ y &= 14 \\ \end{align} </math> jadi panjang BC adalah 14 cm </div></div> <ol start=93> <li>Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kita cari ukuran GC yaitu } \\ \frac{GC}{AD} &= \frac{CE}{DE} \\ \frac{GC}{15} &= \frac{6}{18} \\ GC &= 5 \\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC } \\ &= \frac{1}{2} \cdot BC \cdot CF - \frac{1}{2} \cdot GC \cdot CE \\ &= \frac{1}{2} \cdot 15 \cdot 18 - \frac{1}{2} \cdot 5 \cdot 6 \\ &= 135 - 15 \\ &= 120 \\ \end{align} </math> jadi luas daerah BFEG adalah 120 cm² </div></div> <ol start=94> <li>Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ AI &= \frac{1}{5} AF \\ &= \frac{1}{5} 10 \\ &= 2 \\ IF &= AF-AI \\ &= 10-2 \\ &= 8 \\ \text{luas trapesium AFGD } &= \frac{(AD+EF) \cdot AF}{2} \\ &= \frac{(4+6)10}{2} \\ &= 50 \\ \text{luas segitiga AID } &= \frac{AI \cdot AF}{2} \\ &= \frac{(2)4}{2} \\ &= 4 \\ \text{luas segitiga IFG } &= \frac{IF \cdot FG}{2} \\ &= \frac{(8)6}{2} \\ &= 24 \\ \text{luas daerah segitiga IGD } &= \text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG } \\ &= 50-4-24 \\ &= 22 \\ \end{align} </math> jadi luas daerah segitiga IGD adalah 22 cm² </div></div> <ol start=96> <li>Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-<math>\pi</math>) cm³ maka berapa luas permukaan kerucut tersebut?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ \text{volume balok} \\ V_b &= x^2(12) \\ \text{volume kerucut} \\ V_b &= \frac{1}{3}\pi x^2(12) \\ &= 4\pi x^2 \\ V_{b-k} &= Vb-Vk \\ 100(3-\pi) &= 12x^2-4\pi x^2 \\ 100(3-\pi) &= 4x^2(3-\pi) \\ x^2 &= 25 \\ x &= 5 \\ s &= \sqrt{12^2+5^2} \\ &= \sqrt{144+25} \\ &= \sqrt{169} \\ &= 13 \\ \text{luas permukaan kerucut } &= \pi r(r+s) \\ &= \pi(5)(5+13) \\ &= 90\pi \\ \end{align} </math> jadi luas daerah permukaan kerucut adalah 90<math>\pi</math> cm² </div></div> <ol start=97> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 3a+1 \\ B &= 3b+2 \\ A(A+1)+3B \\ (3a+1)(3a+1+1)+3(3b+2) \\ (3a+1)(3a+2)+9b+6 \\ 9a^2+9a+2+9b+6 \\ 9a^2+9a+9b+8 \\ 9(a^2+a+b)+8 \\ \text{sisa pembagiannya adalah } 8 \\ \end{align} </math> </div></div> <ol start=98> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 9a+7 \\ B &= 9b+8 \\ A(A-5)+9B \\ (9a+7)(9a+7-5)+9(9b+8) \\ (9a+7)(9a+2)+81b+72 \\ 81a^2+81a+14+81b+72 \\ 81a^2+81a+81b+86 \\ 81a^2+81a+81b+81+5 \\ 81(a^2+a+b+1)+5 \\ \text{sisa pembagiannya adalah } 5 \\ \end{align} </math> </div></div> <ol start=99> <li>Jika <math>\begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix}</math> maka berapa hasil dari A²¹+A²⁵+A⁴⁶?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A^2 &= A \cdot A \\ &= \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \\ A^3 &= A^2 \cdot A \\ &= \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ &= - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= -I \\ A^{21}+A^{25}+A^{46} &= A^{21} \cdot (I+A^4+A^{25}) \\ &= A^{21} \cdot (I+A^3 \cdot A +A^{24} \cdot A) \\ &= (A^3)^7 \cdot (I+A^3 \cdot A +(A^3)^8 \cdot A) \\ &= (-I)^7 \cdot (I-I \cdot A +(-I)^8 \cdot A) \\ &= -I \cdot (I-A+A) \\ &= -I \cdot I \\ &= -I \\ &= -\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ \end{align} </math> </div></div> <ol start=100> <li>Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida?</li></ol> : bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15 : ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9). : ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8) : maka Ida menulis bilangan 8. <ol start=101> <li>Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9?</li></ol> : Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu. : untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja. : untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja. : Jadi banyaknya bilangan mungkin 2. <ol start=102> <li>Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli!</li></ol> ;cara 1 # 8¹ = 1 # 8² = 1 (8²=8¹x8¹ sama dengan 1x1) # 8³ = 1 (8³=8¹x8² sama dengan 1x1) # 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2) # 8⁵ = 1 # 8ⁿ = 1 (semua n untuk bilangan asli) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli ;cara 2 # 8ⁿ = b mod 7 # 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil) # (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya) # 8ⁿ = 1ⁿ mod 7 # 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli <ol start=103> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 5?</li></ol> ;cara 1 # 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 # 7¹ = 7 (sisa 1) # 7² = 49 (sisa 2) # 7³ = 343 (sisa 3) # 7⁴ = 2,401 (sisa 0) # 7⁵ = 16,807 # 7⁶ = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3 ;cara 2 :17¹ = 2 :17² = 4 :17³ = 3 :17⁴ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3 :17⁹⁹ = (17⁴)²⁴ x 17³ Untuk 17⁴ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 3. Jadi 17⁹⁹ dibagi 7 bersisa 3 ;cara 3 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 17⁴ ::17⁴ = 1 mod 5 ::(17⁴)²⁴ = 1²⁴ mod 5 ::17⁹⁶ = 1²⁴ mod 5 ::17⁹⁶ = 1 mod 5 ::17⁹⁶ x 17³ = 1 x 17³ mod 5 ::17⁹⁹ = 17³ mod 5 ::17⁹⁹ = 17 x 17 x 17 mod 5 ::17⁹⁹ = 2 x 2 x 2 mod 5 ::17⁹⁹ = 8 mod 5 ::17⁹⁹ = 3 mod 5 Jadi 17⁹⁹ dibagi 5 bersisa 3 <ol start=104> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 7?</li></ol> ;cara 1 :17¹ = 3 :17² = 2 :17³ = 6 :17⁴ = 4 :17⁵ = 5 :17⁶ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3 :17⁹⁹ = (17⁶)¹⁶ x 17³ Untuk 17⁶ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 6. Jadi 17⁹⁹ dibagi 7 bersisa 6 ;cara 2 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 17⁶ ::17⁶ = 1 mod 7 ::(17⁶)¹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1 mod 7 ::17⁹⁶ x 17³ = 1 x 17³ mod 7 ::17⁹⁹ = 17³ mod 7 ::17⁹⁹ = 17 x 17 x 17 mod 7 ::17⁹⁹ = 3 x 3 x 3 mod 7 ::17⁹⁹ = 27 mod 7 ::17⁹⁹ = 6 mod 7 Jadi 17⁹⁹ dibagi 7 bersisa 6 <ol start=105> <li>Berapa hasil sisa dari 41²⁰²⁴ dibagi 33?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 41^{2024} &= 41^{2024} \text{ mod } 33 \\ &= (33 \times 3 + 2)^{2024} \text{ mod } 33 \\ &= 2^{2024} \text{ mod } 33 \\ &= 2^{2020} 2^4 \text{ mod } 33 \\ &= (2^5)^{404} 2^4 \text{ mod } 33 \\ &= (33 - 1)^{404} 2^4 \text{ mod } 33 \\ &= (-1)^{404} 2^4 \text{ mod } 33 \\ &= 2^4 \text{ mod } 33 \\ &= 16 \text{ mod } 33 \\ \text{Jadi hasil sisa adalah } 16 \\ \end{align} </math> </div></div> <ol start=106> <li>Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 99^{99} &= (3^2 \times 11)^{99} \\ &= 3^{198} \times 11^{99} \\ 243^n &= (3^5)^n \\ &= 3^{5n} \\ \text{agar bisa membagi, maka} \\ 5n &= 198 \\ n &= 39.6 \\ \text{jadi bilangan n terbesar adalah } 39 \\ \end{align} </math> </div></div> <ol start=107> <li>Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸?</lu></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 88^{88} &= (8 \times 11)^{88} \\ &= 8^{88} \times 11^{88} \\ &= 8^{87} \times 8 \times 11^{88} \\ &= (8^3)^{29} \times 8 \times 11^{88} \\ &= 512^{29} \times 8 \times 11^{88} \\ 512^n &= 512^{29} \\ \text{jadi bilangan n terbesar adalah } 29 \\ \end{align} </math> </div></div> <ol start=108> <li>Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6!</li></ol> ; cara 1 : KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. : N dibagi 3 sisa 1 : N dibagi 5 sisa 2 : N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a : KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70 : KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21 : KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97 ; cara 2 : Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35 : kemudian ditambahkan sisa masing-masing sesuai dengan KPK. : KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, <b>97</b> : KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, <b>97</b> : KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, <b>97</b> Jadi bilangan bulat positif adalah 97 :: NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 <ol start=109> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |- | 4 || 0 || Semua isi ember B dibuang |- | 1 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. <ol start=110> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |} nah ada ember A berisi 4 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |- | 1 || 0 || Semua isi ember B dibuang |- | 0 || 1 || Tuangkan 1 l dari ember A ke B sehingga ember A kosong |- | 3 || 1 || Isikan 3 l ke ember A |- | 0 || 4 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |} nah ada ember B berisi 4 liter. # Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kemungkinan 1} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 +3 \cdot ((\frac{4}{3})^x)^2 &= 7 \cdot (\frac{4}{3})^x \\ \text{misalkan } (\frac{4}{3})^x = y \\ 4 +3y^2 &= 7y \\ 3y^2-7y+4 &= 0 \\ (3y-4)(y-1) &= 0 \\ y_1 = \frac{4}{3} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{4}{3} \\ (\frac{4}{3})^x &= \frac{4}{3} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{4}{3})^x &= 1 \\ x &= 0 \\ \text{kemungkinan 2} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot ((\frac{3}{4})^x)^2 +3 &= 7 \cdot (\frac{3}{4})^x \\ \text{misalkan } (\frac{3}{4})^x = y \\ 4y^2 +3 &= 7y \\ 4y^2-7y+3 &= 0 \\ (4y-3)(y-1) &= 0 \\ y_1 = \frac{3}{4} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{3}{4} \\ (\frac{3}{4})^x &= \frac{3}{4} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{3}{4})^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-6}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{2x-4}{3}+2x-6 &= \frac{3}{2x-4}+\frac{1}{2x-6} \\ \frac{2x-4}{3}+2x-6 &= \frac{1}{\frac{2x-4}{3}}+\frac{1}{2x-6} \\ \text{misalkan } \frac{2x-4}{3}+2x-6 = y \text{ dan } 2x-6 = z \\ y+z &= \frac{1}{y}+\frac{1}{z} \\ y+z &= \frac{y+z}{xy} \\ y+z-\frac{y+z}{xy} &= 0 \\ (y+z)(1-\frac{1}{xy} &= 0 \\ \end{align} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 2uqgz23a7z8qppbiz2503nu923ejwp6 117492 117491 2026-07-13T04:58:38Z Akuindo 8654 117492 wikitext text/x-wiki contoh soal <ol start=1> <li>Berapa hasil dari <math>\sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Misalkan 2020 = p} \\ \sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64} &= \sqrt{(2020-5) \cdot (2020-3) \cdot (2020+3) \cdot (2020+5) + 64} \\ &= \sqrt{(p-5) \cdot (p-3) \cdot (p+3) \cdot (p+5) + 64} \\ &= \sqrt{(p-5) \cdot (p+5) \cdot (p-3) \cdot (p+3) + 64} \\ &= \sqrt{(p^2-25) \cdot (p^2-9) + 64} \\ &= \sqrt{p^4-34p^2+ 225 + 64} \\ &= \sqrt{p^4-34p^2+ 289} \\ &= \sqrt{(p^2-17)^2} \\ &= p^2-17 \\ &= 2020^2-17 \\ &= (2000+20)^2-17 \\ &= 4.000.000+80.000+400-17 \\ &= 4.080.383 \\ \end{align} </math> </div></div> <ol start=2> <li>Berapa nilai x dari <math>\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} = \frac{9}{10}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} &= \frac{9}{10} \\ \text{misalkan untuk } \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} = p \\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} &= p \\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}} &= p^2 \\ x^2-x-p &= p^2 \\ x^2-2x+1+x-1 &= p^2+p \\ (x-1)^2+(x-1) &= p^2+p \\ x-1 &= p \\ \text{misalkan untuk } \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}} &= q^3 \\ x^2 q &= q^3 \\ x^2 &= q^2 \\ x &= q \\ \frac{x-1}{x} &= \frac{9}{10} \\ x &= 10 \\ \end{align} </math> </div></div> <ol start=3> <li>Berapa nilai x dari <math>(\frac{x}{x+10})^{x+10}=\frac{1}{1024}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{x+10}{x})^{-(x+10)} &= (1024)^{-1} \\ (\frac{x+10}{x})^{x+10} &= 1024 \\ (\frac{x+10}{x})^{x+10} &= 2^{10} \\ (\frac{x+10}{x})^{\frac{x+10}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (\frac{1}{2})^{-1} \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (1+(-\frac{1}{2}))^{(1+(-\frac{2}{1}))} \\ \frac{10}{x} &= -\frac{1}{2} \\ x &= -20 \\ \end{align} </math> </div></div> <ol start=4> <li>Berapa nilai x dari <math>x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{2}+\sqrt{x+\frac{1}{4}} &= (\sqrt{x+\frac{1}{4}})^2+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+(\frac{1}{2})^2 \\ &= (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 \\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}} &= 4 \\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2} &= 4 \\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 4 \\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 &= 4 \\ \sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 2 \\ \sqrt{x+\frac{1}{4}} &= \frac{3}{2} \\ x+\frac{1}{4} &= \frac{9}{4} \\ x &= 2 \\ \end{align} </math> </div></div> <ol start=5> <li>Berapa nilai x dari <math>\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^3}{\sqrt{8-x^2}}+x^2-8 &= 0 \\ \frac{x^3}{\sqrt{8-x^2}} &= 8-x^2 \\ x^3 &= (8-x^2)^{\frac{3}{2}} \\ x &= (8-x^2)^{\frac{1}{2}} \\ x^2 &= 8-x^2 \\ 2x^2-8 &= 0 \\ x^2-4 &= 0 \\ (x-2)(x+2) &= 0 \\ \text{membuktikan } \\ x=2 \text{ maka hasilnya 0 } \\ x=-2 \text{ maka hasilnya -8 } \\ \text{jadi } x=2 \\ \end{align} </math> </div></div> <ol start=6> <li>Berapa nilai x dari <math>\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} = 5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} &= 5 \\ \frac{x^{50}+x^{60}+x^{70}}{31}} &= 5^5 \\ x^{50}+x^{60}+x^{70} &= 5^5 \cdot 31 \\ x^{50}(1+x^{10}+x^{20}) &= 5^5 \cdot 31 \\ (x^{10}^5)(1+x^{10}+(x^{10}^2) &= 5^5 \cdot 31 \\ \text{ misalkan } x^{10} = a \\ a^5(1+a+a^2) &= 5^5 \cdot 31 \\ a &= 5 \\ x^{10} &= 5 \\ x &= ^5 log 10 \\ \end{align} </math> </div></div> <ol start=7> <li>Berapa nilai x dari <math>\sqrt{3x+5+\sqrt{4x+5}} = x</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{3x+5+\sqrt{4x+5}} &= x \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \text{misalkan } \sqrt{4x+5}=y \text{ dan } 4x+5=y^2 \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \sqrt{y^2+y-x} &= x \\ y^2+y &= x^2+x \\ y=x \\ 4x+5 &= y^2 \\ 4x+5 &= x^2 \\ x^2-4x-5 &= 0 \\ (x-5)(x+1) &= 0 \\ x=5 &\text{ atau } x=-1 \text{ (TM) } \\ \end{align} </math> </div></div> <ol start=8> <li>Berapa nilai x dari <math>\sqrt{1+\sqrt{1+x}} = \sqrt[3]{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{1+\sqrt{1+x}} &= \sqrt[3]{x} \\ \sqrt[3]{x} &= n \\ x &= n^3 \\ \sqrt{1+\sqrt{1+n^3}} &= n \\ 1+\sqrt{1+n^3} &= n^2 \\ \sqrt{1+n^3} &= n^2-1 \\ 1+n^3 &= n^4-2n^2+1 \\ n^4-n^3-2n^2 &= 0 \\ n^2(n^2-n-2) &= 0 \\ n^2(n-2)(n+1) &= 0 \\ n=0, n=2 \text{ atau } n=-1 \\ n &= 0 \\ x &= 0^3 \\ &= 0 \\ n &= 2 \\ x &= 2^3 \\ &= 8 \\ n &= -1 \\ x &= (-1)^3 \\ &= -1 \\ \text{yang paling mungkin untuk nilai x adalah } 8 \\ \end{align} </math> </div></div> <ol start=9> <li>Berapa nilai x dari <math>\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} &= \frac{\sqrt{1+x}}{\sqrt{x}} \\ \sqrt{x}(\sqrt{1+x}+\sqrt{x}) &= (\sqrt{1+x}-\sqrt{x})\sqrt{1+x} \\ \sqrt{x(1+x)}+x &= 1+x-\sqrt{x(1+x)} \\ 2\sqrt{x(1+x)} &= 1 \\ \sqrt{x(1+x)} &= \frac{1}{2} \\ x(1+x) &= \frac{1}{4} \\ x^2+x &= \frac{1}{4} \\ 4x^2+4x &= 1 \\ 4x^2+4x-1 &= 0 \\ x &= \frac{-4 \pm \sqrt{4^2-4(4)(-1)}}{2(4)} \\ &= \frac{-4 \pm \sqrt{32}}{8} \\ &= \frac{-4 \pm 4\sqrt{2}}{8} \\ &= \frac{-1 \pm \sqrt{2}}{2} \\ \text{karena akar x harus minimal nol jadi } x = \frac{-1+\sqrt{2}}{2} \\ \end{align} </math> </div></div> <ol start=10> <li>Berapa nilai x dari <math>\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x-\sqrt{x+1}}{x+\sqrt{x+1}} &= \frac{11}{19} \\ \text{misalkan } \sqrt{x+1}=y \text{ dan } x=y^2-1 \\ \frac{y^2-1-y}{y^2-1+y} &= \frac{11}{19} \\ 19(y^2-y-1) &= 11(y^2+y-1) \\ 19y^2-19y-19 &= 11y^2+11y-11 \\ 8y^2-30y-8 &= 0 \\ 4y^2-15y-4 &= 0 \\ (4y+1)(y-4) &= 0 \\ y=-\frac{1}{4} \text{ (TM) atau } & y=4 \\ x &= 4^2-1 \\ &= 15 \\ \end{align} </math> </div></div> <ol start=11> <li>Berapa nilai x dari <math>\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} &= 98 \\ \text{misalkan } \sqrt{x^2-1}=y \\ \frac{x+y}{x-y}+\frac{x-y}{x+y} &= 98 \\ \frac{(x+y)^2+(x-y)^2}{(x-y)(x+y)} &= 98 \\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} &= 98 \\ \frac{2(x^2+y^2)}{x^2-y^2} &= 98 \\ \frac{x^2+y^2}{x^2-y^2} &= 49 \\ x^2+y^2 &= 49(x^2-y^2) \\ x^2+y^2 &= 49x^2-49y^2 \\ 48x^2 &= 50y^2 \\ 24x^2 &= 25y^2 \\ 24x^2 &= 25(\sqrt{x^2-1})^2 \\ 24x^2 &= 25(x^2-1) \\ 24x^2 &= 25x^2-25 \\ x^2 &= 25 \\ x &= \pm 5 \\ \end{align} </math> </div></div> <ol start=12> <li>Berapa nilai x dari <math>\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \sqrt{x}=y \text{ dan } x=y^2 \\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} &= \frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\sqrt{\frac{y^2}{y^2+y}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\frac{y}{\sqrt{y^2+y}} \\ y^2+y-\sqrt{(y^2+y)(y^2-y)} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^4-y^2} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^2(y^2-1)} &= \frac{5}{4}y \\ y(y+1)-y\sqrt{y^2-1} &= \frac{5}{4}y \\ y+1-\sqrt{y^2-1} &= \frac{5}{4} \\ -\sqrt{y^2-1} &= \frac{1}{4}-y \\ y^2-1 &= (\frac{1}{4}-y)^2 \\ y^2-1 &= \frac{1}{16}-\frac{1}{2}y+y^2 \\ -1 &= \frac{1}{16}-\frac{1}{2}y \\ \frac{1}{2}y &= \frac{1}{16}+1 \\ \frac{1}{2}y &= \frac{17}{16} \\ y &= \frac{17}{8} \\ x &= (\frac{17}{8})^2 \\ &= \frac{289}{64} \\ \end{align} </math> </div></div> <ol start=13> <li>Berapa nilai x dari <math>\sqrt[4]{62+x}+\sqrt[4]{275-x}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ misalkan } \sqrt[4]{62+x}=a, 62+x=a^4, \sqrt[4]{275-x}=b \text{ dan } 275-x=b^4 \\ a+b &= 7 \\ (a+b)^2 &= 49 \\ a^2+b^2+2ab &= 49 \\ a^2+b^2 &= 49-2ab \\ a^4+b^4 &= 62+x+275-x \\ (a^2+b^2)^2-2(ab)^2 &= 337 \\ (49-2ab)^2-2(ab)^2 &= 337 \\ 2401-196ab+4(ab)^2-2(ab)^2 &= 337 \\ 2(ab)^2-196ab+2064 &= 0 \\ (ab)^2-98ab+1032 &= 0 \\ (ab-12)(ab-86) &= 0 \\ ab = 12 \text{ atau } & ab = 86 \text{ (TM) karena hasil kali maksimum yaitu 12 } \\ ab =12 \text{ dan } a+b=7 \\ a+b &= 7 \\ b &= 7-a \\ ab &= 12 \\ a(7-a) &= 12 \\ -a^2+7a &= 12 \\ a^2-7a+12 &= 0 \\ (a-3)(a-4) &= 0 \\ a=3 \text{ atau } & a=4 \\ a=3, b=4 \\ 62+x &= a^4 \\ 62+x &= (3)^4 \\ 62+x &= 81 \\ x &= 19 \\ a=4, b=3 \\ 62+x &= a^4 \\ 62+x &= (4)^4 \\ 62+x &= 256 \\ x &= 194 \\ \end{align} </math> </div></div> <ol start=14> <li>Berapa nilai x dari <math>\sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2} &= 7 \\ (\sqrt[3]{8+x})^2-\sqrt[3]{8+x} \sqrt[3]{27-x}+(\sqrt[3]{27-x})^2 &= 7 \\ \text{misalkan } \sqrt[3]{8+x}=a, 8+x=a^3, \sqrt[3]{27-x}=b \text{ dan } 27-x=b^3 \\ a^2-ab+b^2 &= 7 \\ a^3+b^3 &= 8+x+27-x \\ &= 35 \\ a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ 35 &= (a+b)(7) \\ a+b &= 5 \\ b &= 5-a \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ 5^3 &= 35+3ab(5) \\ 125 &= 35+15ab \\ 80 &= 15ab \\ ab &= 6 \\ a(5-a) &= 6 \\ 5a-a^2 &= 6 \\ a^2-5a+6 &= 6 \\ (a-2)(a-3) &= 6 \\ a=2 &\text{ atau } a=3 \\ a=2, b=3 \text{ dan } a=3,b=2 \\ 8+x &= a^3 \\ &= 2^3 \\ &= 8 \\ x &= 0 \\ 8+x &= a^3 \\ &= 3^3 \\ &= 27 \\ x &= 19 \\ \end{align} </math> </div></div> <ol start=15> <li>Berapa nilai x dari <math>3^x+5^x-9^x+15^x-25^x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3^x+5^x-9^x+15^x-25^x &= 1 \\ 3^x+5^x-(3^2)^x+(3 \cdot 5)^x-(5^2)^x &= 1 \\ 3^x+5^x-(3^x)^2+(3^x \cdot 5^x)-(5^x)^2 &= 1 \\ \text{misalkan } 3^x=a \text{ dan } 5^x=b \\ a+b-a^2+ab-b^2 &= 1 \\ a^2-ab+b^2-a-b+1 &= 0 \\ 2a^2-2ab+2b^2-2a-2b+2 &= 0 \\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1 &= 0 \\ (a-b)^2+(a-1)^2+(b-1)^2 &= 0 \\ a-b=0; a-1=0; b-1 &= 0 \\ a=b &= 1 \\ 3^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> <ol start=16> <li>Berapa nilai x dari <math>^6log x^2+^{6x}log \frac{6}{x}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ \text{misalkan } 6x=a \text{ maka } x=\frac{a}{6} \\ ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ ^6log (\frac{a}{6})^2+^{6 \frac{a}{6}}log \frac{6}{\frac{a}{6}} &= 1 \\ ^6log \frac{a^2}{6^2}+^alog \frac{6^2}{a} &= 1 \\ ^6log a^2-^6log 6^2+^alog 6^2-^alog a &= 1 \\ 2 ^6log a-2 ^6log 6+2 ^alog 6-^alog a &= 1 \\ 2 ^6log a-2+2 \frac{1}{^6log a}-1 &= 1 \\ 2 ^6log a+2 \frac{1}{^6log a}-4 &= 0 \\ 2 ^6log^2 a-4 ^6log a+2 &= 0 \\ ^6log^2 a-2 ^6log a+1 &= 0 \\ (^6log a-1)^2 &= 0 \\ ^6log a &= 1 \\ a &= 6 \\ x &= \frac{a}{6} \\ &= \frac{6}{6} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=17> <li>Berapa nilai x dari (x+500)³+x=20?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+500)^3+x &= 20 \\ \text{misalkan } a=x+500 \text{ maka } x=a-500 \\ a^3+a-500 &= 20 \\ a^3+a &= 520 \\ a(a^2+1) &= 8 \cdot 65 \\ a(a^2+1) &= 8(64+1) \\ a(a^2+1) &= 8(8^2+1) \\ a &= 8 \\ x &= 8-500 \\ &= -492 \\ \end{align} </math> </div></div> <ol start=18> <li>Berapa nilai x dari <math>\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2} &= 0 \\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}} &= \frac{1}{2} \\ \frac{x^n+4^n}{x^n+16^n} &= (\frac{1}{2})^n \\ \frac{x^n+4^n}{x^n+16^n} &= \frac{1}{2^n} \\ 2^n(x^n+4^n) &= x^n+16^n \\ 2^n(x^n+2^{2n}) &= x^n+2^{4n} \\ 2^n \cdot x^n+2^{3n} &= x^n+2^{4n} \\ 2^n \cdot x^n-x^n &= 2^{4n}-2^{3n} \\ x^n(2^n-1) &= 2^{3n}(2^n-1) \\ x^n &= 2^{3n} \\ x^n &= (2^3)^n \\ x^n &= 8^n \\ x &= 8 \\ \end{align} </math> </div></div> <ol start=19> <li>Berapa hasil dari <math>\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ x &= \frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5 \cdot 6}+\sqrt{5 \cdot 5}+\sqrt{6 \cdot 4}+\sqrt{5 \cdot 4}}{\sqrt{5 \cdot 4}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5} \cdot \sqrt{6}+\sqrt{5} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{4}}{2 \cdot \sqrt{5}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{6} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{5}+\sqrt{5} \cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}} \\ &= \frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ &= \frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ \frac{1}{x} &= \frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}} \\ &= \frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ &= \frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1} \\ &= \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5} \\ &= \sqrt{6}-\sqrt{4} \\ &= \sqrt{6}-2 \\ x &= \frac{1}{\sqrt{6}-2} \\ &= \frac{\sqrt{6}+2}{6-4} \\ &= \frac{\sqrt{6}+2}{2} \\ &= 1+\frac{\sqrt{6}}{2} \\ \end{align} </math> </div></div> <ol start=20> <li>Berapa hasil dari <math>(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5 \\ \text{misalkan } x=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ x &= \frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ &= \frac{\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}} \\ &= \frac{\sqrt{3}+1}{4} \\ 4x &= \sqrt{3}+1 \\ 4x-1 &= \sqrt{3} \\ (4x-1)^2 &= 3 \\ 16x^2-8x+1 &= 3 \\ 16x^2 &= 8x+2 \\ 8x^2 &= 4x+1 \\ x^2 &= \frac{4x+1}{8} \\ *cara 1 \\ x^3 &= x \cdot x^2 \\ &= x(\frac{4x+1}{8}) \\ &= \frac{4x^2+x}{8} \\ &= \frac{4x^2}{8}+\frac{x}{8} \\ &= \frac{4(\frac{4x+1}{8})}{8}+\frac{x}{8} \\ &= \frac{16x+4}{64}+\frac{x}{8} \\ &= \frac{4x+1}{16}+\frac{x}{8} \\ &= \frac{4x+1+2x}{16} \\ &= \frac{6x+1}{16} \\ x^5 &= x^2 \cdot x^3 \\ &= (\frac{4x+1}{8})(\frac{6x+1}{16}) \\ &= \frac{24x^2+10x+1}{128} \\ &= \frac{24x^2}{128}+\frac{10x+1}{128} \\ &= \frac{24(\frac{4x+1}{8})}{128}+\frac{10x+1}{128} \\ &= \frac{96x+24}{1024}+\frac{10x+1}{128} \\ &= \frac{96x+24+80x+8}{1024} \\ &= \frac{176x+32}{1024} \\ &= \frac{176x}{1024}+\frac{32}{1024} \\ &= \frac{176}{1024}(\frac{\sqrt{3}+1}{4})+\frac{32}{1024} \\ &= \frac{44(\sqrt{3}+1)}{1024}+\frac{32}{1024} \\ &= \frac{44\sqrt{3}+44}{1024}+\frac{32}{1024} \\ &= \frac{76+44\sqrt{3}}{1024} \\ &= \frac{19+11\sqrt{3}}{256} \\ *cara 2 \\ x^4 &= (x^2)^2 \\ &= (\frac{4x+1}{8})^2 \\ &= \frac{16x^2+8x+1}{64} \\ &= \frac{16x^2}{64}+\frac{8x}{64}+\frac{1}{64} \\ &= \frac{x^2}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{\frac{4x+1}{8}}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{4x}{32}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{8}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{4}+\frac{3}{64} \\ x^5 &= x \cdot x^4 \\ &= (\frac{\sqrt{3}+1}{4})(\frac{x}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\frac{\sqrt{3}+1}{4}}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\sqrt{3}+1}{16}+\frac{3}{64}) \\ &= \frac{(\sqrt{3}+1)^2}{64}+(\frac{\sqrt{3}+1}{4})\frac{3}{64} \\ &= \frac{3+2\sqrt{3}+1}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{4+2\sqrt{3}}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{16+8\sqrt{3}}{256}+\frac{3\sqrt{3}+3}{256} \\ &= \frac{19+11\sqrt{3}}{256} \\ \end{align} </math> </div></div> <ol start=21> <li>Berapa hasil dari <math>\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4} &= \frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots) \\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots &= \frac{1}{1-\frac{1}{4}} \\ &= \frac{4}{3} \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{4}{3}) \\ &= \frac{1}{4}+\frac{16}{3} \\ &= \frac{67}{12} \\ x &= \frac{67}{9} \\ \end{align} </math> </div></div> <ol start=22> <li>Berapa nilai y-x jika <math>\frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} = \frac{x}{y}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} &= \frac{x}{y} \\ \frac{\frac{106 \times 107}{2}}{\frac{106}{2}(4+109)} &= \frac{x}{y} \\ \frac{53 \times 107}{53 \times 113} &= \frac{x}{y} \\ y-x &= 113-107 = 6 \\ \end{align} </math> </div></div> <ol start=23> <li>Berapa angka satuan dari hasil 17²⁰²⁴?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan angka satuannya} \\ 17^1 &= 7 \\ 17^2 &= 9 \\ 17^3 &= 3 \\ 17^4 &= 1 \\ 17^5 &= 7 \\ 17^6 &= 9 \\ 17^7 &= 3 \\ 17^8 &= 1 \\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1} \end{align} </math> </div></div> <ol start=24> <li>Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ 1! + 2! + 3! + 4! + \dots + 2024! &= 1 + (1x2) + (1x2x3) + (1x2x3x4) + \dots + 2024! \\ &= 1 + 2 + 6 + 24 + 120 + 720 + \dots + 2024! \\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka } 1+2+6+24 = 33 \text{ jadi angka satuannya adalah } 3 \end{align} </math> </div></div> <ol start=25> <li>Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ \frac{1! + 2! + 3! + 4! + \dots + 2024!}{12} &= \frac{1 + 1x2 + 1x2x3 + 1x2x3x4 + \dots + 2024!}{12} \\ &= \frac{1 + 2 + 6 + 24 + \dots + 2024!}{12} \\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi } 1+2+6 = 9 \end{align} </math> </div></div> <ol start=26> <li>Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{penjumlahan} \\ 1+1+1+1+ \dots \text{ (sebanyak 88 buah) }+x+y &= A \\ 88+x+y &= A \\ \text{perkalian} \\ 1 \times 1 \times 1 \times \dots \text{ (sebanyak 88 buah) }\times x \times y &= A \\ x \times y &= A \\ 88+x+y &= xy \\ xy-y &= 88+x \\ y(x-1) &= 88+x \\ y &= \frac{88+x}{x-1} \\ \text{uji selidiki untuk x=2} \\ y &= \frac{88+2}{2-1} \\ &= 90 \\ \text{buktikan} \\ 88+x+y &= xy \\ 88+2+90 &= 2(90) \\ 180 &= 180 \\ \text{terbukti} \\ \text{nilai A adalah } 180 \\ \end{align} </math> </div></div> <ol start=27> <li>Berapakah nilai x, y dan z dari <math>x+y-z=1, x^2+y^2-z^2=-5 \text{ dan } x^3+y^3-z^3=-53</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+y-z &= 1 \\ x+y &= z+1 \\ x^2+2xy+y^2 &= z^2+2z+1 \\ x^2+y^2-z^2 &= 2z+1-2xy \\ -5 &= 2z+1-2xy \\ 2xy &= 2z+6 \\ xy &= z+3 \\ x^2+y^2-z^2 &= -5 \\ x^2+y^2 &= z^2-5 \\ x^3+y^3-z^3 &= -53 \\ (x+y)(x^2-xy+y^2)-z^3+53 &= 0 \\ (x+y)(x^2+y^2-xy)-z^3+53 &= 0 \\ (z+1)(z^2-5-(z+3))-z^3+53 &= 0 \\ (z+1)(z^2-z-8)-z^3+53 &= 0 \\ z^3-z^2-8z+z^2-z-8-z^3+53 &= 0 \\ -9z+45 &= 0 \\ -9z &= -45 \\ z &= 5 \\ x+y &= 5+1 \\ x+y &= 6 \\ x &= 6-y \\ xy &= 5+3 \\ xy &= 8 \\ (6-y)y &= 8 \\ 6y-y^2 &= 8 \\ y^2-6y+8 &= 0 \\ (y-4)(y-2) &= 0 \\ y=4 \text{ atau } y=2 \\ \text{jika } y=4 \\ x+y &= z+1 \\ x+4 &= 5+1 \\ x &= 2 \\ \text{jika } y=2 \\ x+y &= z+1 \\ x+2 &= 5+1 \\ x &= 4 \\ \end{align} </math> </div></div> <ol start=28> <li>Berapakah nilai titik koordinat (x,y) dari <math>\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}</math> dan <math>\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x+y}+\sqrt{x-y} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+y}-\sqrt{x-y} &= \sqrt{\frac{52y}{3x}} \\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y}) &= \sqrt{\frac{432x}{13y}} \cdot \sqrt{\frac{52y}{3x}} \\ x+y-x+y &= \sqrt{\frac{432x \cdot 52y}{13y \cdot 3x}} \\ 2y &= \sqrt{144 \cdot 4} \\ 2y &= \sqrt{576} \\ 2y &= 24 \\ y &= 12 \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13(12)}} \\ x+12+x-12+2 \cdot \sqrt{x+12} \cdot \sqrt{x-12} &= \frac{36x}{13} \\ 2x+2 \sqrt{x^2-144} &= \frac{36x}{13} \\ 2(x+\sqrt{x^2-144}) &= \frac{36x}{13} \\ x+\sqrt{x^2-144} &= \frac{18x}{13} \\ \sqrt{x^2-144} &= \frac{5x}{13} \\ x^2-144 &= \frac{25x^2}{169} \\ \frac{144x^2}{169}-144 &= 0 \\ \frac{x^2}{169}-1 &= 0 \\ x^2-169 &= 0 \\ (x-13)(x+13) &= 0 \\ x_1=13 &\text{ atau } x_2=-13 \text{ (TM) karena } x>y \\ \end{align} </math> jadi titik koordinat (13,12) </div></div> <ol start=29> <li>Berapakah nilai dari <math>x^2-7x</math> jika <math>(x-2)^2+\frac{1}{(x-2)^2} = 11</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x-2)^2+\frac{1}{(x-2)^2} &= 11 \\ (x-2)^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2)^2} &= 11-2 \\ (x-2-\frac{1}{x-2})^2 &= 9 \\ x-2-\frac{1}{x-2} &= 3 \\ (x-2)^2-1 &= 3(x-2) \\ x^2-4x+4-1 &= 3x-6 \\ x^2-7x &= -9 \\ \end{align} </math> </div></div> <ol start=30> <li>Berapakah nilai dari <math>\frac{(x+y)^2(x+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)}</math> jika xy+yz+xz=1?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy+yz+xz &= 1 \\ x^2+xy+yz+xz &= x^2+1 \\ x(x+y)+z(x+y) &= x^2+1 \\ (x+y)(x+z) &= x^2+1 \\ \text{dengan pola yang sama } \\ (y+x)(y+z) &= y^2+1 \\ (x+z)(y+z) &= z^2+1 \\ \frac{(x+y)^2(y+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)} &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)} \\ &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)^2(y+z)^2(x+z)^2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=31> <li>Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} w+5 &= w+x+y+z+5 \\ x+4 &= w+x+y+z+5 \\ y+3 &= w+x+y+z+5 \\ z+2 &= w+x+y+z+5 \\ \text{jumlahkan keempat persamaan } \\ w+x+y+z+14 &= 4(w+x+y+z+5) \\ w+x+y+z+14 &= 4(w+x+y+z)+20 \\ 3(w+x+y+z) &= -6 \\ w+x+y+z &= -2 \\ \end{align} </math> </div></div> <ol start=32> <li>Berapakah nilai dari <math>\frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2}</math> jika <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3</math> dan x+y+z=xyz?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2} &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2} \\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) \\ 3^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz}) \\ 9 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz}) \\ &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2 \\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} &= 7 \\ \end{align} </math> </div></div> <ol start=33> <li>Berapakah nilai dari <math>\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}</math> jika <math>x^2+y^2+z^2 = -2(ab+bc+ac)</math>?>/li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+z^2 &= -2(xy+yz+xz) \\ x^2+y^2+z^2+2(xy+yz+xz) &= 0 \\ (x+y+z)^2 &= 0 \\ x+y+z &= 0 \\ x+y &= -z \\ x+z &= -y \\ y+z &= -x \\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z} &= \frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x} \\ &= -2-(-5)-(-7) \\ &= 10 \\ \end{align} </math> </div></div> <ol start=34> <li>Berapakah nilai dari <math>\frac{20xyz}{xy+yz+xz}</math> jika <math>16^x = 256^y = 625^z = 40</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 16^x = 256^y = 625^z &= 40 \\ 2^{4x} = 4^{4y} = 5^{4z} &= 40 \\ 2^{4x} &= 40 \\ 2 &= 40^{\frac{1}{4x}} \\ 4^{4y} &= 40 \\ 4 &= 40^{\frac{1}{4y}} \\ 5^{4z} &= 40 \\ 5 &= 40^{\frac{1}{4z}} \\ 2 \cdot 4 \cdot 5 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z}} \\ 1 &= \frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z} \\ 4 &= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{20xyz}{xy+yz+xz} &= 20 \cdot \frac{xyz}{xy+yz+xz} \\ &= 20 \cdot (\frac{xy+yz+xz}{xyz})^{-1} \\ &= 20 \cdot (\frac{1}{z} + \frac{1}{x} + \frac{1}{y})^{-1} \\ &= 20 \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{-1} \\ &= 20 \cdot (4)^{-1} \\ &= 20 \cdot \frac{1}{4} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=35> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+3x^2+1}</math> jika <math>6x^2+25x+6=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 6x^2+25x+6 &= 0 \\ 6x+25+\frac{6}{x} &= 0 \\ 6(x+\frac{1}{x}) &= -25 \\ x+\frac{1}{x} &= \frac{-25}{6} \\ (c+\frac{1}{x})^2 &= (\frac{-25}{6})^2 \\ x^2+2+\frac{1}{x^2} &= \frac{625}{36} \\ x^2+\frac{1}{x^2} &= \frac{625}{36}-2 \\ x^2+\frac{1}{x^2} &= \frac{553}{36} \\ \frac{x^2}{x^4+3x^2+1} &= \frac{1}{x^2+3+\frac{1}{x^2}} \\ &= \frac{1}{a^2+\frac{1}{x^2}+3} \\ &= \frac{1}{\frac{553}{36}+3} \\ &= \frac{1}{\frac{661}{36}} \\ &= \frac{36}{661} \\ \end{align} </math> </div></div> <ol start=36> <li>Berapakah nilai dari <math>\frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5} &= \frac{(9+2\sqrt{20})^{1013}}{((2)^3+3(2)^2(\sqrt{5})+3(2)(\sqrt{5})^2+(\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{((2+\sqrt{5})^2)^{1013}}{((2+\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{(2+\sqrt{5})^{2026}}{(2+\sqrt{5})^{2025}}+6-\sqrt{5} \\ &= 2+\sqrt{5}+6-\sqrt{5} \\ &= 8 \\ \end{align} </math> </div></div> <ol start=37> <li>Berapakah nilai dari <math>27x^3+\frac{8}{x^3}</math> jika <math>3x+\frac{2}{x}=6</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3x+\frac{2}{x} &= 6 \\ (3x+\frac{2}{x})^3 &= 6^3 \\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3} &= 216 \\ 27x^3+18(6)+\frac{8}{x^3} &= 216 \\ 27x^3+108+\frac{8}{x^3} &= 216 \\ 27x^3+\frac{8}{x^3} &= 108 \\ \end{align} </math> </div></div> <ol start=38> <li>Berapakah nilai dari <math>x^6+\frac{8}{x^3}</math> jika <math>x^3+\frac{1}{x^3}=8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3+\frac{1}{x^3} &= 8 \\ x^3 &= 8-\frac{1}{x^3} \\ x^6 &= 8x^3-1 \\ x^6+\frac{8}{x^3} &= 8x^3-1+\frac{8}{x^3} \\ &= 8x^3+\frac{8}{x^3}-1 \\ &= 8(x^3+\frac{1}{x^3})-1 \\ &= 8(8)-1 \\ &= 63 \\ \end{align} </math> </div></div> <ol start=39> <li>Berapakah nilai dari <math>4x+\frac{25}{x}</math> jika <math>2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2\sqrt{x}+\frac{5}{\sqrt{x}} &= 4x-\frac{25}{x} \\ 2\sqrt{x}+\frac{5}{\sqrt{x}} &= (2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}}) \\ 1 &= 2\sqrt{x}-\frac{5}{\sqrt{x}} \\ 1^2 &= (2\sqrt{x}-\frac{5}{\sqrt{x}})^2 \\ 1 &= 4x-20+\frac{25}{x} \\ 4x+\frac{25}{x} &= 21 \\ \end{align} </math> </div></div> <ol start=40> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2-x+1}{x^2+x+1} &= \frac{5}{6} \\ \frac{x^2+1-x}{x^2+1+x} &= \frac{5}{6} \\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} &= \frac{5}{6} \\ \text{ misalkan } x+\frac{1}{x} &= y \\ \frac{y-1}{y+1} &= \frac{5}{6} \\ 6(y-1) &= 5(y+1) \\ 6y-6 &= 5y+5 \\ y &= 11 \\ x+\frac{1}{x} &= 11 \\ \end{align} </math> </div></div> <ol start=41> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt{x}+x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x}+x &= 1 \\ x-1 &= -\sqrt{x} \\ (x-1)^2 &= (-\sqrt{x})^2 \\ x^2-2x+1 &= x \\ x^2-3x+1 &= 0 \\ x-3+\frac{1}{x} &= 0 \\ x+\frac{1}{x} &= 3 \\ \end{align} </math> </div></div> <ol start=42> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt[3]{x}-\sqrt[3]{x-36}=3</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x}-\sqrt[3]{x-36} &= 3 \\ (\sqrt[3]{x}-\sqrt[3]{x-36})^3 &= 3^3 \\ x-(x-36)-3 \sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36}) &= 27 \\ 36-3 \sqrt[3]{x(x-36)}3 &= 27 \\ -9 \sqrt[3]{x(x-36)} &= -9 \\ \sqrt[3]{x(x-36)} &= 1 \\ x(x-36) &= 1 \\ x^2-36x-1 &= 0 \\ x-36-\frac{1}{x} &= 0 \\ x-\frac{1}{x} &= 36 \\ \end{align} </math> </div></div> <ol start=43> <li>Berapakah nilai dari <math>x+\frac{16}{x}</math> jika <math>x-3\sqrt{x}=4</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x-3\sqrt{x} &= 4 \\ x-4 &= 3\sqrt{x} \\ x^2-8x+16 &= 9x \\ x^2-17x+16 &= 0 \\ x-17+\frac{16}{x} &= 0 \\ x+\frac{16}{x} &= 17 \\ \end{align} </math> </div></div> <ol start=44> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+4}</math> jika <math>x^2-7x+2=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2-7x+2 &= 0 \\ x^2+2 &= 7x \\ x+\frac{2}{x} &= 7 \\ x^2+4+\frac{4}{x^2} &= 49 \\ x^2+\frac{4}{x^2} &= 45 \\ \frac{x^4+4}{x^2} &= 45 \\ \frac{x^2}{x^4+4} &= \frac{1}{45} \\ \end{align} </math> </div></div> <ol start=45> <li>Berapakah nilai dari <math>x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}</math> jika <math>x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}} &= 25 \\ x^{\frac{1}{2}}+x^{-\frac{1}{2}} &= 23 \\ x+2+x^{-1} &= 529 \\ x+x^{-1} &= 527 \\ x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+x^{-\frac{3}{4}} &= 110 \\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1} &= x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}} \\ &= 527+110 \\ &= 637 \\ \end{align} </math> </div></div> <ol start=46> <li>Berapakah nilai dari <math>\sqrt{8x^6+x^5+x^4+5x^3+1}</math> jika <math>\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5} &= 0 \\ \frac{x^2+x+1}{x^5} &= 0 \\ x^2+x+1 &= 0 \\ x^2+x+1 &= 0 \\ (x-1)(x^2+x+1) &= 0(x-1) \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ \sqrt{8x^6+x^5+x^4+5x^3+1} &= \sqrt{(2x^3)^2+x^3x^2+x^3x+5x^3+1} \\ &= \sqrt{(2(1))^2+(1)x^2+(1)x+5(1)+1} \\ &= \sqrt{(2)^2+x^2+x+5+1} \\ &= \sqrt{4+x^2+x+1+5} \\ &= \sqrt{4+0+5} \\ &= \sqrt{9} \\ &= 3 \\ \end{align} </math> </div></div> <ol start=47> <li>Berapakah nilai dari <math>f(1)+f(2)+f(3)+ \dots + f(99)</math> jika <math>f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{1}{\sqrt{x+1}+\sqrt{x}} \\ &= \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} \\ &= \sqrt{x+1}-\sqrt{x} \\ f(1)+f(2)+f(3)+ \dots + f(98)+f(99) &= \sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+ \cdot + \sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99} \\ &= \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+ \cdot + \sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ &= \sqrt{100}-\sqrt{1} \\ &= 10-1 \\ &= 9 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots + \frac{2024}{2025})</math> jika <math>h(x)=\frac{3}{3+9^x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} h(x) &= \frac{3}{3+9^x} \\ h(1-x) &= \frac{3}{3+9^{1-x}} \\ &= \frac{3}{3+\frac{9}{9^x}} \\ &= \frac{9^x}{3+9^x} \\ h(x)+h(1-x) &= \frac{3}{3+9^x}+\frac{9^x}{3+9^x} \\ &= \frac{3+9^x}{3+9^x} \\ &= 1 \\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots +(1-\frac{2}{2025})+(1-\frac{1}{2025})) \\ & 5(1+1+1+ \dots +1+1) \text{ sebanyak 1012 kali } \\ & 5(1012) \\ & 5060 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>\frac{7^{2025} - 7^{2023} + 432}{7^{2024} + 7^{2023} + 72}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72} &= \frac{7^{2023}7^{2}-7^{2023} + 48 \times 9}{7^{2023}7^1+7^{2023}+8 \times 9} \\ &= \frac{7^{2023}(7^{2}-1)+48 \times 9}{7^{2023}(7^1+1)+8 \times 9} \\ &= \frac{7^{2023}(49-1)+48 \times 9}{7^{2023}(7+1) + 8 \times 9} \\ &= \frac{7^{2023} \times 48+48 \times 9}{7^{2023} \times 8+8 \times 9} \\ &= \frac{48(7^{2023}+9)}{8(7^{2023}+9)} \\ &= \frac{48}{8} \\ &= 6 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>tan (x+\frac{\pi}{4})</math> jika <math>\frac{1}{cos x}-tan x = \frac{4}{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{cos x}-tan x &= \frac{4}{5} \\ sec x-tan x &= \frac{4}{5} \\ sec^2 x-tan^2 x &= 1 \\ (sec x+tan x)(sec x-tan x) &= 1 \\ (sec x+tan x)\frac{4}{5} &= 1 \\ sec x+tan x &= \frac{5}{4} \\ \text{kedua persamaan dengan cara metode eliminasi } \\ 2 tan x &= \frac{5}{4}-\frac{4}{5} \\ 2 tan x &= \frac{9}{20} \\ tan x &= \frac{9}{40} \\ tan (x+\frac{\pi}{4}) &= \frac{tan x+tan \frac{\pi}{4}}{1-tan x \cdot tan \frac{\pi}{4}} \\ &= \frac{\frac{9}{40}+1}{1-\frac{9}{40} \cdot 1} \\ &= \frac{\frac{49}{40}}{\frac{31}{40}} \\ &= \frac{49}{31} \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>sin^3 x+csc^3 x</math> jika <math>sin x-csc x = 8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ Dengan menggunakan rumus: } (a-b)^3 &= a^3-b^3-3ab(a-b) \\ (sin x-csc x)^3 &= sin^3 x-csc^3 x-3sin x csc x(sin x-csc x) \\ 8^3 &= sin^3 x-csc^3 x-3sin x (\frac{1}{sin x})(8) \\ 512 &= sin^3 x-csc^3 x-24 \\ sin^3 x-csc^3 x &= 512+24 \\ sin^3 x-csc^3 x &= 536 \\ \end{align} </math> </div></div> <ol start=52> <li>Berapakah nilai dari <math>(sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2</math> jika <math>\frac{1}{sin x}+\frac{1}{cos x} = 10</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{sin x}+\frac{1}{cos x} &= 10 \\ \frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} &= 100 \\ (sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2 &= sin^2 x+\frac{2sin x}{cos x}+\frac{1}{cos^2 x}+cos^2 x+\frac{2cos x}{sin x}+\frac{1}{sin^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2(sin^2 x+cos^2 x)}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+100 \\ &= 101 \\ \end{align} </math> </div></div> <ol start=53> <li>Berapakah nilai dari (x-1)⁶ jika <math>x=\frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin 80^\circ &= cos 10^\circ \\ sin 80^\circ-cos 10^\circ &= 0 \\ x &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+2 sin 20^\circ cos 20^\circ}{cos 10^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+4 sin 10^\circ cos 10^\circ cos 20^\circ}{cos 10^\circ} \\ &= 4 cos 55^\circ cos 25^\circ+4 sin 10^\circ cos 20^\circ \\ &= 2(2 cos 55^\circ cos 25^\circ+2 sin 10^\circ cos 20^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ+sin (-10)^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ-sin 10^\circ) \\ &= 2(cos 80^\circ-sin 10^\circ+cos 30^\circ+sin 30^\circ) \\ &= 2(cos 80^\circ-sin (90^\circ-80^\circ)+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(cos 80^\circ-cos 80^\circ+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= \sqrt{3}+1 \\ x-1 &= \sqrt{3} \\ (x-1)^6 &= (\sqrt{3})^6 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=54> <li>Berapakah nilai dari x jika <math>x=\frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ} \\ 2x sin 20^\circ-x sin 40 ^\circ &= x sin 20^\circ-x^2 sin 10^\circ \\ x^2 sin 10^\circ+x sin 20^\circ-x sin 40 ^\circ &= 0 \\ x(x sin 10^\circ+sin 20^\circ-sin 40 ^\circ) &= 0 \\ x = 0 &\text{ atau } x sin 10^\circ+sin 20^\circ-sin 40 ^\circ = 0 \\ x sin 10^\circ+sin 20^\circ-sin 40 ^\circ &= 0 \\ x sin 10^\circ &= sin 40 ^\circ-sin 20^\circ \\ x &= \frac{sin 40 ^\circ-sin 20^\circ}{sin 10^\circ} \\ &= \frac{2 cos 30 ^\circ sin 10^\circ}{sin 10^\circ} \\ &= 2 cos 30 ^\circ \\ &= \frac{2 \sqrt{3}}{2} \\ &= \sqrt{3} \\ \end{align} </math> </div></div> <ol start=55> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x^2}{x^2-16y^2} = \frac{625}{49}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2}{x^2-16y^2} &= \frac{625}{49} \\ \frac{x^2-16y^2}{x^2} &= \frac{49}{625} \text{ (terbalik posisinya)} \\ 1-\frac{16y^2}{x^2} &= \frac{49}{625} \\ \frac{16y^2}{x^2} &= 1 - \frac{49}{625} \\ (\frac{4y}{x})^2 &= \frac{576}{625} \\ (\frac{4y}{x})^2 &= (\frac{24}{25})^2 \\ \frac{4y}{x} &= \frac{24}{25} \\ \frac{y}{x} &= \frac{6}{25} \\ \frac{x}{y} &= \frac{25}{6} \\ \end{align} </math> </div></div> <ol start=56> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x}{y}+\frac{x+10y}{y+10x} = 2</math> serta bilangan real untuk x dan y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x}{y}+\frac{x+10y}{y+10x} &= 2 \\ \frac{x}{y}+\frac{\frac{x}{y}+10}{1+10\frac{x}{y}} &= 2 \\ \text{misalkan } \frac{x}{y} = a \\ a+\frac{a+10}{1+10a} &= 2 \\ a(1+10a)+a+10 &= 2(1+10a) \\ 10a^2+a+a+10 &= 2+20a \\ 10a^2-18a+8 &= 0 \\ 5a^2-9a+4 &= 0 \\ (5a-4)(a-1) &= 0 \\ a = \frac{4}{5} &\text{ atau } a = 1 \\ \text{jadi } \frac{x}{y} = {\frac{4}{5}, 1} \\ \end{align} </math> </div></div> <ol start=57> <li>Berapakah nilai dari xy jika <math>x^4+y^4+x^2y^2=15 \text{ dan } x^2+y^2+xy=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+xy &= 5 \\ x^2+y^2 &= 5-xy \\ x^4+y^4+x^2y^2 &= 15 \\ (x^2)^2+(y^2)^2+2x^2y^2-x^2y^2 &= 15 \\ (x^2+y^2)^2-x^2y^2 &= 15 \\ (5-xy)^2-x^2y^2 &= 15 \\ 25-10xy+x^2y^2-x^2y^2 &= 15 \\ 25-10xy &= 15 \\ 10xy &= 10 \\ xy &= 1 \\ \end{align} </math> </div></div> <ol start=58> <li>Berapakah nilai dari x jika <math>4^x = 63(4^3+1)(4^6+1)(4^{12}+1)+1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 4^x &= 63(4^3+1)(4^6+1)(4^{12}+1)+1 \\ 4^x-1 &= 63(4^3+1)(4^6+1)(4^{12}+1) \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{4^3-1} \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{63} \\ &= (4^3+1)(4^6+1)(4^{12}+1)(4^3-1) \\ &= (4^3-1)(4^3+1)(4^6+1)(4^{12}+1) \\ &= (4^6-1)(4^6+1)(4^{12}+1) \\ &= (4^{12}-1)(4^{12}+1) \\ &= 4^{24}-1 \\ 4^x &= 4^{24} \\ x &= 24 \\ \end{align} </math> </div></div> <ol start=59> <li>Berapakah nilai dari <math>\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}</math> jika <math>x=\sqrt{9+4\sqrt{5}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \sqrt{9+4\sqrt{5}} \\ x &= 2+\sqrt{5} \\ x^2 &= 9+4\sqrt{5} \\ x^2-4x &= 9+4\sqrt{5}-4(2+\sqrt{5}) \\ x^2-4x &= 1 \\ x^2 &= 4x+1 \\ x^3 &= x \cdot x^2 \\ &= x(4x+1) \\ &= 4x^2+x \\ &= 4(4x+1)+x \\ &= 16x+4+x \\ &= 17x+4 \\ x^4 &= x \cdot x^3 \\ &= x(17x+4) \\ &= 17x^2+4x \\ &= 17(4x+1)+4x \\ &= 68x+17+4x \\ &= 72x+17 \\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1} &= \frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1} \\ &= \frac{72x+17-85x-20+8x+2+5x+3}{2} \\ &= \frac{2}{2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=60> <li>Berapakah nilai dari <math>\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}</math> jika 2x-1=<math>\sqrt{61}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \frac{x^3+1}{x^5-x^4-x^3+x^2} = p \\ p &= \frac{x^3+1}{x^5-x^4-x^3+x^2} \\ &= \frac{x^3+1}{x^5-x^4-(x^3-x^2)} \\ &= \frac{x^3+1}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)} \\ &= \frac{x^2-x+1}{x^2(x-1)^2} \\ &= \frac{x^2-x+1}{(x(x-1))^2} \\ &= \frac{x(x-1)+1}{(x(x-1))^2} \\ 2x-1 &= \sqrt{61} \\ x &= \frac{\sqrt{61}+1}{2} \\ x-1 &= \frac{\sqrt{61}-1}{2} \\ x(x-1) &= (\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2}) \\ &= \frac{61-1}{4} \\ &= \frac{60}{4} \\ &= 15 \\ p &= \frac{x(x-1)+1}{(x(x-1))^2} \\ &= \frac{15+1}{15^2} \\ &= \frac{16}{15^2} \\ \sqrt{p} &= \sqrt{\frac{16}{15^2}} \\ &= \frac{4}{15} \\ \end{align} </math> </div></div> <ol start=61> <li>Berapakah nilai dari <math>(\frac{x-3}{x})^{25}</math> jika <math>x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\sqrt[5]{8}+\sqrt[5]{2} &= 1+\sqrt[5]{16}+\sqrt[5]{4} \\ x+(\sqrt[5]{2})^3+\sqrt[5]{2} &= 1+(\sqrt[5]{2})^4+(\sqrt[5]{2})^2 \\ x &= (\sqrt[5]{2})^4-(\sqrt[5]{2})^3+(\sqrt[5]{2})^2-\sqrt[5]{2}+1 \\ \text{misalkan } \sqrt[5]{2} = p \\ x &= p^4-p^3+p^2-p+1 \\ x &= \frac{p^5+1}{p+1} \\ (\frac{x-3}{x})^{25} &= (1-\frac{3}{x})^{25} \\ &= (1-\frac{3}{\frac{p^5+1}{p+1}})^{25} \\ &= (1-\frac{3(p+1)}{p^5+1})^{25} \\ &= (1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2})^5+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{2+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-3\sqrt[5]{2}-3)}{3})^{25} \\ &= (-\sqrt[5]{2})^{25} \\ &= (-2)^5 \\ &= -32 \\ \end{align} </math> </div></div> <ol start=62> <li>Berapakah nilai dari <math>x^{50}+x^{49}+x^{48}+x^{47}+x^{46}</math> jika <math>x^2+x+1=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+x+1 &= 0 \\ x^2+x &= -1 \\ \frac{x^3-1}{x-1} &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46} &= x^{48}(x^2+x+1)+x^{45}(x^2+x) \\ &= x^{48}(0)+(x^3)^{15}(-1) \\ &= 0+(1)^{15}(-1) \\ &= -1 \\ \end{align} </math> </div></div> <ol start=63> <li>Berapakah 2²⁴ dari <math>8^7+8^6+8^5+8^4+8^3+8^2+8+1=A</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= A \\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1) &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8 &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= 8A+1 \\ 8^8+A &= 8A+1 \\ 8^8 &= 7A+1 \\ (2^3)^8 &= 7A+1 \\ 2^{24} &= 7A+1 \\ \end{align} </math> </div></div> <ol start=64> <li>Berapakah nilai dari <math>x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1</math> jika <math>x+\frac{1}{x}=\sqrt{3}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{x} &= \sqrt{3} \\ x^2+2+\frac{1}{x^2} &= 3 \\ x^2-1+\frac{1}{x^2} &= 0 \\ x^2(x^2-1+\frac{1}{x^2}) &= x^2(0) \\ x^4-x^2+1 &= 0 \\ (x^2+1)(x^4-x^2+1) &= (x^2+1)0 \\ x^6-x^4+x^2+x^4-x^2+1 &= 0 \\ x^6+1 &= 0 \\ x^6 &= -1 \\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1 &= {x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1 \\ &= (-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2-1+1 \\ &= -1+1-1+1-1+1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=65> <li>Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari <math>\frac{c-b}{a-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= ax^2+bx+c \\ f(2) &= a(2)^2+2b+c = 4 \\ &= 4a+2b+c = 4 \\ f(7) &= a(7)^2+7b+c = 49 \\ &= 49a+7b+c = 49 \\ 49a+7b+c &= 49 \\ 4a+2b+c &= 4 \\ 45a+5b &= 45 \text{ (f(7) dikurangi f(2)) } \\ 9a+b &= 9 \\ b &= -9a+9 \\ 4a+2b+c &= 4 \\ 4a+2(-9a+9)+c &= 4 \\ 4a-18a+18+c &= 4 \\ -14a+18+c &= 4 \\ c &= 14a-14 \\ \frac{c-b}{a-1} &= \frac{14a-14-(-9a+9)}{a-1} \\ &= \frac{14(a-1)+9(a-1)}{a-1} \\ &= \frac{(14+9)(a-1)}{a-1} \\ &= 23 \\ \end{align} </math> </div></div> <ol start=66> <li>Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+y)^3 &= x^3+y^3+3xy(x+y) \\ 11^3 &= 242+3xy(11) \text{ (dibagi 11)} \\ 11^2 &= 22+3xy \\ 121 &= 22+3xy \\ 99 &= 3xy \\ xy &= 33 \\ (x-y)^2 &= x^2+y^2-2xy \\ &= ((x+y)^2-2xy)-2xy \\ &= (x+y)^2-4xy \\ &= 11^2-4(33) \\ &= 121-132 \\ &= -11 \\ \end{align} </math> </div></div> <ol start=67> <li>Berapa f(1)+f(-1) jika <math>f(\frac{ax-b}{bx-a})</math>=x²-5x+6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ jika} f(1) = f(\frac{ax-b}{bx-a}) \\ 1 &= \frac{ax-b}{bx-a} \\ bx-a &= ax-b \\ (b-a)x &= -b+a \\ &= -(b-a) \\ &= -1 \\ f(1) &= x^2-5x+6 \\ &= (-1)^2-5(-1)+6 \\ &= 12 \\ \text{ jika} f(-1) = f(\frac{ax-b}{bx-a}) \\ -1 &= \frac{ax-b}{bx-a} \\ -(bx-a) &= ax-b \\ -bx+a &= ax-b \\ (-b-a)x &= -b-a \\ &= 1 \\ f(-1) &= x^2-5x+6 \\ &= (1)^2-5(1)+6 \\ &= 2 \\ f(1)+f(-1) &= 12+2 \\ &= 14 \\ \end{align} </math> </div></div> <ol start=68> <li>berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)-x &= f(x-1) \\ f(x)-f(x-1) &= x \\ x=1 ; f(1)-f(0) &= 1 \\ x=2 ; f(2)-f(1) &= 2 \\ x=3 ; f(3)-f(2) &= 3 \\ x=4 ; f(4)-f(3) &= 4 \\ \dots \\ x=200 ; f(200)-f(199) &= 200 \\ \text{ jumlahkan tersebut menjadi } \\ f(200)-f(0) &= 1+2+3+4+\dots+200 \\ &= \frac{200 \cdot 201}{2} \\ &= 20.100 \\ f(200)-1 &= 20.100 \\ &= 20.101 \\ \end{align} </math> </div></div> <ol start=69> <li>Misalkan f(x) adalah fungsi rekursif yang berlaku ∀x ∈ R sebagai berikut: : f(x)+f(15-x) = 2024 : f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)+f(15-x) &= 2024 \\ f(15+x) &= f(x)+2020 \\ *cara 1 \\ \text{ganti x dengan 15+x } \\ f(15+x)+f(-x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ *cara 2 \\ \text{ganti x dengan -x } \\ f(-x)+f(15+x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ \end{align} </math> </div></div> <ol start=70> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>2f(\frac{2002}{x}) + f(x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2f(\frac{2002}{x}) + f(x) &= 3x \\ \text{ganti x dengan 2 } \\ 2f(\frac{2002}{2}) + f(2) &= 3(2) \\ 2f(1001) + f(2) &= 6 \\ \text{ganti x dengan 1001 } \\ 2f(\frac{2002}{1001}) + f(1001) &= 3(1001) \\ 2f(2) + f(1001) &= 3003 \\ 2f(2) + f(1001) &= 3003 \\ f(1001) &= 3003 - 2f(2) \\ 2f(1001) + f(2) &= 6 \\ 2(3003 - 2f(2)) + f(2) &= 6 \\ 6006 - 4f(2) + f(2) &= 6 \\ 3f(2) &= 6000 \\ f(2) &= 2000 \\ \end{align} </math> </div></div> <ol start=71> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>f(\frac{1}{x}) + \frac{1}{x}f(-x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{1}{x})+\frac{1}{x}f(-x) &= 3x \\ \text{ganti x dengan 1/3 } \\ f(3)+3f(-\frac{1}{3}) &= 1 \\ \text{ganti x dengan -3 } \\ f(-\frac{1}{3}) - \frac{1}{3}f(3) &= -9 \\ \text{dikalikan 3 } \\ 3f(-\frac{1}{3})-f(3) &= -27 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ 2f(3) &= 28 \\ f(3) &= 14 \\ \end{align} </math> </div></div> <ol start=72> <li>Diketahui polinom <math>f(7^b-1)=7^{3b}-10</math>. tentukan nilai f(5)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(5) &= f(7^b-1) \\ 5 &= 7^b-1 \\ 7^b &= 6 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(6-1) &= 6^3-10 \\ f(5) &= 216-10 \\ &= 206 \\ *cara 2 \\ \text{misalkan } 7^b-1=a \text{ maka } 7^b=a+1 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(a) &= (a+1)^3-10 \\ f(5) &= (5+1)^3-10 \\ &= 6^3-10 \\ &= 216-10 \\ &= 206 \\ \end{align} </math> </div></div> <ol start=73> <li>Diketahui polinom <math>f(6^b-7)=6^{3b}-2 \cdot 6^{2b}-4</math>. tentukan nilai f(-2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(-2) &= f(6^b-7) \\ -2 &= 6^b-7 \\ 6^b &= 5 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(5-7) &= 5^3-2 \cdot 5^2-4 \\ f(-2) &= 125-50-4 \\ &= 71 \\ *cara 2 \\ \text{misalkan } 6^b-7=a \text{ maka } 6^b=a+7 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(a) &= (a+7)^3-2(a+7)^2-4 \\ f(-2) &= (-2+7)^3-2(-2+7)^2-4 \\ &= 5^3-2(5)^2-4 \\ &= 125-50-4 \\ &= 71 \\ \end{align} </math> </div></div> <ol start=74> <li>Jika <math>f(xy)=\frac{f(x)}{y}</math> dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(10) &= 7 \\ f(2 \cdot 5) &= 7 \\ f(xy) &= \frac{f(x)}{y} \\ f(2 \cdot 5) &= \frac{f(2)}{5} \\ 7 &= \frac{f(2)}{5} \\ f(2) &= 35 \\ \end{align} </math> </div></div> <ol start=75> <li>Jika <math>f(xy)=\frac{f(x+y)}{xy}</math> dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(15) &= 16 \\ f(3 \cdot 5) &= 16 \\ f(xy) &= \frac{f(x+y)}{xy} \\ f(3 \cdot 5) &= \frac{f(3+5)}{3 \cdot 5} \\ f(15) &= \frac{f(8)}{15} \\ 16 &= \frac{f(8)}{15} \\ f(8) &= 240 \\ \end{align} </math> </div></div> <ol start=76> <li>Jika <math>f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15</math> maka tentukan nilai f(16)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x+\frac{1}{x}+6) &= x^2+\frac{1}{x^2}+15 \\ &= (x+\frac{1}{x})^2-2+15 \\ &= (x+\frac{1}{x})^2+13 \\ \text{misalkan } x+\frac{1}{x} &= p \\ f(x+\frac{1}{x}+6) &= (x+\frac{1}{x})^2+13 \\ f(p+6) &= p^2+13 \\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 } \\ f(p+6) &= p^2+13 \\ f(10+6) &= 10^2+13 \\ f(16) &= 100+13 \\ &= 113 \\ \end{align} </math> </div></div> <ol start=77> <li>Tentukan nilai x jika <math>f(x)=\frac{4}{4-x}</math> dan <math>f(x \cdot f(x))^{\frac{f(4x)}{f(x)}}=256</math>!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{4}{4-x} \\ f(4x) &= \frac{4}{4-4x} \\ \frac{f(4x)}{f(x)} &= \frac{\frac{4}{4-4x}}{\frac{4}{4-x}} \\ &= \frac{4-x}{4-4x} \\ f(x \cdot f(x)) &= f(x(\frac{4}{4-x})) \\ &= f(\frac{4x}{4-x}) \\ &= \frac{4}{4-(\frac{4x}{4-x})} \\ &= \frac{4}{\frac{16-4x-4x}{4-x}} \\ &= \frac{4}{\frac{16-8x}{4-x}} \\ &= \frac{4(4-x)}{4(4-4x)} \\ &= \frac{4-x}{4-4x} \\ \text{misalkan } \frac{4-x}{4-4x} &= a \\ f(x \cdot f(x))^{\frac{f(4x)}{f(x)}} &= 256 \\ a^a &= 256 \\ a^a &= 4^4 \\ a &= 4 \\ \frac{4-x}{4-4x} &= 4 \\ 4-x &= 16-16x \\ 15x &= 12 \\ x &= \frac{4}{5} \\ \end{align} </math> </div></div> <ol start=78> <li>Fungsi <math>f(x) = \frac{kx}{2x+1} \text{dengan } x \neq -\frac{1}{2}</math>. Dengan f(f(x)) = x maka tentukan nilai k!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{kx}{2x+1} \\ f(f(x)) &= x \\ f(\frac{kx}{2x+1}) &= x \\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1} &= x \\ \frac{\frac{k^2x}{2x+1}}{\frac{2kx+2x+1}{2x+1}} &= x \\ \frac{k^2x}{2kx+2x+1} &= x \\ \frac{k^2}{2kx+2x+1} &= 1 \\ k^2 &= 2kx+2x+1 \\ k^2-2kx &= 2x+1 \\ k^2-2kx+x^2 &= x^2+2x+1 \\ (k-x)^2 &= (x+1)^2 \\ (k-x)^2-(x+1)^2 &= 0 \\ (k-x+x+1)(k-x-(x+1)) &= 0 \\ k=-1 &\text{ atau } k=2x+1 &\text{ (TM) } \\ \end{align} </math> </div></div> <ol start=79> <li>Jika n = 2023²+2024² maka berapa hasil dari <math>\sqrt{2n-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n &= 2023^2+2024^2 \\ &= 2023^2+(2023+1)^2 \\ \text{misalkan 2023 = p} \\ n &= p^2+(p+1)^2 \\ &= p^2+p^2+2p+1 \\ &= 2p^2+2p+1 \\ \sqrt{2n-1} &= \sqrt{2(2p^2+2p+1)-1} \\ &= \sqrt{4p^2+4p+2-1} \\ &= \sqrt{4p^2+4p+1} \\ &= \sqrt{(2p+1)^2} \\ &= 2p+1 \\ &= 2(2023)+1 \\ &= 4046+1 \\ &= 4047 \\ \end{align} </math> </div></div> <ol start=80> <li>Tentukan nilai dari a+b+c merupakan bilangan bulat positif jika ab = 2, bc = 3 dan ac = 6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab \cdot bc \cdot ac &= 2 \cdot 3 \cdot 6 \\ (abc)^2 &= 36 \\ abc &= \pm 6 \\ abc &= 6 \\ \frac{abc}{ab} &= c = \frac{6}{2} = 3 \\ \frac{abc}{bc} &= a = \frac{6}{3} = 2 \\ \frac{abc}{ac} &= b = \frac{6}{6} = 1 \\ a+b+c &= 6 \\ \end{align} </math> </div></div> <ol start=81> <li>Tentukan nilai dari (a-c)^b jika <math>\frac{ab}{a+b} = \frac{1}{3}</math>, <math>\frac{bc}{b+c} = \frac{1}{4}</math> dan <math>\frac{ac}{a+c} = \frac{1}{9}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{ab}{a+b} &= \frac{1}{3} \\ \frac{a+b}{ab} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{b} + \frac{1}{a} &= 3 \\ \frac{bc}{b+c} &= \frac{1}{4} \\ \frac{b+c}{bc} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{b} &= 4 \\ \frac{ac}{a+c} &= \frac{1}{9} \\ \frac{a+c}{ac} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{a} &= 9 \\ \text{Misalkan 1/a = x, 1/b = y dan 1/c = z} \\ x+y &= 3 \\ y+z &= 4 \\ x+z &= 9 \\ x+y &= 3 \\ y+z &= 4 \\ x-z &= -1 \\ x-z &= -1 \\ x+z &= 9 \\ 2x &= 8 \\ x &= 4 \\ x-z &= -1 \\ 4-z &= -1 \\ z &= 5 \\ x+y &= 3 \\ 4+y &= 3 \\ y &= -1 \\ \frac{1}{a} &= 4 \\ a &= \frac{1}{4} \\ \frac{1}{b} &= -1 \\ b &= -1 \\ \frac{1}{c} &= 5 \\ c &= \frac{1}{5} \\ (a-c)^b &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=82> <li>Tentukan nilai dari a, b dan c jika <math>\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}</math> dan a+2b+3c=28?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan k untuk semua ketiga persamaan tersebut } \\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5} &= k \\ a+b &= 2k \\ a+c &= 4k \\ b+c &= 5k \\ 2a+b+c &= 6k \\ 2a+5k &= 6k \\ k &= 2a \\ a &= \frac{k}{2} \\ b &= \frac{3k}{2} \\ c &= \frac{7k}{2} \\ a+2b+3c &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ a &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ b &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ c &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=83> <li>Tentukan nilai dari (b+c)^a jika <math>\frac{a+b+c}{2} = \sqrt{a-2}+\sqrt{b-1}+\sqrt{c}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{a+b+c}{2} &= \sqrt{a-2}+\sqrt{b-1}+\sqrt{c} \\ a+b+c &= 2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}) \\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c} &= 0 \\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1 &= 0 \\ (\sqrt{a-2}-1)^2+(\sqrt{b-1}-1)^2+(\sqrt{c}-1)^2 &= 0 \\ (\sqrt{a-2}-1)^2 &= 0 \\ \sqrt{a-2}-1 &= 0 \\ \sqrt{a-2} &= 1 \\ a-2 &= 1 \\ a &= 3 \\ (\sqrt{b-1}-1)^2 &= 0 \\ \sqrt{b-1}-1 &= 0 \\ \sqrt{b-1} &= 1 \\ b-1 &= 1 \\ b &= 1 \\ (\sqrt{c}-1)^2 &= 0 \\ \sqrt{c}-1 &= 0 \\ \sqrt{c} &= 1 \\ c &= 1 \\ (b+c)^a &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=84> <li>x dan y merupakan bilangan tak nol. Jika xy = <math>\frac{x}{y}</math> = x-y maka berapa nilai x+y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy &= \frac{x}{y} \\ y^2 &= 1 \\ y^2 - 1 &= 0 \\ (y-1)(y+1) &= 0 \\ y = 1 &\text{ atau } y = -1 \\ \frac{x}{y} &= x-y \\ x &= xy-y^2 \\ x-xy &= -y^2 \\ x(1-y) &= -y^2 \\ x &= \frac{-y^2}{1-y} \\ \text{cek y=1 } \\ x &= \frac{-1^2}{1-1} \\ \text{tidak memenuhi syarat } \\ \text{cek y=-1 } \\ x &= \frac{-(-1)^2}{1-(-1)} \\ &= \frac{-1}{2} \\ x+y &= -1-\frac{1}{2} \\ &= -\frac{3}{2} \\ \end{align} </math> </div></div> <ol start=86> <li>Berapa nilai x dari <math>(\frac{a}{b})^3+(\frac{b}{a})^3 = 2\sqrt{x}</math> jika <math>\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b} \\ \frac{a+b}{ab} &= \frac{1}{a+b} \\ (a+b)^2 &= ab \\ a^2+2ab+b^2 &= ab \\ a^2+b^2 &= -ab \\ \text{misalkan } \frac{a}{b}+\frac{b}{a} = n \\ \frac{a}{b}+\frac{b}{a} &= n \\ \frac{a^2+b^2}{ab} &= n \\ a^2+b^2 &= nab \\ n &= -1 \\ \frac{a}{b}+\frac{b}{a} &= n \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3(\frac{a}{b}+\frac{b}{a}) &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3n &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3 &= n^3-3n \\ &= (-1)^3-3(-1) \\ &= 2 \\ 2\sqrt{x} &= 2 \\ \sqrt{x} &= 1 \\ x &= 1 \\ \end{align} </math> </div></div> <ol start=87> <li>Berapa nilai m dari <math>x^2-mx-1=0</math> jika <math>\sqrt[3]{x_1}+\sqrt[3]{x_2}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x_1} &= a \\ x_1 &= a^3 \\ \sqrt[3]{x_2} &= b \\ x_2 &= b^3 \\ \sqrt[3]{x_1}+\sqrt[3]{x_2} &= 1 \\ a+b &= 1 \\ x^2-mx-1 &= 0 \\ x_1+x_2 &= m \\ x_1 \cdot x_2 &= -1 \\ x_1+x_2 &= m \\ a^3+b^3 &= m \\ x_1 \cdot x_2 &= -1 \\ a^3 \cdot b^3 &= -1 \\ (ab)^2 &= (-1)^3 \\ ab &= -1 \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ (1)^3 &= m+3(-1)(1) \\ 1 &= m-3 \\ m &= 4 \\ \end{align} </math> </div></div> <ol start=88> <li>Berapa nilai <math>\frac{x_1}{x_2}</math> dari <math>ax^2-18x-b=0</math> jika <math>ab=45</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab &= 45 \\ b &= \frac{45}{a} \\ ax^2-18x-b &= 0 \\ ax^2-18x-\frac{45}{a} &= 0 \\ a^2x^2-18ax-45 &= 0 \\ (ax-3)(ax-15) &= 0 \\ ax-3 &= 0 \\ x &= \frac{3}{a} \\ ax-15 &= 0 \\ x &= \frac{15}{a} \\ \frac{x_1}{x_2} &= \frac{\frac{3}{a}}{\frac{15}{a}} \\ &= \frac{3}{15} \\ &= \frac{1}{5} \\ \frac{x_1}{x_2} &= \frac{\frac{15}{a}}{\frac{3}{a}} \\ &= \frac{15}{3} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=89> <li>Jika <math>\frac{u_3}{u_1+u_2} = \frac{7}{8}</math> merupakan barisan aritmetika maka berapa dari <math>\frac{u_2+u_3}{u_1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{u_3}{u_1+u_2} &= \frac{7}{8} \\ \frac{a+2b}{a+a+b} &= \frac{7}{8} \\ \frac{a+2b}{2a+b} &= \frac{7}{8} \\ 8(a+2b) &= 7(2a+b) \\ 8a+16b &= 14a+7b \\ 9b &= 6a \\ b &= \frac{2a}{3} \\ \frac{u_2+u_3}{u_1} &= \frac{a+b+a+2b}{a} \\ &= \frac{2a+3b}{a} \\ &= \frac{2a+3(\frac{2a}{3})}{a} \\ &= \frac{2a+2a}{a} \\ &= \frac{4a}{a} \\ &= 4 \\ \end{align} </math> </div></div> <ol start=90> <li>Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7p+q}{2p+q} &= \frac{17p+q}{7p+q} \\ (7p+q)^2 &= (17p+q)(2p+q) \\ 49p^2+14pq+q^2 &= 34p^2+19pq+q^2 \\ 15p^2 &= 5pq \\ 3p &= q \\ \frac{7p+q}{2p+q} &= \frac{7p+3p}{2p+3p} \\ &= \frac{10p}{5p} \\ &= 2 \\ \end{align} </math> </div></div> <ol start=91> <li>Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{rataan geometris } \\ \sqrt{a \cdot b} &= b-24 \\ a \cdot b &= (b-24)^2 \\ \text{rataan aritmatik } \\ \frac{a+b}{2} &= a+15 \\ a+b &= 2(a+15) \\ a+b &= 2a+30 \\ a &= b-30 \\ a \cdot b &= (b-24)^2 \\ (b-30)b &= (b-24)^2 \\ b^2-30b &= b^2-48b+576 \\ 18b &= 576 \\ b &= 32 \\ a &= b-30 \\ &= 32-30 \\ &= 2 \\ a+b &= 32+2 \\ &= 34 \\ \end{align} </math> </div></div> <ol start=91> <li>Segitiga lancip ABC dengan <math>\frac{a^4+b^4+c^4+a^2b^2}{c^2(a^2+b^2)}=2</math>. tentukan nilai sudut C?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{syarat segitiga lancip semua sudut masing-masing kurang dari } 90^\circ \\ c^2 &= a^2+b^2-2ab cos C \\ cos C &= \frac{a^2+b^2-c^2}{2ab} \\ a^4+b^4+c^4+a^2b^2 &= 2c^2(a^2+b^2) \\ a^4+b^4+a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-2c^2(a^2+b^2)+(c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= (ab)^2 \\ a^2+b^2-c^2 &= \pm ab \\ cos C &= \pm \frac{ab}{2ab} \\ &= \pm \frac{1}{2} \\ &= \frac{1}{2} \text{ (karena sudut harus kurang dari } 90^\circ) \\ C &= 60^\circ \\ \end{align} </math> </div></div> <ol start=92> <li>Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan panjang DA dan EA } = x \text{ dan panjang AB } = y \\ \text{luas segitiga CAB } &= \frac{CA \cdot AB}{2} \\ 45 &= \frac{(x+9)(x+5)}{2} \\ 90 &= x^2+14x+45 \\ x^2+14x &= 45 \\ y^2 &= (x+9)^2+(x+5)^2 \\ &= x^2+18x+81+x^2+10x+25 \\ &= 2x^2+28x+106 \\ &= 2(x^2+14x)+106 \\ &= 2(45)+106 \\ &= 196 \\ y &= 14 \\ \end{align} </math> jadi panjang BC adalah 14 cm </div></div> <ol start=93> <li>Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kita cari ukuran GC yaitu } \\ \frac{GC}{AD} &= \frac{CE}{DE} \\ \frac{GC}{15} &= \frac{6}{18} \\ GC &= 5 \\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC } \\ &= \frac{1}{2} \cdot BC \cdot CF - \frac{1}{2} \cdot GC \cdot CE \\ &= \frac{1}{2} \cdot 15 \cdot 18 - \frac{1}{2} \cdot 5 \cdot 6 \\ &= 135 - 15 \\ &= 120 \\ \end{align} </math> jadi luas daerah BFEG adalah 120 cm² </div></div> <ol start=94> <li>Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ AI &= \frac{1}{5} AF \\ &= \frac{1}{5} 10 \\ &= 2 \\ IF &= AF-AI \\ &= 10-2 \\ &= 8 \\ \text{luas trapesium AFGD } &= \frac{(AD+EF) \cdot AF}{2} \\ &= \frac{(4+6)10}{2} \\ &= 50 \\ \text{luas segitiga AID } &= \frac{AI \cdot AF}{2} \\ &= \frac{(2)4}{2} \\ &= 4 \\ \text{luas segitiga IFG } &= \frac{IF \cdot FG}{2} \\ &= \frac{(8)6}{2} \\ &= 24 \\ \text{luas daerah segitiga IGD } &= \text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG } \\ &= 50-4-24 \\ &= 22 \\ \end{align} </math> jadi luas daerah segitiga IGD adalah 22 cm² </div></div> <ol start=96> <li>Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-<math>\pi</math>) cm³ maka berapa luas permukaan kerucut tersebut?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ \text{volume balok} \\ V_b &= x^2(12) \\ \text{volume kerucut} \\ V_b &= \frac{1}{3}\pi x^2(12) \\ &= 4\pi x^2 \\ V_{b-k} &= Vb-Vk \\ 100(3-\pi) &= 12x^2-4\pi x^2 \\ 100(3-\pi) &= 4x^2(3-\pi) \\ x^2 &= 25 \\ x &= 5 \\ s &= \sqrt{12^2+5^2} \\ &= \sqrt{144+25} \\ &= \sqrt{169} \\ &= 13 \\ \text{luas permukaan kerucut } &= \pi r(r+s) \\ &= \pi(5)(5+13) \\ &= 90\pi \\ \end{align} </math> jadi luas daerah permukaan kerucut adalah 90<math>\pi</math> cm² </div></div> <ol start=97> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 3a+1 \\ B &= 3b+2 \\ A(A+1)+3B \\ (3a+1)(3a+1+1)+3(3b+2) \\ (3a+1)(3a+2)+9b+6 \\ 9a^2+9a+2+9b+6 \\ 9a^2+9a+9b+8 \\ 9(a^2+a+b)+8 \\ \text{sisa pembagiannya adalah } 8 \\ \end{align} </math> </div></div> <ol start=98> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 9a+7 \\ B &= 9b+8 \\ A(A-5)+9B \\ (9a+7)(9a+7-5)+9(9b+8) \\ (9a+7)(9a+2)+81b+72 \\ 81a^2+81a+14+81b+72 \\ 81a^2+81a+81b+86 \\ 81a^2+81a+81b+81+5 \\ 81(a^2+a+b+1)+5 \\ \text{sisa pembagiannya adalah } 5 \\ \end{align} </math> </div></div> <ol start=99> <li>Jika <math>\begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix}</math> maka berapa hasil dari A²¹+A²⁵+A⁴⁶?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A^2 &= A \cdot A \\ &= \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \\ A^3 &= A^2 \cdot A \\ &= \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ &= - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= -I \\ A^{21}+A^{25}+A^{46} &= A^{21} \cdot (I+A^4+A^{25}) \\ &= A^{21} \cdot (I+A^3 \cdot A +A^{24} \cdot A) \\ &= (A^3)^7 \cdot (I+A^3 \cdot A +(A^3)^8 \cdot A) \\ &= (-I)^7 \cdot (I-I \cdot A +(-I)^8 \cdot A) \\ &= -I \cdot (I-A+A) \\ &= -I \cdot I \\ &= -I \\ &= -\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ \end{align} </math> </div></div> <ol start=100> <li>Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida?</li></ol> : bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15 : ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9). : ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8) : maka Ida menulis bilangan 8. <ol start=101> <li>Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9?</li></ol> : Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu. : untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja. : untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja. : Jadi banyaknya bilangan mungkin 2. <ol start=102> <li>Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli!</li></ol> ;cara 1 # 8¹ = 1 # 8² = 1 (8²=8¹x8¹ sama dengan 1x1) # 8³ = 1 (8³=8¹x8² sama dengan 1x1) # 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2) # 8⁵ = 1 # 8ⁿ = 1 (semua n untuk bilangan asli) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli ;cara 2 # 8ⁿ = b mod 7 # 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil) # (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya) # 8ⁿ = 1ⁿ mod 7 # 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli <ol start=103> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 5?</li></ol> ;cara 1 # 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 # 7¹ = 7 (sisa 1) # 7² = 49 (sisa 2) # 7³ = 343 (sisa 3) # 7⁴ = 2,401 (sisa 0) # 7⁵ = 16,807 # 7⁶ = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3 ;cara 2 :17¹ = 2 :17² = 4 :17³ = 3 :17⁴ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3 :17⁹⁹ = (17⁴)²⁴ x 17³ Untuk 17⁴ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 3. Jadi 17⁹⁹ dibagi 7 bersisa 3 ;cara 3 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 17⁴ ::17⁴ = 1 mod 5 ::(17⁴)²⁴ = 1²⁴ mod 5 ::17⁹⁶ = 1²⁴ mod 5 ::17⁹⁶ = 1 mod 5 ::17⁹⁶ x 17³ = 1 x 17³ mod 5 ::17⁹⁹ = 17³ mod 5 ::17⁹⁹ = 17 x 17 x 17 mod 5 ::17⁹⁹ = 2 x 2 x 2 mod 5 ::17⁹⁹ = 8 mod 5 ::17⁹⁹ = 3 mod 5 Jadi 17⁹⁹ dibagi 5 bersisa 3 <ol start=104> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 7?</li></ol> ;cara 1 :17¹ = 3 :17² = 2 :17³ = 6 :17⁴ = 4 :17⁵ = 5 :17⁶ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3 :17⁹⁹ = (17⁶)¹⁶ x 17³ Untuk 17⁶ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 6. Jadi 17⁹⁹ dibagi 7 bersisa 6 ;cara 2 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 17⁶ ::17⁶ = 1 mod 7 ::(17⁶)¹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1 mod 7 ::17⁹⁶ x 17³ = 1 x 17³ mod 7 ::17⁹⁹ = 17³ mod 7 ::17⁹⁹ = 17 x 17 x 17 mod 7 ::17⁹⁹ = 3 x 3 x 3 mod 7 ::17⁹⁹ = 27 mod 7 ::17⁹⁹ = 6 mod 7 Jadi 17⁹⁹ dibagi 7 bersisa 6 <ol start=105> <li>Berapa hasil sisa dari 41²⁰²⁴ dibagi 33?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 41^{2024} &= 41^{2024} \text{ mod } 33 \\ &= (33 \times 3 + 2)^{2024} \text{ mod } 33 \\ &= 2^{2024} \text{ mod } 33 \\ &= 2^{2020} 2^4 \text{ mod } 33 \\ &= (2^5)^{404} 2^4 \text{ mod } 33 \\ &= (33 - 1)^{404} 2^4 \text{ mod } 33 \\ &= (-1)^{404} 2^4 \text{ mod } 33 \\ &= 2^4 \text{ mod } 33 \\ &= 16 \text{ mod } 33 \\ \text{Jadi hasil sisa adalah } 16 \\ \end{align} </math> </div></div> <ol start=106> <li>Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 99^{99} &= (3^2 \times 11)^{99} \\ &= 3^{198} \times 11^{99} \\ 243^n &= (3^5)^n \\ &= 3^{5n} \\ \text{agar bisa membagi, maka} \\ 5n &= 198 \\ n &= 39.6 \\ \text{jadi bilangan n terbesar adalah } 39 \\ \end{align} </math> </div></div> <ol start=107> <li>Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸?</lu></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 88^{88} &= (8 \times 11)^{88} \\ &= 8^{88} \times 11^{88} \\ &= 8^{87} \times 8 \times 11^{88} \\ &= (8^3)^{29} \times 8 \times 11^{88} \\ &= 512^{29} \times 8 \times 11^{88} \\ 512^n &= 512^{29} \\ \text{jadi bilangan n terbesar adalah } 29 \\ \end{align} </math> </div></div> <ol start=108> <li>Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6!</li></ol> ; cara 1 : KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. : N dibagi 3 sisa 1 : N dibagi 5 sisa 2 : N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a : KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70 : KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21 : KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97 ; cara 2 : Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35 : kemudian ditambahkan sisa masing-masing sesuai dengan KPK. : KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, <b>97</b> : KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, <b>97</b> : KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, <b>97</b> Jadi bilangan bulat positif adalah 97 :: NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 <ol start=109> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |- | 4 || 0 || Semua isi ember B dibuang |- | 1 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. <ol start=110> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |} nah ada ember A berisi 4 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |- | 1 || 0 || Semua isi ember B dibuang |- | 0 || 1 || Tuangkan 1 l dari ember A ke B sehingga ember A kosong |- | 3 || 1 || Isikan 3 l ke ember A |- | 0 || 4 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |} nah ada ember B berisi 4 liter. # Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kemungkinan 1} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 +3 \cdot ((\frac{4}{3})^x)^2 &= 7 \cdot (\frac{4}{3})^x \\ \text{misalkan } (\frac{4}{3})^x = y \\ 4 +3y^2 &= 7y \\ 3y^2-7y+4 &= 0 \\ (3y-4)(y-1) &= 0 \\ y_1 = \frac{4}{3} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{4}{3} \\ (\frac{4}{3})^x &= \frac{4}{3} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{4}{3})^x &= 1 \\ x &= 0 \\ \text{kemungkinan 2} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot ((\frac{3}{4})^x)^2 +3 &= 7 \cdot (\frac{3}{4})^x \\ \text{misalkan } (\frac{3}{4})^x = y \\ 4y^2 +3 &= 7y \\ 4y^2-7y+3 &= 0 \\ (4y-3)(y-1) &= 0 \\ y_1 = \frac{3}{4} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{3}{4} \\ (\frac{3}{4})^x &= \frac{3}{4} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{3}{4})^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-6}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{2x-4}{3}+2x-6 &= \frac{3}{2x-4}+\frac{1}{2x-6} \\ \frac{2x-4}{3}+2x-6 &= \frac{1}{\frac{2x-4}{3}}+\frac{1}{2x-6} \\ \text{misalkan } \frac{2x-4}{3} = y \text{ dan } 2x-6 = z \\ y+z &= \frac{1}{y}+\frac{1}{z} \\ y+z &= \frac{y+z}{xy} \\ y+z-\frac{y+z}{xy} &= 0 \\ (y+z)(1-\frac{1}{xy}) &= 0 \\ *y+z=0 \\ y+z &= 0 \\ y &= -z \\ \frac{2x-4}{3} &= -(2x-6) \\ \frac{2(x-2)}{3} &= -2(x-3) \\ \frac{x-2}{3} &= -x+3 \\ x-2 &= -3x+9 \\ 4x &= 11 \\ x &= \frac{11}{4} \\ \end{align} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 7ga2aoqk8t1vq49sx8u0t577zoasxyt 117493 117492 2026-07-13T05:04:53Z Akuindo 8654 117493 wikitext text/x-wiki contoh soal <ol start=1> <li>Berapa hasil dari <math>\sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Misalkan 2020 = p} \\ \sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64} &= \sqrt{(2020-5) \cdot (2020-3) \cdot (2020+3) \cdot (2020+5) + 64} \\ &= \sqrt{(p-5) \cdot (p-3) \cdot (p+3) \cdot (p+5) + 64} \\ &= \sqrt{(p-5) \cdot (p+5) \cdot (p-3) \cdot (p+3) + 64} \\ &= \sqrt{(p^2-25) \cdot (p^2-9) + 64} \\ &= \sqrt{p^4-34p^2+ 225 + 64} \\ &= \sqrt{p^4-34p^2+ 289} \\ &= \sqrt{(p^2-17)^2} \\ &= p^2-17 \\ &= 2020^2-17 \\ &= (2000+20)^2-17 \\ &= 4.000.000+80.000+400-17 \\ &= 4.080.383 \\ \end{align} </math> </div></div> <ol start=2> <li>Berapa nilai x dari <math>\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} = \frac{9}{10}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} &= \frac{9}{10} \\ \text{misalkan untuk } \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} = p \\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} &= p \\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}} &= p^2 \\ x^2-x-p &= p^2 \\ x^2-2x+1+x-1 &= p^2+p \\ (x-1)^2+(x-1) &= p^2+p \\ x-1 &= p \\ \text{misalkan untuk } \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}} &= q^3 \\ x^2 q &= q^3 \\ x^2 &= q^2 \\ x &= q \\ \frac{x-1}{x} &= \frac{9}{10} \\ x &= 10 \\ \end{align} </math> </div></div> <ol start=3> <li>Berapa nilai x dari <math>(\frac{x}{x+10})^{x+10}=\frac{1}{1024}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{x+10}{x})^{-(x+10)} &= (1024)^{-1} \\ (\frac{x+10}{x})^{x+10} &= 1024 \\ (\frac{x+10}{x})^{x+10} &= 2^{10} \\ (\frac{x+10}{x})^{\frac{x+10}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (\frac{1}{2})^{-1} \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (1+(-\frac{1}{2}))^{(1+(-\frac{2}{1}))} \\ \frac{10}{x} &= -\frac{1}{2} \\ x &= -20 \\ \end{align} </math> </div></div> <ol start=4> <li>Berapa nilai x dari <math>x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{2}+\sqrt{x+\frac{1}{4}} &= (\sqrt{x+\frac{1}{4}})^2+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+(\frac{1}{2})^2 \\ &= (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 \\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}} &= 4 \\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2} &= 4 \\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 4 \\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 &= 4 \\ \sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 2 \\ \sqrt{x+\frac{1}{4}} &= \frac{3}{2} \\ x+\frac{1}{4} &= \frac{9}{4} \\ x &= 2 \\ \end{align} </math> </div></div> <ol start=5> <li>Berapa nilai x dari <math>\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^3}{\sqrt{8-x^2}}+x^2-8 &= 0 \\ \frac{x^3}{\sqrt{8-x^2}} &= 8-x^2 \\ x^3 &= (8-x^2)^{\frac{3}{2}} \\ x &= (8-x^2)^{\frac{1}{2}} \\ x^2 &= 8-x^2 \\ 2x^2-8 &= 0 \\ x^2-4 &= 0 \\ (x-2)(x+2) &= 0 \\ \text{membuktikan } \\ x=2 \text{ maka hasilnya 0 } \\ x=-2 \text{ maka hasilnya -8 } \\ \text{jadi } x=2 \\ \end{align} </math> </div></div> <ol start=6> <li>Berapa nilai x dari <math>\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} = 5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} &= 5 \\ \frac{x^{50}+x^{60}+x^{70}}{31}} &= 5^5 \\ x^{50}+x^{60}+x^{70} &= 5^5 \cdot 31 \\ x^{50}(1+x^{10}+x^{20}) &= 5^5 \cdot 31 \\ (x^{10}^5)(1+x^{10}+(x^{10}^2) &= 5^5 \cdot 31 \\ \text{ misalkan } x^{10} = a \\ a^5(1+a+a^2) &= 5^5 \cdot 31 \\ a &= 5 \\ x^{10} &= 5 \\ x &= ^5 log 10 \\ \end{align} </math> </div></div> <ol start=7> <li>Berapa nilai x dari <math>\sqrt{3x+5+\sqrt{4x+5}} = x</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{3x+5+\sqrt{4x+5}} &= x \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \text{misalkan } \sqrt{4x+5}=y \text{ dan } 4x+5=y^2 \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \sqrt{y^2+y-x} &= x \\ y^2+y &= x^2+x \\ y=x \\ 4x+5 &= y^2 \\ 4x+5 &= x^2 \\ x^2-4x-5 &= 0 \\ (x-5)(x+1) &= 0 \\ x=5 &\text{ atau } x=-1 \text{ (TM) } \\ \end{align} </math> </div></div> <ol start=8> <li>Berapa nilai x dari <math>\sqrt{1+\sqrt{1+x}} = \sqrt[3]{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{1+\sqrt{1+x}} &= \sqrt[3]{x} \\ \sqrt[3]{x} &= n \\ x &= n^3 \\ \sqrt{1+\sqrt{1+n^3}} &= n \\ 1+\sqrt{1+n^3} &= n^2 \\ \sqrt{1+n^3} &= n^2-1 \\ 1+n^3 &= n^4-2n^2+1 \\ n^4-n^3-2n^2 &= 0 \\ n^2(n^2-n-2) &= 0 \\ n^2(n-2)(n+1) &= 0 \\ n=0, n=2 \text{ atau } n=-1 \\ n &= 0 \\ x &= 0^3 \\ &= 0 \\ n &= 2 \\ x &= 2^3 \\ &= 8 \\ n &= -1 \\ x &= (-1)^3 \\ &= -1 \\ \text{yang paling mungkin untuk nilai x adalah } 8 \\ \end{align} </math> </div></div> <ol start=9> <li>Berapa nilai x dari <math>\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} &= \frac{\sqrt{1+x}}{\sqrt{x}} \\ \sqrt{x}(\sqrt{1+x}+\sqrt{x}) &= (\sqrt{1+x}-\sqrt{x})\sqrt{1+x} \\ \sqrt{x(1+x)}+x &= 1+x-\sqrt{x(1+x)} \\ 2\sqrt{x(1+x)} &= 1 \\ \sqrt{x(1+x)} &= \frac{1}{2} \\ x(1+x) &= \frac{1}{4} \\ x^2+x &= \frac{1}{4} \\ 4x^2+4x &= 1 \\ 4x^2+4x-1 &= 0 \\ x &= \frac{-4 \pm \sqrt{4^2-4(4)(-1)}}{2(4)} \\ &= \frac{-4 \pm \sqrt{32}}{8} \\ &= \frac{-4 \pm 4\sqrt{2}}{8} \\ &= \frac{-1 \pm \sqrt{2}}{2} \\ \text{karena akar x harus minimal nol jadi } x = \frac{-1+\sqrt{2}}{2} \\ \end{align} </math> </div></div> <ol start=10> <li>Berapa nilai x dari <math>\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x-\sqrt{x+1}}{x+\sqrt{x+1}} &= \frac{11}{19} \\ \text{misalkan } \sqrt{x+1}=y \text{ dan } x=y^2-1 \\ \frac{y^2-1-y}{y^2-1+y} &= \frac{11}{19} \\ 19(y^2-y-1) &= 11(y^2+y-1) \\ 19y^2-19y-19 &= 11y^2+11y-11 \\ 8y^2-30y-8 &= 0 \\ 4y^2-15y-4 &= 0 \\ (4y+1)(y-4) &= 0 \\ y=-\frac{1}{4} \text{ (TM) atau } & y=4 \\ x &= 4^2-1 \\ &= 15 \\ \end{align} </math> </div></div> <ol start=11> <li>Berapa nilai x dari <math>\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} &= 98 \\ \text{misalkan } \sqrt{x^2-1}=y \\ \frac{x+y}{x-y}+\frac{x-y}{x+y} &= 98 \\ \frac{(x+y)^2+(x-y)^2}{(x-y)(x+y)} &= 98 \\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} &= 98 \\ \frac{2(x^2+y^2)}{x^2-y^2} &= 98 \\ \frac{x^2+y^2}{x^2-y^2} &= 49 \\ x^2+y^2 &= 49(x^2-y^2) \\ x^2+y^2 &= 49x^2-49y^2 \\ 48x^2 &= 50y^2 \\ 24x^2 &= 25y^2 \\ 24x^2 &= 25(\sqrt{x^2-1})^2 \\ 24x^2 &= 25(x^2-1) \\ 24x^2 &= 25x^2-25 \\ x^2 &= 25 \\ x &= \pm 5 \\ \end{align} </math> </div></div> <ol start=12> <li>Berapa nilai x dari <math>\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \sqrt{x}=y \text{ dan } x=y^2 \\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} &= \frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\sqrt{\frac{y^2}{y^2+y}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\frac{y}{\sqrt{y^2+y}} \\ y^2+y-\sqrt{(y^2+y)(y^2-y)} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^4-y^2} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^2(y^2-1)} &= \frac{5}{4}y \\ y(y+1)-y\sqrt{y^2-1} &= \frac{5}{4}y \\ y+1-\sqrt{y^2-1} &= \frac{5}{4} \\ -\sqrt{y^2-1} &= \frac{1}{4}-y \\ y^2-1 &= (\frac{1}{4}-y)^2 \\ y^2-1 &= \frac{1}{16}-\frac{1}{2}y+y^2 \\ -1 &= \frac{1}{16}-\frac{1}{2}y \\ \frac{1}{2}y &= \frac{1}{16}+1 \\ \frac{1}{2}y &= \frac{17}{16} \\ y &= \frac{17}{8} \\ x &= (\frac{17}{8})^2 \\ &= \frac{289}{64} \\ \end{align} </math> </div></div> <ol start=13> <li>Berapa nilai x dari <math>\sqrt[4]{62+x}+\sqrt[4]{275-x}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ misalkan } \sqrt[4]{62+x}=a, 62+x=a^4, \sqrt[4]{275-x}=b \text{ dan } 275-x=b^4 \\ a+b &= 7 \\ (a+b)^2 &= 49 \\ a^2+b^2+2ab &= 49 \\ a^2+b^2 &= 49-2ab \\ a^4+b^4 &= 62+x+275-x \\ (a^2+b^2)^2-2(ab)^2 &= 337 \\ (49-2ab)^2-2(ab)^2 &= 337 \\ 2401-196ab+4(ab)^2-2(ab)^2 &= 337 \\ 2(ab)^2-196ab+2064 &= 0 \\ (ab)^2-98ab+1032 &= 0 \\ (ab-12)(ab-86) &= 0 \\ ab = 12 \text{ atau } & ab = 86 \text{ (TM) karena hasil kali maksimum yaitu 12 } \\ ab =12 \text{ dan } a+b=7 \\ a+b &= 7 \\ b &= 7-a \\ ab &= 12 \\ a(7-a) &= 12 \\ -a^2+7a &= 12 \\ a^2-7a+12 &= 0 \\ (a-3)(a-4) &= 0 \\ a=3 \text{ atau } & a=4 \\ a=3, b=4 \\ 62+x &= a^4 \\ 62+x &= (3)^4 \\ 62+x &= 81 \\ x &= 19 \\ a=4, b=3 \\ 62+x &= a^4 \\ 62+x &= (4)^4 \\ 62+x &= 256 \\ x &= 194 \\ \end{align} </math> </div></div> <ol start=14> <li>Berapa nilai x dari <math>\sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2} &= 7 \\ (\sqrt[3]{8+x})^2-\sqrt[3]{8+x} \sqrt[3]{27-x}+(\sqrt[3]{27-x})^2 &= 7 \\ \text{misalkan } \sqrt[3]{8+x}=a, 8+x=a^3, \sqrt[3]{27-x}=b \text{ dan } 27-x=b^3 \\ a^2-ab+b^2 &= 7 \\ a^3+b^3 &= 8+x+27-x \\ &= 35 \\ a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ 35 &= (a+b)(7) \\ a+b &= 5 \\ b &= 5-a \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ 5^3 &= 35+3ab(5) \\ 125 &= 35+15ab \\ 80 &= 15ab \\ ab &= 6 \\ a(5-a) &= 6 \\ 5a-a^2 &= 6 \\ a^2-5a+6 &= 6 \\ (a-2)(a-3) &= 6 \\ a=2 &\text{ atau } a=3 \\ a=2, b=3 \text{ dan } a=3,b=2 \\ 8+x &= a^3 \\ &= 2^3 \\ &= 8 \\ x &= 0 \\ 8+x &= a^3 \\ &= 3^3 \\ &= 27 \\ x &= 19 \\ \end{align} </math> </div></div> <ol start=15> <li>Berapa nilai x dari <math>3^x+5^x-9^x+15^x-25^x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3^x+5^x-9^x+15^x-25^x &= 1 \\ 3^x+5^x-(3^2)^x+(3 \cdot 5)^x-(5^2)^x &= 1 \\ 3^x+5^x-(3^x)^2+(3^x \cdot 5^x)-(5^x)^2 &= 1 \\ \text{misalkan } 3^x=a \text{ dan } 5^x=b \\ a+b-a^2+ab-b^2 &= 1 \\ a^2-ab+b^2-a-b+1 &= 0 \\ 2a^2-2ab+2b^2-2a-2b+2 &= 0 \\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1 &= 0 \\ (a-b)^2+(a-1)^2+(b-1)^2 &= 0 \\ a-b=0; a-1=0; b-1 &= 0 \\ a=b &= 1 \\ 3^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> <ol start=16> <li>Berapa nilai x dari <math>^6log x^2+^{6x}log \frac{6}{x}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ \text{misalkan } 6x=a \text{ maka } x=\frac{a}{6} \\ ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ ^6log (\frac{a}{6})^2+^{6 \frac{a}{6}}log \frac{6}{\frac{a}{6}} &= 1 \\ ^6log \frac{a^2}{6^2}+^alog \frac{6^2}{a} &= 1 \\ ^6log a^2-^6log 6^2+^alog 6^2-^alog a &= 1 \\ 2 ^6log a-2 ^6log 6+2 ^alog 6-^alog a &= 1 \\ 2 ^6log a-2+2 \frac{1}{^6log a}-1 &= 1 \\ 2 ^6log a+2 \frac{1}{^6log a}-4 &= 0 \\ 2 ^6log^2 a-4 ^6log a+2 &= 0 \\ ^6log^2 a-2 ^6log a+1 &= 0 \\ (^6log a-1)^2 &= 0 \\ ^6log a &= 1 \\ a &= 6 \\ x &= \frac{a}{6} \\ &= \frac{6}{6} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=17> <li>Berapa nilai x dari (x+500)³+x=20?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+500)^3+x &= 20 \\ \text{misalkan } a=x+500 \text{ maka } x=a-500 \\ a^3+a-500 &= 20 \\ a^3+a &= 520 \\ a(a^2+1) &= 8 \cdot 65 \\ a(a^2+1) &= 8(64+1) \\ a(a^2+1) &= 8(8^2+1) \\ a &= 8 \\ x &= 8-500 \\ &= -492 \\ \end{align} </math> </div></div> <ol start=18> <li>Berapa nilai x dari <math>\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2} &= 0 \\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}} &= \frac{1}{2} \\ \frac{x^n+4^n}{x^n+16^n} &= (\frac{1}{2})^n \\ \frac{x^n+4^n}{x^n+16^n} &= \frac{1}{2^n} \\ 2^n(x^n+4^n) &= x^n+16^n \\ 2^n(x^n+2^{2n}) &= x^n+2^{4n} \\ 2^n \cdot x^n+2^{3n} &= x^n+2^{4n} \\ 2^n \cdot x^n-x^n &= 2^{4n}-2^{3n} \\ x^n(2^n-1) &= 2^{3n}(2^n-1) \\ x^n &= 2^{3n} \\ x^n &= (2^3)^n \\ x^n &= 8^n \\ x &= 8 \\ \end{align} </math> </div></div> <ol start=19> <li>Berapa hasil dari <math>\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ x &= \frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5 \cdot 6}+\sqrt{5 \cdot 5}+\sqrt{6 \cdot 4}+\sqrt{5 \cdot 4}}{\sqrt{5 \cdot 4}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5} \cdot \sqrt{6}+\sqrt{5} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{4}}{2 \cdot \sqrt{5}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{6} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{5}+\sqrt{5} \cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}} \\ &= \frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ &= \frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ \frac{1}{x} &= \frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}} \\ &= \frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ &= \frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1} \\ &= \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5} \\ &= \sqrt{6}-\sqrt{4} \\ &= \sqrt{6}-2 \\ x &= \frac{1}{\sqrt{6}-2} \\ &= \frac{\sqrt{6}+2}{6-4} \\ &= \frac{\sqrt{6}+2}{2} \\ &= 1+\frac{\sqrt{6}}{2} \\ \end{align} </math> </div></div> <ol start=20> <li>Berapa hasil dari <math>(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5 \\ \text{misalkan } x=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ x &= \frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ &= \frac{\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}} \\ &= \frac{\sqrt{3}+1}{4} \\ 4x &= \sqrt{3}+1 \\ 4x-1 &= \sqrt{3} \\ (4x-1)^2 &= 3 \\ 16x^2-8x+1 &= 3 \\ 16x^2 &= 8x+2 \\ 8x^2 &= 4x+1 \\ x^2 &= \frac{4x+1}{8} \\ *cara 1 \\ x^3 &= x \cdot x^2 \\ &= x(\frac{4x+1}{8}) \\ &= \frac{4x^2+x}{8} \\ &= \frac{4x^2}{8}+\frac{x}{8} \\ &= \frac{4(\frac{4x+1}{8})}{8}+\frac{x}{8} \\ &= \frac{16x+4}{64}+\frac{x}{8} \\ &= \frac{4x+1}{16}+\frac{x}{8} \\ &= \frac{4x+1+2x}{16} \\ &= \frac{6x+1}{16} \\ x^5 &= x^2 \cdot x^3 \\ &= (\frac{4x+1}{8})(\frac{6x+1}{16}) \\ &= \frac{24x^2+10x+1}{128} \\ &= \frac{24x^2}{128}+\frac{10x+1}{128} \\ &= \frac{24(\frac{4x+1}{8})}{128}+\frac{10x+1}{128} \\ &= \frac{96x+24}{1024}+\frac{10x+1}{128} \\ &= \frac{96x+24+80x+8}{1024} \\ &= \frac{176x+32}{1024} \\ &= \frac{176x}{1024}+\frac{32}{1024} \\ &= \frac{176}{1024}(\frac{\sqrt{3}+1}{4})+\frac{32}{1024} \\ &= \frac{44(\sqrt{3}+1)}{1024}+\frac{32}{1024} \\ &= \frac{44\sqrt{3}+44}{1024}+\frac{32}{1024} \\ &= \frac{76+44\sqrt{3}}{1024} \\ &= \frac{19+11\sqrt{3}}{256} \\ *cara 2 \\ x^4 &= (x^2)^2 \\ &= (\frac{4x+1}{8})^2 \\ &= \frac{16x^2+8x+1}{64} \\ &= \frac{16x^2}{64}+\frac{8x}{64}+\frac{1}{64} \\ &= \frac{x^2}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{\frac{4x+1}{8}}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{4x}{32}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{8}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{4}+\frac{3}{64} \\ x^5 &= x \cdot x^4 \\ &= (\frac{\sqrt{3}+1}{4})(\frac{x}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\frac{\sqrt{3}+1}{4}}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\sqrt{3}+1}{16}+\frac{3}{64}) \\ &= \frac{(\sqrt{3}+1)^2}{64}+(\frac{\sqrt{3}+1}{4})\frac{3}{64} \\ &= \frac{3+2\sqrt{3}+1}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{4+2\sqrt{3}}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{16+8\sqrt{3}}{256}+\frac{3\sqrt{3}+3}{256} \\ &= \frac{19+11\sqrt{3}}{256} \\ \end{align} </math> </div></div> <ol start=21> <li>Berapa hasil dari <math>\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4} &= \frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots) \\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots &= \frac{1}{1-\frac{1}{4}} \\ &= \frac{4}{3} \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{4}{3}) \\ &= \frac{1}{4}+\frac{16}{3} \\ &= \frac{67}{12} \\ x &= \frac{67}{9} \\ \end{align} </math> </div></div> <ol start=22> <li>Berapa nilai y-x jika <math>\frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} = \frac{x}{y}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} &= \frac{x}{y} \\ \frac{\frac{106 \times 107}{2}}{\frac{106}{2}(4+109)} &= \frac{x}{y} \\ \frac{53 \times 107}{53 \times 113} &= \frac{x}{y} \\ y-x &= 113-107 = 6 \\ \end{align} </math> </div></div> <ol start=23> <li>Berapa angka satuan dari hasil 17²⁰²⁴?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan angka satuannya} \\ 17^1 &= 7 \\ 17^2 &= 9 \\ 17^3 &= 3 \\ 17^4 &= 1 \\ 17^5 &= 7 \\ 17^6 &= 9 \\ 17^7 &= 3 \\ 17^8 &= 1 \\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1} \end{align} </math> </div></div> <ol start=24> <li>Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ 1! + 2! + 3! + 4! + \dots + 2024! &= 1 + (1x2) + (1x2x3) + (1x2x3x4) + \dots + 2024! \\ &= 1 + 2 + 6 + 24 + 120 + 720 + \dots + 2024! \\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka } 1+2+6+24 = 33 \text{ jadi angka satuannya adalah } 3 \end{align} </math> </div></div> <ol start=25> <li>Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ \frac{1! + 2! + 3! + 4! + \dots + 2024!}{12} &= \frac{1 + 1x2 + 1x2x3 + 1x2x3x4 + \dots + 2024!}{12} \\ &= \frac{1 + 2 + 6 + 24 + \dots + 2024!}{12} \\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi } 1+2+6 = 9 \end{align} </math> </div></div> <ol start=26> <li>Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{penjumlahan} \\ 1+1+1+1+ \dots \text{ (sebanyak 88 buah) }+x+y &= A \\ 88+x+y &= A \\ \text{perkalian} \\ 1 \times 1 \times 1 \times \dots \text{ (sebanyak 88 buah) }\times x \times y &= A \\ x \times y &= A \\ 88+x+y &= xy \\ xy-y &= 88+x \\ y(x-1) &= 88+x \\ y &= \frac{88+x}{x-1} \\ \text{uji selidiki untuk x=2} \\ y &= \frac{88+2}{2-1} \\ &= 90 \\ \text{buktikan} \\ 88+x+y &= xy \\ 88+2+90 &= 2(90) \\ 180 &= 180 \\ \text{terbukti} \\ \text{nilai A adalah } 180 \\ \end{align} </math> </div></div> <ol start=27> <li>Berapakah nilai x, y dan z dari <math>x+y-z=1, x^2+y^2-z^2=-5 \text{ dan } x^3+y^3-z^3=-53</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+y-z &= 1 \\ x+y &= z+1 \\ x^2+2xy+y^2 &= z^2+2z+1 \\ x^2+y^2-z^2 &= 2z+1-2xy \\ -5 &= 2z+1-2xy \\ 2xy &= 2z+6 \\ xy &= z+3 \\ x^2+y^2-z^2 &= -5 \\ x^2+y^2 &= z^2-5 \\ x^3+y^3-z^3 &= -53 \\ (x+y)(x^2-xy+y^2)-z^3+53 &= 0 \\ (x+y)(x^2+y^2-xy)-z^3+53 &= 0 \\ (z+1)(z^2-5-(z+3))-z^3+53 &= 0 \\ (z+1)(z^2-z-8)-z^3+53 &= 0 \\ z^3-z^2-8z+z^2-z-8-z^3+53 &= 0 \\ -9z+45 &= 0 \\ -9z &= -45 \\ z &= 5 \\ x+y &= 5+1 \\ x+y &= 6 \\ x &= 6-y \\ xy &= 5+3 \\ xy &= 8 \\ (6-y)y &= 8 \\ 6y-y^2 &= 8 \\ y^2-6y+8 &= 0 \\ (y-4)(y-2) &= 0 \\ y=4 \text{ atau } y=2 \\ \text{jika } y=4 \\ x+y &= z+1 \\ x+4 &= 5+1 \\ x &= 2 \\ \text{jika } y=2 \\ x+y &= z+1 \\ x+2 &= 5+1 \\ x &= 4 \\ \end{align} </math> </div></div> <ol start=28> <li>Berapakah nilai titik koordinat (x,y) dari <math>\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}</math> dan <math>\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x+y}+\sqrt{x-y} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+y}-\sqrt{x-y} &= \sqrt{\frac{52y}{3x}} \\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y}) &= \sqrt{\frac{432x}{13y}} \cdot \sqrt{\frac{52y}{3x}} \\ x+y-x+y &= \sqrt{\frac{432x \cdot 52y}{13y \cdot 3x}} \\ 2y &= \sqrt{144 \cdot 4} \\ 2y &= \sqrt{576} \\ 2y &= 24 \\ y &= 12 \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13(12)}} \\ x+12+x-12+2 \cdot \sqrt{x+12} \cdot \sqrt{x-12} &= \frac{36x}{13} \\ 2x+2 \sqrt{x^2-144} &= \frac{36x}{13} \\ 2(x+\sqrt{x^2-144}) &= \frac{36x}{13} \\ x+\sqrt{x^2-144} &= \frac{18x}{13} \\ \sqrt{x^2-144} &= \frac{5x}{13} \\ x^2-144 &= \frac{25x^2}{169} \\ \frac{144x^2}{169}-144 &= 0 \\ \frac{x^2}{169}-1 &= 0 \\ x^2-169 &= 0 \\ (x-13)(x+13) &= 0 \\ x_1=13 &\text{ atau } x_2=-13 \text{ (TM) karena } x>y \\ \end{align} </math> jadi titik koordinat (13,12) </div></div> <ol start=29> <li>Berapakah nilai dari <math>x^2-7x</math> jika <math>(x-2)^2+\frac{1}{(x-2)^2} = 11</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x-2)^2+\frac{1}{(x-2)^2} &= 11 \\ (x-2)^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2)^2} &= 11-2 \\ (x-2-\frac{1}{x-2})^2 &= 9 \\ x-2-\frac{1}{x-2} &= 3 \\ (x-2)^2-1 &= 3(x-2) \\ x^2-4x+4-1 &= 3x-6 \\ x^2-7x &= -9 \\ \end{align} </math> </div></div> <ol start=30> <li>Berapakah nilai dari <math>\frac{(x+y)^2(x+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)}</math> jika xy+yz+xz=1?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy+yz+xz &= 1 \\ x^2+xy+yz+xz &= x^2+1 \\ x(x+y)+z(x+y) &= x^2+1 \\ (x+y)(x+z) &= x^2+1 \\ \text{dengan pola yang sama } \\ (y+x)(y+z) &= y^2+1 \\ (x+z)(y+z) &= z^2+1 \\ \frac{(x+y)^2(y+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)} &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)} \\ &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)^2(y+z)^2(x+z)^2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=31> <li>Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} w+5 &= w+x+y+z+5 \\ x+4 &= w+x+y+z+5 \\ y+3 &= w+x+y+z+5 \\ z+2 &= w+x+y+z+5 \\ \text{jumlahkan keempat persamaan } \\ w+x+y+z+14 &= 4(w+x+y+z+5) \\ w+x+y+z+14 &= 4(w+x+y+z)+20 \\ 3(w+x+y+z) &= -6 \\ w+x+y+z &= -2 \\ \end{align} </math> </div></div> <ol start=32> <li>Berapakah nilai dari <math>\frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2}</math> jika <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3</math> dan x+y+z=xyz?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2} &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2} \\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) \\ 3^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz}) \\ 9 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz}) \\ &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2 \\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} &= 7 \\ \end{align} </math> </div></div> <ol start=33> <li>Berapakah nilai dari <math>\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}</math> jika <math>x^2+y^2+z^2 = -2(ab+bc+ac)</math>?>/li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+z^2 &= -2(xy+yz+xz) \\ x^2+y^2+z^2+2(xy+yz+xz) &= 0 \\ (x+y+z)^2 &= 0 \\ x+y+z &= 0 \\ x+y &= -z \\ x+z &= -y \\ y+z &= -x \\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z} &= \frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x} \\ &= -2-(-5)-(-7) \\ &= 10 \\ \end{align} </math> </div></div> <ol start=34> <li>Berapakah nilai dari <math>\frac{20xyz}{xy+yz+xz}</math> jika <math>16^x = 256^y = 625^z = 40</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 16^x = 256^y = 625^z &= 40 \\ 2^{4x} = 4^{4y} = 5^{4z} &= 40 \\ 2^{4x} &= 40 \\ 2 &= 40^{\frac{1}{4x}} \\ 4^{4y} &= 40 \\ 4 &= 40^{\frac{1}{4y}} \\ 5^{4z} &= 40 \\ 5 &= 40^{\frac{1}{4z}} \\ 2 \cdot 4 \cdot 5 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z}} \\ 1 &= \frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z} \\ 4 &= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{20xyz}{xy+yz+xz} &= 20 \cdot \frac{xyz}{xy+yz+xz} \\ &= 20 \cdot (\frac{xy+yz+xz}{xyz})^{-1} \\ &= 20 \cdot (\frac{1}{z} + \frac{1}{x} + \frac{1}{y})^{-1} \\ &= 20 \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{-1} \\ &= 20 \cdot (4)^{-1} \\ &= 20 \cdot \frac{1}{4} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=35> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+3x^2+1}</math> jika <math>6x^2+25x+6=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 6x^2+25x+6 &= 0 \\ 6x+25+\frac{6}{x} &= 0 \\ 6(x+\frac{1}{x}) &= -25 \\ x+\frac{1}{x} &= \frac{-25}{6} \\ (c+\frac{1}{x})^2 &= (\frac{-25}{6})^2 \\ x^2+2+\frac{1}{x^2} &= \frac{625}{36} \\ x^2+\frac{1}{x^2} &= \frac{625}{36}-2 \\ x^2+\frac{1}{x^2} &= \frac{553}{36} \\ \frac{x^2}{x^4+3x^2+1} &= \frac{1}{x^2+3+\frac{1}{x^2}} \\ &= \frac{1}{a^2+\frac{1}{x^2}+3} \\ &= \frac{1}{\frac{553}{36}+3} \\ &= \frac{1}{\frac{661}{36}} \\ &= \frac{36}{661} \\ \end{align} </math> </div></div> <ol start=36> <li>Berapakah nilai dari <math>\frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5} &= \frac{(9+2\sqrt{20})^{1013}}{((2)^3+3(2)^2(\sqrt{5})+3(2)(\sqrt{5})^2+(\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{((2+\sqrt{5})^2)^{1013}}{((2+\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{(2+\sqrt{5})^{2026}}{(2+\sqrt{5})^{2025}}+6-\sqrt{5} \\ &= 2+\sqrt{5}+6-\sqrt{5} \\ &= 8 \\ \end{align} </math> </div></div> <ol start=37> <li>Berapakah nilai dari <math>27x^3+\frac{8}{x^3}</math> jika <math>3x+\frac{2}{x}=6</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3x+\frac{2}{x} &= 6 \\ (3x+\frac{2}{x})^3 &= 6^3 \\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3} &= 216 \\ 27x^3+18(6)+\frac{8}{x^3} &= 216 \\ 27x^3+108+\frac{8}{x^3} &= 216 \\ 27x^3+\frac{8}{x^3} &= 108 \\ \end{align} </math> </div></div> <ol start=38> <li>Berapakah nilai dari <math>x^6+\frac{8}{x^3}</math> jika <math>x^3+\frac{1}{x^3}=8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3+\frac{1}{x^3} &= 8 \\ x^3 &= 8-\frac{1}{x^3} \\ x^6 &= 8x^3-1 \\ x^6+\frac{8}{x^3} &= 8x^3-1+\frac{8}{x^3} \\ &= 8x^3+\frac{8}{x^3}-1 \\ &= 8(x^3+\frac{1}{x^3})-1 \\ &= 8(8)-1 \\ &= 63 \\ \end{align} </math> </div></div> <ol start=39> <li>Berapakah nilai dari <math>4x+\frac{25}{x}</math> jika <math>2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2\sqrt{x}+\frac{5}{\sqrt{x}} &= 4x-\frac{25}{x} \\ 2\sqrt{x}+\frac{5}{\sqrt{x}} &= (2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}}) \\ 1 &= 2\sqrt{x}-\frac{5}{\sqrt{x}} \\ 1^2 &= (2\sqrt{x}-\frac{5}{\sqrt{x}})^2 \\ 1 &= 4x-20+\frac{25}{x} \\ 4x+\frac{25}{x} &= 21 \\ \end{align} </math> </div></div> <ol start=40> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2-x+1}{x^2+x+1} &= \frac{5}{6} \\ \frac{x^2+1-x}{x^2+1+x} &= \frac{5}{6} \\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} &= \frac{5}{6} \\ \text{ misalkan } x+\frac{1}{x} &= y \\ \frac{y-1}{y+1} &= \frac{5}{6} \\ 6(y-1) &= 5(y+1) \\ 6y-6 &= 5y+5 \\ y &= 11 \\ x+\frac{1}{x} &= 11 \\ \end{align} </math> </div></div> <ol start=41> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt{x}+x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x}+x &= 1 \\ x-1 &= -\sqrt{x} \\ (x-1)^2 &= (-\sqrt{x})^2 \\ x^2-2x+1 &= x \\ x^2-3x+1 &= 0 \\ x-3+\frac{1}{x} &= 0 \\ x+\frac{1}{x} &= 3 \\ \end{align} </math> </div></div> <ol start=42> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt[3]{x}-\sqrt[3]{x-36}=3</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x}-\sqrt[3]{x-36} &= 3 \\ (\sqrt[3]{x}-\sqrt[3]{x-36})^3 &= 3^3 \\ x-(x-36)-3 \sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36}) &= 27 \\ 36-3 \sqrt[3]{x(x-36)}3 &= 27 \\ -9 \sqrt[3]{x(x-36)} &= -9 \\ \sqrt[3]{x(x-36)} &= 1 \\ x(x-36) &= 1 \\ x^2-36x-1 &= 0 \\ x-36-\frac{1}{x} &= 0 \\ x-\frac{1}{x} &= 36 \\ \end{align} </math> </div></div> <ol start=43> <li>Berapakah nilai dari <math>x+\frac{16}{x}</math> jika <math>x-3\sqrt{x}=4</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x-3\sqrt{x} &= 4 \\ x-4 &= 3\sqrt{x} \\ x^2-8x+16 &= 9x \\ x^2-17x+16 &= 0 \\ x-17+\frac{16}{x} &= 0 \\ x+\frac{16}{x} &= 17 \\ \end{align} </math> </div></div> <ol start=44> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+4}</math> jika <math>x^2-7x+2=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2-7x+2 &= 0 \\ x^2+2 &= 7x \\ x+\frac{2}{x} &= 7 \\ x^2+4+\frac{4}{x^2} &= 49 \\ x^2+\frac{4}{x^2} &= 45 \\ \frac{x^4+4}{x^2} &= 45 \\ \frac{x^2}{x^4+4} &= \frac{1}{45} \\ \end{align} </math> </div></div> <ol start=45> <li>Berapakah nilai dari <math>x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}</math> jika <math>x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}} &= 25 \\ x^{\frac{1}{2}}+x^{-\frac{1}{2}} &= 23 \\ x+2+x^{-1} &= 529 \\ x+x^{-1} &= 527 \\ x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+x^{-\frac{3}{4}} &= 110 \\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1} &= x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}} \\ &= 527+110 \\ &= 637 \\ \end{align} </math> </div></div> <ol start=46> <li>Berapakah nilai dari <math>\sqrt{8x^6+x^5+x^4+5x^3+1}</math> jika <math>\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5} &= 0 \\ \frac{x^2+x+1}{x^5} &= 0 \\ x^2+x+1 &= 0 \\ x^2+x+1 &= 0 \\ (x-1)(x^2+x+1) &= 0(x-1) \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ \sqrt{8x^6+x^5+x^4+5x^3+1} &= \sqrt{(2x^3)^2+x^3x^2+x^3x+5x^3+1} \\ &= \sqrt{(2(1))^2+(1)x^2+(1)x+5(1)+1} \\ &= \sqrt{(2)^2+x^2+x+5+1} \\ &= \sqrt{4+x^2+x+1+5} \\ &= \sqrt{4+0+5} \\ &= \sqrt{9} \\ &= 3 \\ \end{align} </math> </div></div> <ol start=47> <li>Berapakah nilai dari <math>f(1)+f(2)+f(3)+ \dots + f(99)</math> jika <math>f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{1}{\sqrt{x+1}+\sqrt{x}} \\ &= \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} \\ &= \sqrt{x+1}-\sqrt{x} \\ f(1)+f(2)+f(3)+ \dots + f(98)+f(99) &= \sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+ \cdot + \sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99} \\ &= \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+ \cdot + \sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ &= \sqrt{100}-\sqrt{1} \\ &= 10-1 \\ &= 9 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots + \frac{2024}{2025})</math> jika <math>h(x)=\frac{3}{3+9^x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} h(x) &= \frac{3}{3+9^x} \\ h(1-x) &= \frac{3}{3+9^{1-x}} \\ &= \frac{3}{3+\frac{9}{9^x}} \\ &= \frac{9^x}{3+9^x} \\ h(x)+h(1-x) &= \frac{3}{3+9^x}+\frac{9^x}{3+9^x} \\ &= \frac{3+9^x}{3+9^x} \\ &= 1 \\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots +(1-\frac{2}{2025})+(1-\frac{1}{2025})) \\ & 5(1+1+1+ \dots +1+1) \text{ sebanyak 1012 kali } \\ & 5(1012) \\ & 5060 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>\frac{7^{2025} - 7^{2023} + 432}{7^{2024} + 7^{2023} + 72}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72} &= \frac{7^{2023}7^{2}-7^{2023} + 48 \times 9}{7^{2023}7^1+7^{2023}+8 \times 9} \\ &= \frac{7^{2023}(7^{2}-1)+48 \times 9}{7^{2023}(7^1+1)+8 \times 9} \\ &= \frac{7^{2023}(49-1)+48 \times 9}{7^{2023}(7+1) + 8 \times 9} \\ &= \frac{7^{2023} \times 48+48 \times 9}{7^{2023} \times 8+8 \times 9} \\ &= \frac{48(7^{2023}+9)}{8(7^{2023}+9)} \\ &= \frac{48}{8} \\ &= 6 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>tan (x+\frac{\pi}{4})</math> jika <math>\frac{1}{cos x}-tan x = \frac{4}{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{cos x}-tan x &= \frac{4}{5} \\ sec x-tan x &= \frac{4}{5} \\ sec^2 x-tan^2 x &= 1 \\ (sec x+tan x)(sec x-tan x) &= 1 \\ (sec x+tan x)\frac{4}{5} &= 1 \\ sec x+tan x &= \frac{5}{4} \\ \text{kedua persamaan dengan cara metode eliminasi } \\ 2 tan x &= \frac{5}{4}-\frac{4}{5} \\ 2 tan x &= \frac{9}{20} \\ tan x &= \frac{9}{40} \\ tan (x+\frac{\pi}{4}) &= \frac{tan x+tan \frac{\pi}{4}}{1-tan x \cdot tan \frac{\pi}{4}} \\ &= \frac{\frac{9}{40}+1}{1-\frac{9}{40} \cdot 1} \\ &= \frac{\frac{49}{40}}{\frac{31}{40}} \\ &= \frac{49}{31} \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>sin^3 x+csc^3 x</math> jika <math>sin x-csc x = 8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ Dengan menggunakan rumus: } (a-b)^3 &= a^3-b^3-3ab(a-b) \\ (sin x-csc x)^3 &= sin^3 x-csc^3 x-3sin x csc x(sin x-csc x) \\ 8^3 &= sin^3 x-csc^3 x-3sin x (\frac{1}{sin x})(8) \\ 512 &= sin^3 x-csc^3 x-24 \\ sin^3 x-csc^3 x &= 512+24 \\ sin^3 x-csc^3 x &= 536 \\ \end{align} </math> </div></div> <ol start=52> <li>Berapakah nilai dari <math>(sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2</math> jika <math>\frac{1}{sin x}+\frac{1}{cos x} = 10</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{sin x}+\frac{1}{cos x} &= 10 \\ \frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} &= 100 \\ (sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2 &= sin^2 x+\frac{2sin x}{cos x}+\frac{1}{cos^2 x}+cos^2 x+\frac{2cos x}{sin x}+\frac{1}{sin^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2(sin^2 x+cos^2 x)}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+100 \\ &= 101 \\ \end{align} </math> </div></div> <ol start=53> <li>Berapakah nilai dari (x-1)⁶ jika <math>x=\frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin 80^\circ &= cos 10^\circ \\ sin 80^\circ-cos 10^\circ &= 0 \\ x &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+2 sin 20^\circ cos 20^\circ}{cos 10^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+4 sin 10^\circ cos 10^\circ cos 20^\circ}{cos 10^\circ} \\ &= 4 cos 55^\circ cos 25^\circ+4 sin 10^\circ cos 20^\circ \\ &= 2(2 cos 55^\circ cos 25^\circ+2 sin 10^\circ cos 20^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ+sin (-10)^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ-sin 10^\circ) \\ &= 2(cos 80^\circ-sin 10^\circ+cos 30^\circ+sin 30^\circ) \\ &= 2(cos 80^\circ-sin (90^\circ-80^\circ)+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(cos 80^\circ-cos 80^\circ+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= \sqrt{3}+1 \\ x-1 &= \sqrt{3} \\ (x-1)^6 &= (\sqrt{3})^6 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=54> <li>Berapakah nilai dari x jika <math>x=\frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ} \\ 2x sin 20^\circ-x sin 40 ^\circ &= x sin 20^\circ-x^2 sin 10^\circ \\ x^2 sin 10^\circ+x sin 20^\circ-x sin 40 ^\circ &= 0 \\ x(x sin 10^\circ+sin 20^\circ-sin 40 ^\circ) &= 0 \\ x = 0 &\text{ atau } x sin 10^\circ+sin 20^\circ-sin 40 ^\circ = 0 \\ x sin 10^\circ+sin 20^\circ-sin 40 ^\circ &= 0 \\ x sin 10^\circ &= sin 40 ^\circ-sin 20^\circ \\ x &= \frac{sin 40 ^\circ-sin 20^\circ}{sin 10^\circ} \\ &= \frac{2 cos 30 ^\circ sin 10^\circ}{sin 10^\circ} \\ &= 2 cos 30 ^\circ \\ &= \frac{2 \sqrt{3}}{2} \\ &= \sqrt{3} \\ \end{align} </math> </div></div> <ol start=55> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x^2}{x^2-16y^2} = \frac{625}{49}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2}{x^2-16y^2} &= \frac{625}{49} \\ \frac{x^2-16y^2}{x^2} &= \frac{49}{625} \text{ (terbalik posisinya)} \\ 1-\frac{16y^2}{x^2} &= \frac{49}{625} \\ \frac{16y^2}{x^2} &= 1 - \frac{49}{625} \\ (\frac{4y}{x})^2 &= \frac{576}{625} \\ (\frac{4y}{x})^2 &= (\frac{24}{25})^2 \\ \frac{4y}{x} &= \frac{24}{25} \\ \frac{y}{x} &= \frac{6}{25} \\ \frac{x}{y} &= \frac{25}{6} \\ \end{align} </math> </div></div> <ol start=56> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x}{y}+\frac{x+10y}{y+10x} = 2</math> serta bilangan real untuk x dan y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x}{y}+\frac{x+10y}{y+10x} &= 2 \\ \frac{x}{y}+\frac{\frac{x}{y}+10}{1+10\frac{x}{y}} &= 2 \\ \text{misalkan } \frac{x}{y} = a \\ a+\frac{a+10}{1+10a} &= 2 \\ a(1+10a)+a+10 &= 2(1+10a) \\ 10a^2+a+a+10 &= 2+20a \\ 10a^2-18a+8 &= 0 \\ 5a^2-9a+4 &= 0 \\ (5a-4)(a-1) &= 0 \\ a = \frac{4}{5} &\text{ atau } a = 1 \\ \text{jadi } \frac{x}{y} = {\frac{4}{5}, 1} \\ \end{align} </math> </div></div> <ol start=57> <li>Berapakah nilai dari xy jika <math>x^4+y^4+x^2y^2=15 \text{ dan } x^2+y^2+xy=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+xy &= 5 \\ x^2+y^2 &= 5-xy \\ x^4+y^4+x^2y^2 &= 15 \\ (x^2)^2+(y^2)^2+2x^2y^2-x^2y^2 &= 15 \\ (x^2+y^2)^2-x^2y^2 &= 15 \\ (5-xy)^2-x^2y^2 &= 15 \\ 25-10xy+x^2y^2-x^2y^2 &= 15 \\ 25-10xy &= 15 \\ 10xy &= 10 \\ xy &= 1 \\ \end{align} </math> </div></div> <ol start=58> <li>Berapakah nilai dari x jika <math>4^x = 63(4^3+1)(4^6+1)(4^{12}+1)+1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 4^x &= 63(4^3+1)(4^6+1)(4^{12}+1)+1 \\ 4^x-1 &= 63(4^3+1)(4^6+1)(4^{12}+1) \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{4^3-1} \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{63} \\ &= (4^3+1)(4^6+1)(4^{12}+1)(4^3-1) \\ &= (4^3-1)(4^3+1)(4^6+1)(4^{12}+1) \\ &= (4^6-1)(4^6+1)(4^{12}+1) \\ &= (4^{12}-1)(4^{12}+1) \\ &= 4^{24}-1 \\ 4^x &= 4^{24} \\ x &= 24 \\ \end{align} </math> </div></div> <ol start=59> <li>Berapakah nilai dari <math>\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}</math> jika <math>x=\sqrt{9+4\sqrt{5}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \sqrt{9+4\sqrt{5}} \\ x &= 2+\sqrt{5} \\ x^2 &= 9+4\sqrt{5} \\ x^2-4x &= 9+4\sqrt{5}-4(2+\sqrt{5}) \\ x^2-4x &= 1 \\ x^2 &= 4x+1 \\ x^3 &= x \cdot x^2 \\ &= x(4x+1) \\ &= 4x^2+x \\ &= 4(4x+1)+x \\ &= 16x+4+x \\ &= 17x+4 \\ x^4 &= x \cdot x^3 \\ &= x(17x+4) \\ &= 17x^2+4x \\ &= 17(4x+1)+4x \\ &= 68x+17+4x \\ &= 72x+17 \\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1} &= \frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1} \\ &= \frac{72x+17-85x-20+8x+2+5x+3}{2} \\ &= \frac{2}{2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=60> <li>Berapakah nilai dari <math>\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}</math> jika 2x-1=<math>\sqrt{61}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \frac{x^3+1}{x^5-x^4-x^3+x^2} = p \\ p &= \frac{x^3+1}{x^5-x^4-x^3+x^2} \\ &= \frac{x^3+1}{x^5-x^4-(x^3-x^2)} \\ &= \frac{x^3+1}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)} \\ &= \frac{x^2-x+1}{x^2(x-1)^2} \\ &= \frac{x^2-x+1}{(x(x-1))^2} \\ &= \frac{x(x-1)+1}{(x(x-1))^2} \\ 2x-1 &= \sqrt{61} \\ x &= \frac{\sqrt{61}+1}{2} \\ x-1 &= \frac{\sqrt{61}-1}{2} \\ x(x-1) &= (\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2}) \\ &= \frac{61-1}{4} \\ &= \frac{60}{4} \\ &= 15 \\ p &= \frac{x(x-1)+1}{(x(x-1))^2} \\ &= \frac{15+1}{15^2} \\ &= \frac{16}{15^2} \\ \sqrt{p} &= \sqrt{\frac{16}{15^2}} \\ &= \frac{4}{15} \\ \end{align} </math> </div></div> <ol start=61> <li>Berapakah nilai dari <math>(\frac{x-3}{x})^{25}</math> jika <math>x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\sqrt[5]{8}+\sqrt[5]{2} &= 1+\sqrt[5]{16}+\sqrt[5]{4} \\ x+(\sqrt[5]{2})^3+\sqrt[5]{2} &= 1+(\sqrt[5]{2})^4+(\sqrt[5]{2})^2 \\ x &= (\sqrt[5]{2})^4-(\sqrt[5]{2})^3+(\sqrt[5]{2})^2-\sqrt[5]{2}+1 \\ \text{misalkan } \sqrt[5]{2} = p \\ x &= p^4-p^3+p^2-p+1 \\ x &= \frac{p^5+1}{p+1} \\ (\frac{x-3}{x})^{25} &= (1-\frac{3}{x})^{25} \\ &= (1-\frac{3}{\frac{p^5+1}{p+1}})^{25} \\ &= (1-\frac{3(p+1)}{p^5+1})^{25} \\ &= (1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2})^5+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{2+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-3\sqrt[5]{2}-3)}{3})^{25} \\ &= (-\sqrt[5]{2})^{25} \\ &= (-2)^5 \\ &= -32 \\ \end{align} </math> </div></div> <ol start=62> <li>Berapakah nilai dari <math>x^{50}+x^{49}+x^{48}+x^{47}+x^{46}</math> jika <math>x^2+x+1=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+x+1 &= 0 \\ x^2+x &= -1 \\ \frac{x^3-1}{x-1} &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46} &= x^{48}(x^2+x+1)+x^{45}(x^2+x) \\ &= x^{48}(0)+(x^3)^{15}(-1) \\ &= 0+(1)^{15}(-1) \\ &= -1 \\ \end{align} </math> </div></div> <ol start=63> <li>Berapakah 2²⁴ dari <math>8^7+8^6+8^5+8^4+8^3+8^2+8+1=A</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= A \\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1) &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8 &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= 8A+1 \\ 8^8+A &= 8A+1 \\ 8^8 &= 7A+1 \\ (2^3)^8 &= 7A+1 \\ 2^{24} &= 7A+1 \\ \end{align} </math> </div></div> <ol start=64> <li>Berapakah nilai dari <math>x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1</math> jika <math>x+\frac{1}{x}=\sqrt{3}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{x} &= \sqrt{3} \\ x^2+2+\frac{1}{x^2} &= 3 \\ x^2-1+\frac{1}{x^2} &= 0 \\ x^2(x^2-1+\frac{1}{x^2}) &= x^2(0) \\ x^4-x^2+1 &= 0 \\ (x^2+1)(x^4-x^2+1) &= (x^2+1)0 \\ x^6-x^4+x^2+x^4-x^2+1 &= 0 \\ x^6+1 &= 0 \\ x^6 &= -1 \\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1 &= {x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1 \\ &= (-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2-1+1 \\ &= -1+1-1+1-1+1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=65> <li>Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari <math>\frac{c-b}{a-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= ax^2+bx+c \\ f(2) &= a(2)^2+2b+c = 4 \\ &= 4a+2b+c = 4 \\ f(7) &= a(7)^2+7b+c = 49 \\ &= 49a+7b+c = 49 \\ 49a+7b+c &= 49 \\ 4a+2b+c &= 4 \\ 45a+5b &= 45 \text{ (f(7) dikurangi f(2)) } \\ 9a+b &= 9 \\ b &= -9a+9 \\ 4a+2b+c &= 4 \\ 4a+2(-9a+9)+c &= 4 \\ 4a-18a+18+c &= 4 \\ -14a+18+c &= 4 \\ c &= 14a-14 \\ \frac{c-b}{a-1} &= \frac{14a-14-(-9a+9)}{a-1} \\ &= \frac{14(a-1)+9(a-1)}{a-1} \\ &= \frac{(14+9)(a-1)}{a-1} \\ &= 23 \\ \end{align} </math> </div></div> <ol start=66> <li>Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+y)^3 &= x^3+y^3+3xy(x+y) \\ 11^3 &= 242+3xy(11) \text{ (dibagi 11)} \\ 11^2 &= 22+3xy \\ 121 &= 22+3xy \\ 99 &= 3xy \\ xy &= 33 \\ (x-y)^2 &= x^2+y^2-2xy \\ &= ((x+y)^2-2xy)-2xy \\ &= (x+y)^2-4xy \\ &= 11^2-4(33) \\ &= 121-132 \\ &= -11 \\ \end{align} </math> </div></div> <ol start=67> <li>Berapa f(1)+f(-1) jika <math>f(\frac{ax-b}{bx-a})</math>=x²-5x+6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ jika} f(1) = f(\frac{ax-b}{bx-a}) \\ 1 &= \frac{ax-b}{bx-a} \\ bx-a &= ax-b \\ (b-a)x &= -b+a \\ &= -(b-a) \\ &= -1 \\ f(1) &= x^2-5x+6 \\ &= (-1)^2-5(-1)+6 \\ &= 12 \\ \text{ jika} f(-1) = f(\frac{ax-b}{bx-a}) \\ -1 &= \frac{ax-b}{bx-a} \\ -(bx-a) &= ax-b \\ -bx+a &= ax-b \\ (-b-a)x &= -b-a \\ &= 1 \\ f(-1) &= x^2-5x+6 \\ &= (1)^2-5(1)+6 \\ &= 2 \\ f(1)+f(-1) &= 12+2 \\ &= 14 \\ \end{align} </math> </div></div> <ol start=68> <li>berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)-x &= f(x-1) \\ f(x)-f(x-1) &= x \\ x=1 ; f(1)-f(0) &= 1 \\ x=2 ; f(2)-f(1) &= 2 \\ x=3 ; f(3)-f(2) &= 3 \\ x=4 ; f(4)-f(3) &= 4 \\ \dots \\ x=200 ; f(200)-f(199) &= 200 \\ \text{ jumlahkan tersebut menjadi } \\ f(200)-f(0) &= 1+2+3+4+\dots+200 \\ &= \frac{200 \cdot 201}{2} \\ &= 20.100 \\ f(200)-1 &= 20.100 \\ &= 20.101 \\ \end{align} </math> </div></div> <ol start=69> <li>Misalkan f(x) adalah fungsi rekursif yang berlaku ∀x ∈ R sebagai berikut: : f(x)+f(15-x) = 2024 : f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)+f(15-x) &= 2024 \\ f(15+x) &= f(x)+2020 \\ *cara 1 \\ \text{ganti x dengan 15+x } \\ f(15+x)+f(-x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ *cara 2 \\ \text{ganti x dengan -x } \\ f(-x)+f(15+x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ \end{align} </math> </div></div> <ol start=70> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>2f(\frac{2002}{x}) + f(x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2f(\frac{2002}{x}) + f(x) &= 3x \\ \text{ganti x dengan 2 } \\ 2f(\frac{2002}{2}) + f(2) &= 3(2) \\ 2f(1001) + f(2) &= 6 \\ \text{ganti x dengan 1001 } \\ 2f(\frac{2002}{1001}) + f(1001) &= 3(1001) \\ 2f(2) + f(1001) &= 3003 \\ 2f(2) + f(1001) &= 3003 \\ f(1001) &= 3003 - 2f(2) \\ 2f(1001) + f(2) &= 6 \\ 2(3003 - 2f(2)) + f(2) &= 6 \\ 6006 - 4f(2) + f(2) &= 6 \\ 3f(2) &= 6000 \\ f(2) &= 2000 \\ \end{align} </math> </div></div> <ol start=71> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>f(\frac{1}{x}) + \frac{1}{x}f(-x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{1}{x})+\frac{1}{x}f(-x) &= 3x \\ \text{ganti x dengan 1/3 } \\ f(3)+3f(-\frac{1}{3}) &= 1 \\ \text{ganti x dengan -3 } \\ f(-\frac{1}{3}) - \frac{1}{3}f(3) &= -9 \\ \text{dikalikan 3 } \\ 3f(-\frac{1}{3})-f(3) &= -27 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ 2f(3) &= 28 \\ f(3) &= 14 \\ \end{align} </math> </div></div> <ol start=72> <li>Diketahui polinom <math>f(7^b-1)=7^{3b}-10</math>. tentukan nilai f(5)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(5) &= f(7^b-1) \\ 5 &= 7^b-1 \\ 7^b &= 6 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(6-1) &= 6^3-10 \\ f(5) &= 216-10 \\ &= 206 \\ *cara 2 \\ \text{misalkan } 7^b-1=a \text{ maka } 7^b=a+1 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(a) &= (a+1)^3-10 \\ f(5) &= (5+1)^3-10 \\ &= 6^3-10 \\ &= 216-10 \\ &= 206 \\ \end{align} </math> </div></div> <ol start=73> <li>Diketahui polinom <math>f(6^b-7)=6^{3b}-2 \cdot 6^{2b}-4</math>. tentukan nilai f(-2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(-2) &= f(6^b-7) \\ -2 &= 6^b-7 \\ 6^b &= 5 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(5-7) &= 5^3-2 \cdot 5^2-4 \\ f(-2) &= 125-50-4 \\ &= 71 \\ *cara 2 \\ \text{misalkan } 6^b-7=a \text{ maka } 6^b=a+7 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(a) &= (a+7)^3-2(a+7)^2-4 \\ f(-2) &= (-2+7)^3-2(-2+7)^2-4 \\ &= 5^3-2(5)^2-4 \\ &= 125-50-4 \\ &= 71 \\ \end{align} </math> </div></div> <ol start=74> <li>Jika <math>f(xy)=\frac{f(x)}{y}</math> dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(10) &= 7 \\ f(2 \cdot 5) &= 7 \\ f(xy) &= \frac{f(x)}{y} \\ f(2 \cdot 5) &= \frac{f(2)}{5} \\ 7 &= \frac{f(2)}{5} \\ f(2) &= 35 \\ \end{align} </math> </div></div> <ol start=75> <li>Jika <math>f(xy)=\frac{f(x+y)}{xy}</math> dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(15) &= 16 \\ f(3 \cdot 5) &= 16 \\ f(xy) &= \frac{f(x+y)}{xy} \\ f(3 \cdot 5) &= \frac{f(3+5)}{3 \cdot 5} \\ f(15) &= \frac{f(8)}{15} \\ 16 &= \frac{f(8)}{15} \\ f(8) &= 240 \\ \end{align} </math> </div></div> <ol start=76> <li>Jika <math>f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15</math> maka tentukan nilai f(16)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x+\frac{1}{x}+6) &= x^2+\frac{1}{x^2}+15 \\ &= (x+\frac{1}{x})^2-2+15 \\ &= (x+\frac{1}{x})^2+13 \\ \text{misalkan } x+\frac{1}{x} &= p \\ f(x+\frac{1}{x}+6) &= (x+\frac{1}{x})^2+13 \\ f(p+6) &= p^2+13 \\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 } \\ f(p+6) &= p^2+13 \\ f(10+6) &= 10^2+13 \\ f(16) &= 100+13 \\ &= 113 \\ \end{align} </math> </div></div> <ol start=77> <li>Tentukan nilai x jika <math>f(x)=\frac{4}{4-x}</math> dan <math>f(x \cdot f(x))^{\frac{f(4x)}{f(x)}}=256</math>!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{4}{4-x} \\ f(4x) &= \frac{4}{4-4x} \\ \frac{f(4x)}{f(x)} &= \frac{\frac{4}{4-4x}}{\frac{4}{4-x}} \\ &= \frac{4-x}{4-4x} \\ f(x \cdot f(x)) &= f(x(\frac{4}{4-x})) \\ &= f(\frac{4x}{4-x}) \\ &= \frac{4}{4-(\frac{4x}{4-x})} \\ &= \frac{4}{\frac{16-4x-4x}{4-x}} \\ &= \frac{4}{\frac{16-8x}{4-x}} \\ &= \frac{4(4-x)}{4(4-4x)} \\ &= \frac{4-x}{4-4x} \\ \text{misalkan } \frac{4-x}{4-4x} &= a \\ f(x \cdot f(x))^{\frac{f(4x)}{f(x)}} &= 256 \\ a^a &= 256 \\ a^a &= 4^4 \\ a &= 4 \\ \frac{4-x}{4-4x} &= 4 \\ 4-x &= 16-16x \\ 15x &= 12 \\ x &= \frac{4}{5} \\ \end{align} </math> </div></div> <ol start=78> <li>Fungsi <math>f(x) = \frac{kx}{2x+1} \text{dengan } x \neq -\frac{1}{2}</math>. Dengan f(f(x)) = x maka tentukan nilai k!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{kx}{2x+1} \\ f(f(x)) &= x \\ f(\frac{kx}{2x+1}) &= x \\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1} &= x \\ \frac{\frac{k^2x}{2x+1}}{\frac{2kx+2x+1}{2x+1}} &= x \\ \frac{k^2x}{2kx+2x+1} &= x \\ \frac{k^2}{2kx+2x+1} &= 1 \\ k^2 &= 2kx+2x+1 \\ k^2-2kx &= 2x+1 \\ k^2-2kx+x^2 &= x^2+2x+1 \\ (k-x)^2 &= (x+1)^2 \\ (k-x)^2-(x+1)^2 &= 0 \\ (k-x+x+1)(k-x-(x+1)) &= 0 \\ k=-1 &\text{ atau } k=2x+1 &\text{ (TM) } \\ \end{align} </math> </div></div> <ol start=79> <li>Jika n = 2023²+2024² maka berapa hasil dari <math>\sqrt{2n-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n &= 2023^2+2024^2 \\ &= 2023^2+(2023+1)^2 \\ \text{misalkan 2023 = p} \\ n &= p^2+(p+1)^2 \\ &= p^2+p^2+2p+1 \\ &= 2p^2+2p+1 \\ \sqrt{2n-1} &= \sqrt{2(2p^2+2p+1)-1} \\ &= \sqrt{4p^2+4p+2-1} \\ &= \sqrt{4p^2+4p+1} \\ &= \sqrt{(2p+1)^2} \\ &= 2p+1 \\ &= 2(2023)+1 \\ &= 4046+1 \\ &= 4047 \\ \end{align} </math> </div></div> <ol start=80> <li>Tentukan nilai dari a+b+c merupakan bilangan bulat positif jika ab = 2, bc = 3 dan ac = 6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab \cdot bc \cdot ac &= 2 \cdot 3 \cdot 6 \\ (abc)^2 &= 36 \\ abc &= \pm 6 \\ abc &= 6 \\ \frac{abc}{ab} &= c = \frac{6}{2} = 3 \\ \frac{abc}{bc} &= a = \frac{6}{3} = 2 \\ \frac{abc}{ac} &= b = \frac{6}{6} = 1 \\ a+b+c &= 6 \\ \end{align} </math> </div></div> <ol start=81> <li>Tentukan nilai dari (a-c)^b jika <math>\frac{ab}{a+b} = \frac{1}{3}</math>, <math>\frac{bc}{b+c} = \frac{1}{4}</math> dan <math>\frac{ac}{a+c} = \frac{1}{9}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{ab}{a+b} &= \frac{1}{3} \\ \frac{a+b}{ab} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{b} + \frac{1}{a} &= 3 \\ \frac{bc}{b+c} &= \frac{1}{4} \\ \frac{b+c}{bc} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{b} &= 4 \\ \frac{ac}{a+c} &= \frac{1}{9} \\ \frac{a+c}{ac} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{a} &= 9 \\ \text{Misalkan 1/a = x, 1/b = y dan 1/c = z} \\ x+y &= 3 \\ y+z &= 4 \\ x+z &= 9 \\ x+y &= 3 \\ y+z &= 4 \\ x-z &= -1 \\ x-z &= -1 \\ x+z &= 9 \\ 2x &= 8 \\ x &= 4 \\ x-z &= -1 \\ 4-z &= -1 \\ z &= 5 \\ x+y &= 3 \\ 4+y &= 3 \\ y &= -1 \\ \frac{1}{a} &= 4 \\ a &= \frac{1}{4} \\ \frac{1}{b} &= -1 \\ b &= -1 \\ \frac{1}{c} &= 5 \\ c &= \frac{1}{5} \\ (a-c)^b &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=82> <li>Tentukan nilai dari a, b dan c jika <math>\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}</math> dan a+2b+3c=28?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan k untuk semua ketiga persamaan tersebut } \\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5} &= k \\ a+b &= 2k \\ a+c &= 4k \\ b+c &= 5k \\ 2a+b+c &= 6k \\ 2a+5k &= 6k \\ k &= 2a \\ a &= \frac{k}{2} \\ b &= \frac{3k}{2} \\ c &= \frac{7k}{2} \\ a+2b+3c &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ a &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ b &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ c &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=83> <li>Tentukan nilai dari (b+c)^a jika <math>\frac{a+b+c}{2} = \sqrt{a-2}+\sqrt{b-1}+\sqrt{c}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{a+b+c}{2} &= \sqrt{a-2}+\sqrt{b-1}+\sqrt{c} \\ a+b+c &= 2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}) \\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c} &= 0 \\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1 &= 0 \\ (\sqrt{a-2}-1)^2+(\sqrt{b-1}-1)^2+(\sqrt{c}-1)^2 &= 0 \\ (\sqrt{a-2}-1)^2 &= 0 \\ \sqrt{a-2}-1 &= 0 \\ \sqrt{a-2} &= 1 \\ a-2 &= 1 \\ a &= 3 \\ (\sqrt{b-1}-1)^2 &= 0 \\ \sqrt{b-1}-1 &= 0 \\ \sqrt{b-1} &= 1 \\ b-1 &= 1 \\ b &= 1 \\ (\sqrt{c}-1)^2 &= 0 \\ \sqrt{c}-1 &= 0 \\ \sqrt{c} &= 1 \\ c &= 1 \\ (b+c)^a &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=84> <li>x dan y merupakan bilangan tak nol. Jika xy = <math>\frac{x}{y}</math> = x-y maka berapa nilai x+y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy &= \frac{x}{y} \\ y^2 &= 1 \\ y^2 - 1 &= 0 \\ (y-1)(y+1) &= 0 \\ y = 1 &\text{ atau } y = -1 \\ \frac{x}{y} &= x-y \\ x &= xy-y^2 \\ x-xy &= -y^2 \\ x(1-y) &= -y^2 \\ x &= \frac{-y^2}{1-y} \\ \text{cek y=1 } \\ x &= \frac{-1^2}{1-1} \\ \text{tidak memenuhi syarat } \\ \text{cek y=-1 } \\ x &= \frac{-(-1)^2}{1-(-1)} \\ &= \frac{-1}{2} \\ x+y &= -1-\frac{1}{2} \\ &= -\frac{3}{2} \\ \end{align} </math> </div></div> <ol start=86> <li>Berapa nilai x dari <math>(\frac{a}{b})^3+(\frac{b}{a})^3 = 2\sqrt{x}</math> jika <math>\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b} \\ \frac{a+b}{ab} &= \frac{1}{a+b} \\ (a+b)^2 &= ab \\ a^2+2ab+b^2 &= ab \\ a^2+b^2 &= -ab \\ \text{misalkan } \frac{a}{b}+\frac{b}{a} = n \\ \frac{a}{b}+\frac{b}{a} &= n \\ \frac{a^2+b^2}{ab} &= n \\ a^2+b^2 &= nab \\ n &= -1 \\ \frac{a}{b}+\frac{b}{a} &= n \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3(\frac{a}{b}+\frac{b}{a}) &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3n &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3 &= n^3-3n \\ &= (-1)^3-3(-1) \\ &= 2 \\ 2\sqrt{x} &= 2 \\ \sqrt{x} &= 1 \\ x &= 1 \\ \end{align} </math> </div></div> <ol start=87> <li>Berapa nilai m dari <math>x^2-mx-1=0</math> jika <math>\sqrt[3]{x_1}+\sqrt[3]{x_2}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x_1} &= a \\ x_1 &= a^3 \\ \sqrt[3]{x_2} &= b \\ x_2 &= b^3 \\ \sqrt[3]{x_1}+\sqrt[3]{x_2} &= 1 \\ a+b &= 1 \\ x^2-mx-1 &= 0 \\ x_1+x_2 &= m \\ x_1 \cdot x_2 &= -1 \\ x_1+x_2 &= m \\ a^3+b^3 &= m \\ x_1 \cdot x_2 &= -1 \\ a^3 \cdot b^3 &= -1 \\ (ab)^2 &= (-1)^3 \\ ab &= -1 \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ (1)^3 &= m+3(-1)(1) \\ 1 &= m-3 \\ m &= 4 \\ \end{align} </math> </div></div> <ol start=88> <li>Berapa nilai <math>\frac{x_1}{x_2}</math> dari <math>ax^2-18x-b=0</math> jika <math>ab=45</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab &= 45 \\ b &= \frac{45}{a} \\ ax^2-18x-b &= 0 \\ ax^2-18x-\frac{45}{a} &= 0 \\ a^2x^2-18ax-45 &= 0 \\ (ax-3)(ax-15) &= 0 \\ ax-3 &= 0 \\ x &= \frac{3}{a} \\ ax-15 &= 0 \\ x &= \frac{15}{a} \\ \frac{x_1}{x_2} &= \frac{\frac{3}{a}}{\frac{15}{a}} \\ &= \frac{3}{15} \\ &= \frac{1}{5} \\ \frac{x_1}{x_2} &= \frac{\frac{15}{a}}{\frac{3}{a}} \\ &= \frac{15}{3} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=89> <li>Jika <math>\frac{u_3}{u_1+u_2} = \frac{7}{8}</math> merupakan barisan aritmetika maka berapa dari <math>\frac{u_2+u_3}{u_1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{u_3}{u_1+u_2} &= \frac{7}{8} \\ \frac{a+2b}{a+a+b} &= \frac{7}{8} \\ \frac{a+2b}{2a+b} &= \frac{7}{8} \\ 8(a+2b) &= 7(2a+b) \\ 8a+16b &= 14a+7b \\ 9b &= 6a \\ b &= \frac{2a}{3} \\ \frac{u_2+u_3}{u_1} &= \frac{a+b+a+2b}{a} \\ &= \frac{2a+3b}{a} \\ &= \frac{2a+3(\frac{2a}{3})}{a} \\ &= \frac{2a+2a}{a} \\ &= \frac{4a}{a} \\ &= 4 \\ \end{align} </math> </div></div> <ol start=90> <li>Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7p+q}{2p+q} &= \frac{17p+q}{7p+q} \\ (7p+q)^2 &= (17p+q)(2p+q) \\ 49p^2+14pq+q^2 &= 34p^2+19pq+q^2 \\ 15p^2 &= 5pq \\ 3p &= q \\ \frac{7p+q}{2p+q} &= \frac{7p+3p}{2p+3p} \\ &= \frac{10p}{5p} \\ &= 2 \\ \end{align} </math> </div></div> <ol start=91> <li>Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{rataan geometris } \\ \sqrt{a \cdot b} &= b-24 \\ a \cdot b &= (b-24)^2 \\ \text{rataan aritmatik } \\ \frac{a+b}{2} &= a+15 \\ a+b &= 2(a+15) \\ a+b &= 2a+30 \\ a &= b-30 \\ a \cdot b &= (b-24)^2 \\ (b-30)b &= (b-24)^2 \\ b^2-30b &= b^2-48b+576 \\ 18b &= 576 \\ b &= 32 \\ a &= b-30 \\ &= 32-30 \\ &= 2 \\ a+b &= 32+2 \\ &= 34 \\ \end{align} </math> </div></div> <ol start=91> <li>Segitiga lancip ABC dengan <math>\frac{a^4+b^4+c^4+a^2b^2}{c^2(a^2+b^2)}=2</math>. tentukan nilai sudut C?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{syarat segitiga lancip semua sudut masing-masing kurang dari } 90^\circ \\ c^2 &= a^2+b^2-2ab cos C \\ cos C &= \frac{a^2+b^2-c^2}{2ab} \\ a^4+b^4+c^4+a^2b^2 &= 2c^2(a^2+b^2) \\ a^4+b^4+a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-2c^2(a^2+b^2)+(c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= (ab)^2 \\ a^2+b^2-c^2 &= \pm ab \\ cos C &= \pm \frac{ab}{2ab} \\ &= \pm \frac{1}{2} \\ &= \frac{1}{2} \text{ (karena sudut harus kurang dari } 90^\circ) \\ C &= 60^\circ \\ \end{align} </math> </div></div> <ol start=92> <li>Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan panjang DA dan EA } = x \text{ dan panjang AB } = y \\ \text{luas segitiga CAB } &= \frac{CA \cdot AB}{2} \\ 45 &= \frac{(x+9)(x+5)}{2} \\ 90 &= x^2+14x+45 \\ x^2+14x &= 45 \\ y^2 &= (x+9)^2+(x+5)^2 \\ &= x^2+18x+81+x^2+10x+25 \\ &= 2x^2+28x+106 \\ &= 2(x^2+14x)+106 \\ &= 2(45)+106 \\ &= 196 \\ y &= 14 \\ \end{align} </math> jadi panjang BC adalah 14 cm </div></div> <ol start=93> <li>Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kita cari ukuran GC yaitu } \\ \frac{GC}{AD} &= \frac{CE}{DE} \\ \frac{GC}{15} &= \frac{6}{18} \\ GC &= 5 \\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC } \\ &= \frac{1}{2} \cdot BC \cdot CF - \frac{1}{2} \cdot GC \cdot CE \\ &= \frac{1}{2} \cdot 15 \cdot 18 - \frac{1}{2} \cdot 5 \cdot 6 \\ &= 135 - 15 \\ &= 120 \\ \end{align} </math> jadi luas daerah BFEG adalah 120 cm² </div></div> <ol start=94> <li>Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ AI &= \frac{1}{5} AF \\ &= \frac{1}{5} 10 \\ &= 2 \\ IF &= AF-AI \\ &= 10-2 \\ &= 8 \\ \text{luas trapesium AFGD } &= \frac{(AD+EF) \cdot AF}{2} \\ &= \frac{(4+6)10}{2} \\ &= 50 \\ \text{luas segitiga AID } &= \frac{AI \cdot AF}{2} \\ &= \frac{(2)4}{2} \\ &= 4 \\ \text{luas segitiga IFG } &= \frac{IF \cdot FG}{2} \\ &= \frac{(8)6}{2} \\ &= 24 \\ \text{luas daerah segitiga IGD } &= \text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG } \\ &= 50-4-24 \\ &= 22 \\ \end{align} </math> jadi luas daerah segitiga IGD adalah 22 cm² </div></div> <ol start=96> <li>Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-<math>\pi</math>) cm³ maka berapa luas permukaan kerucut tersebut?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ \text{volume balok} \\ V_b &= x^2(12) \\ \text{volume kerucut} \\ V_b &= \frac{1}{3}\pi x^2(12) \\ &= 4\pi x^2 \\ V_{b-k} &= Vb-Vk \\ 100(3-\pi) &= 12x^2-4\pi x^2 \\ 100(3-\pi) &= 4x^2(3-\pi) \\ x^2 &= 25 \\ x &= 5 \\ s &= \sqrt{12^2+5^2} \\ &= \sqrt{144+25} \\ &= \sqrt{169} \\ &= 13 \\ \text{luas permukaan kerucut } &= \pi r(r+s) \\ &= \pi(5)(5+13) \\ &= 90\pi \\ \end{align} </math> jadi luas daerah permukaan kerucut adalah 90<math>\pi</math> cm² </div></div> <ol start=97> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 3a+1 \\ B &= 3b+2 \\ A(A+1)+3B \\ (3a+1)(3a+1+1)+3(3b+2) \\ (3a+1)(3a+2)+9b+6 \\ 9a^2+9a+2+9b+6 \\ 9a^2+9a+9b+8 \\ 9(a^2+a+b)+8 \\ \text{sisa pembagiannya adalah } 8 \\ \end{align} </math> </div></div> <ol start=98> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 9a+7 \\ B &= 9b+8 \\ A(A-5)+9B \\ (9a+7)(9a+7-5)+9(9b+8) \\ (9a+7)(9a+2)+81b+72 \\ 81a^2+81a+14+81b+72 \\ 81a^2+81a+81b+86 \\ 81a^2+81a+81b+81+5 \\ 81(a^2+a+b+1)+5 \\ \text{sisa pembagiannya adalah } 5 \\ \end{align} </math> </div></div> <ol start=99> <li>Jika <math>\begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix}</math> maka berapa hasil dari A²¹+A²⁵+A⁴⁶?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A^2 &= A \cdot A \\ &= \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \\ A^3 &= A^2 \cdot A \\ &= \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ &= - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= -I \\ A^{21}+A^{25}+A^{46} &= A^{21} \cdot (I+A^4+A^{25}) \\ &= A^{21} \cdot (I+A^3 \cdot A +A^{24} \cdot A) \\ &= (A^3)^7 \cdot (I+A^3 \cdot A +(A^3)^8 \cdot A) \\ &= (-I)^7 \cdot (I-I \cdot A +(-I)^8 \cdot A) \\ &= -I \cdot (I-A+A) \\ &= -I \cdot I \\ &= -I \\ &= -\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ \end{align} </math> </div></div> <ol start=100> <li>Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida?</li></ol> : bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15 : ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9). : ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8) : maka Ida menulis bilangan 8. <ol start=101> <li>Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9?</li></ol> : Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu. : untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja. : untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja. : Jadi banyaknya bilangan mungkin 2. <ol start=102> <li>Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli!</li></ol> ;cara 1 # 8¹ = 1 # 8² = 1 (8²=8¹x8¹ sama dengan 1x1) # 8³ = 1 (8³=8¹x8² sama dengan 1x1) # 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2) # 8⁵ = 1 # 8ⁿ = 1 (semua n untuk bilangan asli) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli ;cara 2 # 8ⁿ = b mod 7 # 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil) # (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya) # 8ⁿ = 1ⁿ mod 7 # 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli <ol start=103> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 5?</li></ol> ;cara 1 # 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 # 7¹ = 7 (sisa 1) # 7² = 49 (sisa 2) # 7³ = 343 (sisa 3) # 7⁴ = 2,401 (sisa 0) # 7⁵ = 16,807 # 7⁶ = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3 ;cara 2 :17¹ = 2 :17² = 4 :17³ = 3 :17⁴ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3 :17⁹⁹ = (17⁴)²⁴ x 17³ Untuk 17⁴ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 3. Jadi 17⁹⁹ dibagi 7 bersisa 3 ;cara 3 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 17⁴ ::17⁴ = 1 mod 5 ::(17⁴)²⁴ = 1²⁴ mod 5 ::17⁹⁶ = 1²⁴ mod 5 ::17⁹⁶ = 1 mod 5 ::17⁹⁶ x 17³ = 1 x 17³ mod 5 ::17⁹⁹ = 17³ mod 5 ::17⁹⁹ = 17 x 17 x 17 mod 5 ::17⁹⁹ = 2 x 2 x 2 mod 5 ::17⁹⁹ = 8 mod 5 ::17⁹⁹ = 3 mod 5 Jadi 17⁹⁹ dibagi 5 bersisa 3 <ol start=104> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 7?</li></ol> ;cara 1 :17¹ = 3 :17² = 2 :17³ = 6 :17⁴ = 4 :17⁵ = 5 :17⁶ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3 :17⁹⁹ = (17⁶)¹⁶ x 17³ Untuk 17⁶ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 6. Jadi 17⁹⁹ dibagi 7 bersisa 6 ;cara 2 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 17⁶ ::17⁶ = 1 mod 7 ::(17⁶)¹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1 mod 7 ::17⁹⁶ x 17³ = 1 x 17³ mod 7 ::17⁹⁹ = 17³ mod 7 ::17⁹⁹ = 17 x 17 x 17 mod 7 ::17⁹⁹ = 3 x 3 x 3 mod 7 ::17⁹⁹ = 27 mod 7 ::17⁹⁹ = 6 mod 7 Jadi 17⁹⁹ dibagi 7 bersisa 6 <ol start=105> <li>Berapa hasil sisa dari 41²⁰²⁴ dibagi 33?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 41^{2024} &= 41^{2024} \text{ mod } 33 \\ &= (33 \times 3 + 2)^{2024} \text{ mod } 33 \\ &= 2^{2024} \text{ mod } 33 \\ &= 2^{2020} 2^4 \text{ mod } 33 \\ &= (2^5)^{404} 2^4 \text{ mod } 33 \\ &= (33 - 1)^{404} 2^4 \text{ mod } 33 \\ &= (-1)^{404} 2^4 \text{ mod } 33 \\ &= 2^4 \text{ mod } 33 \\ &= 16 \text{ mod } 33 \\ \text{Jadi hasil sisa adalah } 16 \\ \end{align} </math> </div></div> <ol start=106> <li>Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 99^{99} &= (3^2 \times 11)^{99} \\ &= 3^{198} \times 11^{99} \\ 243^n &= (3^5)^n \\ &= 3^{5n} \\ \text{agar bisa membagi, maka} \\ 5n &= 198 \\ n &= 39.6 \\ \text{jadi bilangan n terbesar adalah } 39 \\ \end{align} </math> </div></div> <ol start=107> <li>Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸?</lu></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 88^{88} &= (8 \times 11)^{88} \\ &= 8^{88} \times 11^{88} \\ &= 8^{87} \times 8 \times 11^{88} \\ &= (8^3)^{29} \times 8 \times 11^{88} \\ &= 512^{29} \times 8 \times 11^{88} \\ 512^n &= 512^{29} \\ \text{jadi bilangan n terbesar adalah } 29 \\ \end{align} </math> </div></div> <ol start=108> <li>Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6!</li></ol> ; cara 1 : KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. : N dibagi 3 sisa 1 : N dibagi 5 sisa 2 : N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a : KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70 : KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21 : KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97 ; cara 2 : Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35 : kemudian ditambahkan sisa masing-masing sesuai dengan KPK. : KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, <b>97</b> : KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, <b>97</b> : KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, <b>97</b> Jadi bilangan bulat positif adalah 97 :: NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 <ol start=109> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |- | 4 || 0 || Semua isi ember B dibuang |- | 1 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. <ol start=110> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |} nah ada ember A berisi 4 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |- | 1 || 0 || Semua isi ember B dibuang |- | 0 || 1 || Tuangkan 1 l dari ember A ke B sehingga ember A kosong |- | 3 || 1 || Isikan 3 l ke ember A |- | 0 || 4 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |} nah ada ember B berisi 4 liter. # Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kemungkinan 1} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 +3 \cdot ((\frac{4}{3})^x)^2 &= 7 \cdot (\frac{4}{3})^x \\ \text{misalkan } (\frac{4}{3})^x = y \\ 4 +3y^2 &= 7y \\ 3y^2-7y+4 &= 0 \\ (3y-4)(y-1) &= 0 \\ y_1 = \frac{4}{3} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{4}{3} \\ (\frac{4}{3})^x &= \frac{4}{3} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{4}{3})^x &= 1 \\ x &= 0 \\ \text{kemungkinan 2} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot ((\frac{3}{4})^x)^2 +3 &= 7 \cdot (\frac{3}{4})^x \\ \text{misalkan } (\frac{3}{4})^x = y \\ 4y^2 +3 &= 7y \\ 4y^2-7y+3 &= 0 \\ (4y-3)(y-1) &= 0 \\ y_1 = \frac{3}{4} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{3}{4} \\ (\frac{3}{4})^x &= \frac{3}{4} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{3}{4})^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-6}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{2x-4}{3}+2x-6 &= \frac{3}{2x-4}+\frac{1}{2x-6} \\ \frac{2x-4}{3}+2x-6 &= \frac{1}{\frac{2x-4}{3}}+\frac{1}{2x-6} \\ \text{misalkan } \frac{2x-4}{3} = y \text{ dan } 2x-6 = z \\ y+z &= \frac{1}{y}+\frac{1}{z} \\ y+z &= \frac{y+z}{xy} \\ y+z-\frac{y+z}{xy} &= 0 \\ (y+z)(1-\frac{1}{xy}) &= 0 \\ *y+z=0 \\ y+z &= 0 \\ y &= -z \\ \frac{2x-4}{3} &= -(2x-6) \\ \frac{2(x-2)}{3} &= -2(x-3) \\ \frac{x-2}{3} &= -x+3 \\ x-2 &= -3x+9 \\ 4x &= 11 \\ x &= \frac{11}{4} \\ *1-\frac{1}{xy}=0 \\ 1-\frac{1}{xy} &= 0 \\ xy-1 &= 0 \\ xy &= 0 \\ (\frac{2x-4}{3})(2x-6) &= 0 \\ \frac{4x^2-20x+24}{3} &= 0 \\ 4x^2-20x+24 &= 0 \\ x^2-5x+6 &= 0 \\ (x-2)(x-3) &= 0 \\ x = 2 &\text{ atau } x = 3 \\ \end{align} </math> </div></div> [[Kategori:Soal-Soal Matematika]] t1zsga2z2niqk5mqahti8ediyf41mie 117494 117493 2026-07-13T05:07:48Z Akuindo 8654 117494 wikitext text/x-wiki contoh soal <ol start=1> <li>Berapa hasil dari <math>\sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Misalkan 2020 = p} \\ \sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64} &= \sqrt{(2020-5) \cdot (2020-3) \cdot (2020+3) \cdot (2020+5) + 64} \\ &= \sqrt{(p-5) \cdot (p-3) \cdot (p+3) \cdot (p+5) + 64} \\ &= \sqrt{(p-5) \cdot (p+5) \cdot (p-3) \cdot (p+3) + 64} \\ &= \sqrt{(p^2-25) \cdot (p^2-9) + 64} \\ &= \sqrt{p^4-34p^2+ 225 + 64} \\ &= \sqrt{p^4-34p^2+ 289} \\ &= \sqrt{(p^2-17)^2} \\ &= p^2-17 \\ &= 2020^2-17 \\ &= (2000+20)^2-17 \\ &= 4.000.000+80.000+400-17 \\ &= 4.080.383 \\ \end{align} </math> </div></div> <ol start=2> <li>Berapa nilai x dari <math>\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} = \frac{9}{10}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} &= \frac{9}{10} \\ \text{misalkan untuk } \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} = p \\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} &= p \\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}} &= p^2 \\ x^2-x-p &= p^2 \\ x^2-2x+1+x-1 &= p^2+p \\ (x-1)^2+(x-1) &= p^2+p \\ x-1 &= p \\ \text{misalkan untuk } \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}} &= q^3 \\ x^2 q &= q^3 \\ x^2 &= q^2 \\ x &= q \\ \frac{x-1}{x} &= \frac{9}{10} \\ x &= 10 \\ \end{align} </math> </div></div> <ol start=3> <li>Berapa nilai x dari <math>(\frac{x}{x+10})^{x+10}=\frac{1}{1024}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{x+10}{x})^{-(x+10)} &= (1024)^{-1} \\ (\frac{x+10}{x})^{x+10} &= 1024 \\ (\frac{x+10}{x})^{x+10} &= 2^{10} \\ (\frac{x+10}{x})^{\frac{x+10}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (\frac{1}{2})^{-1} \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (1+(-\frac{1}{2}))^{(1+(-\frac{2}{1}))} \\ \frac{10}{x} &= -\frac{1}{2} \\ x &= -20 \\ \end{align} </math> </div></div> <ol start=4> <li>Berapa nilai x dari <math>x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{2}+\sqrt{x+\frac{1}{4}} &= (\sqrt{x+\frac{1}{4}})^2+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+(\frac{1}{2})^2 \\ &= (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 \\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}} &= 4 \\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2} &= 4 \\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 4 \\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 &= 4 \\ \sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 2 \\ \sqrt{x+\frac{1}{4}} &= \frac{3}{2} \\ x+\frac{1}{4} &= \frac{9}{4} \\ x &= 2 \\ \end{align} </math> </div></div> <ol start=5> <li>Berapa nilai x dari <math>\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^3}{\sqrt{8-x^2}}+x^2-8 &= 0 \\ \frac{x^3}{\sqrt{8-x^2}} &= 8-x^2 \\ x^3 &= (8-x^2)^{\frac{3}{2}} \\ x &= (8-x^2)^{\frac{1}{2}} \\ x^2 &= 8-x^2 \\ 2x^2-8 &= 0 \\ x^2-4 &= 0 \\ (x-2)(x+2) &= 0 \\ \text{membuktikan } \\ x=2 \text{ maka hasilnya 0 } \\ x=-2 \text{ maka hasilnya -8 } \\ \text{jadi } x=2 \\ \end{align} </math> </div></div> <ol start=6> <li>Berapa nilai x dari <math>\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} = 5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} &= 5 \\ \frac{x^{50}+x^{60}+x^{70}}{31}} &= 5^5 \\ x^{50}+x^{60}+x^{70} &= 5^5 \cdot 31 \\ x^{50}(1+x^{10}+x^{20}) &= 5^5 \cdot 31 \\ (x^{10}^5)(1+x^{10}+(x^{10}^2) &= 5^5 \cdot 31 \\ \text{ misalkan } x^{10} = a \\ a^5(1+a+a^2) &= 5^5 \cdot 31 \\ a &= 5 \\ x^{10} &= 5 \\ x &= ^5 log 10 \\ \end{align} </math> </div></div> <ol start=7> <li>Berapa nilai x dari <math>\sqrt{3x+5+\sqrt{4x+5}} = x</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{3x+5+\sqrt{4x+5}} &= x \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \text{misalkan } \sqrt{4x+5}=y \text{ dan } 4x+5=y^2 \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \sqrt{y^2+y-x} &= x \\ y^2+y &= x^2+x \\ y=x \\ 4x+5 &= y^2 \\ 4x+5 &= x^2 \\ x^2-4x-5 &= 0 \\ (x-5)(x+1) &= 0 \\ x=5 &\text{ atau } x=-1 \text{ (TM) } \\ \end{align} </math> </div></div> <ol start=8> <li>Berapa nilai x dari <math>\sqrt{1+\sqrt{1+x}} = \sqrt[3]{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{1+\sqrt{1+x}} &= \sqrt[3]{x} \\ \sqrt[3]{x} &= n \\ x &= n^3 \\ \sqrt{1+\sqrt{1+n^3}} &= n \\ 1+\sqrt{1+n^3} &= n^2 \\ \sqrt{1+n^3} &= n^2-1 \\ 1+n^3 &= n^4-2n^2+1 \\ n^4-n^3-2n^2 &= 0 \\ n^2(n^2-n-2) &= 0 \\ n^2(n-2)(n+1) &= 0 \\ n=0, n=2 \text{ atau } n=-1 \\ n &= 0 \\ x &= 0^3 \\ &= 0 \\ n &= 2 \\ x &= 2^3 \\ &= 8 \\ n &= -1 \\ x &= (-1)^3 \\ &= -1 \\ \text{yang paling mungkin untuk nilai x adalah } 8 \\ \end{align} </math> </div></div> <ol start=9> <li>Berapa nilai x dari <math>\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} &= \frac{\sqrt{1+x}}{\sqrt{x}} \\ \sqrt{x}(\sqrt{1+x}+\sqrt{x}) &= (\sqrt{1+x}-\sqrt{x})\sqrt{1+x} \\ \sqrt{x(1+x)}+x &= 1+x-\sqrt{x(1+x)} \\ 2\sqrt{x(1+x)} &= 1 \\ \sqrt{x(1+x)} &= \frac{1}{2} \\ x(1+x) &= \frac{1}{4} \\ x^2+x &= \frac{1}{4} \\ 4x^2+4x &= 1 \\ 4x^2+4x-1 &= 0 \\ x &= \frac{-4 \pm \sqrt{4^2-4(4)(-1)}}{2(4)} \\ &= \frac{-4 \pm \sqrt{32}}{8} \\ &= \frac{-4 \pm 4\sqrt{2}}{8} \\ &= \frac{-1 \pm \sqrt{2}}{2} \\ \text{karena akar x harus minimal nol jadi } x = \frac{-1+\sqrt{2}}{2} \\ \end{align} </math> </div></div> <ol start=10> <li>Berapa nilai x dari <math>\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x-\sqrt{x+1}}{x+\sqrt{x+1}} &= \frac{11}{19} \\ \text{misalkan } \sqrt{x+1}=y \text{ dan } x=y^2-1 \\ \frac{y^2-1-y}{y^2-1+y} &= \frac{11}{19} \\ 19(y^2-y-1) &= 11(y^2+y-1) \\ 19y^2-19y-19 &= 11y^2+11y-11 \\ 8y^2-30y-8 &= 0 \\ 4y^2-15y-4 &= 0 \\ (4y+1)(y-4) &= 0 \\ y=-\frac{1}{4} \text{ (TM) atau } & y=4 \\ x &= 4^2-1 \\ &= 15 \\ \end{align} </math> </div></div> <ol start=11> <li>Berapa nilai x dari <math>\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} &= 98 \\ \text{misalkan } \sqrt{x^2-1}=y \\ \frac{x+y}{x-y}+\frac{x-y}{x+y} &= 98 \\ \frac{(x+y)^2+(x-y)^2}{(x-y)(x+y)} &= 98 \\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} &= 98 \\ \frac{2(x^2+y^2)}{x^2-y^2} &= 98 \\ \frac{x^2+y^2}{x^2-y^2} &= 49 \\ x^2+y^2 &= 49(x^2-y^2) \\ x^2+y^2 &= 49x^2-49y^2 \\ 48x^2 &= 50y^2 \\ 24x^2 &= 25y^2 \\ 24x^2 &= 25(\sqrt{x^2-1})^2 \\ 24x^2 &= 25(x^2-1) \\ 24x^2 &= 25x^2-25 \\ x^2 &= 25 \\ x &= \pm 5 \\ \end{align} </math> </div></div> <ol start=12> <li>Berapa nilai x dari <math>\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \sqrt{x}=y \text{ dan } x=y^2 \\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} &= \frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\sqrt{\frac{y^2}{y^2+y}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\frac{y}{\sqrt{y^2+y}} \\ y^2+y-\sqrt{(y^2+y)(y^2-y)} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^4-y^2} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^2(y^2-1)} &= \frac{5}{4}y \\ y(y+1)-y\sqrt{y^2-1} &= \frac{5}{4}y \\ y+1-\sqrt{y^2-1} &= \frac{5}{4} \\ -\sqrt{y^2-1} &= \frac{1}{4}-y \\ y^2-1 &= (\frac{1}{4}-y)^2 \\ y^2-1 &= \frac{1}{16}-\frac{1}{2}y+y^2 \\ -1 &= \frac{1}{16}-\frac{1}{2}y \\ \frac{1}{2}y &= \frac{1}{16}+1 \\ \frac{1}{2}y &= \frac{17}{16} \\ y &= \frac{17}{8} \\ x &= (\frac{17}{8})^2 \\ &= \frac{289}{64} \\ \end{align} </math> </div></div> <ol start=13> <li>Berapa nilai x dari <math>\sqrt[4]{62+x}+\sqrt[4]{275-x}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ misalkan } \sqrt[4]{62+x}=a, 62+x=a^4, \sqrt[4]{275-x}=b \text{ dan } 275-x=b^4 \\ a+b &= 7 \\ (a+b)^2 &= 49 \\ a^2+b^2+2ab &= 49 \\ a^2+b^2 &= 49-2ab \\ a^4+b^4 &= 62+x+275-x \\ (a^2+b^2)^2-2(ab)^2 &= 337 \\ (49-2ab)^2-2(ab)^2 &= 337 \\ 2401-196ab+4(ab)^2-2(ab)^2 &= 337 \\ 2(ab)^2-196ab+2064 &= 0 \\ (ab)^2-98ab+1032 &= 0 \\ (ab-12)(ab-86) &= 0 \\ ab = 12 \text{ atau } & ab = 86 \text{ (TM) karena hasil kali maksimum yaitu 12 } \\ ab =12 \text{ dan } a+b=7 \\ a+b &= 7 \\ b &= 7-a \\ ab &= 12 \\ a(7-a) &= 12 \\ -a^2+7a &= 12 \\ a^2-7a+12 &= 0 \\ (a-3)(a-4) &= 0 \\ a=3 \text{ atau } & a=4 \\ a=3, b=4 \\ 62+x &= a^4 \\ 62+x &= (3)^4 \\ 62+x &= 81 \\ x &= 19 \\ a=4, b=3 \\ 62+x &= a^4 \\ 62+x &= (4)^4 \\ 62+x &= 256 \\ x &= 194 \\ \end{align} </math> </div></div> <ol start=14> <li>Berapa nilai x dari <math>\sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2} &= 7 \\ (\sqrt[3]{8+x})^2-\sqrt[3]{8+x} \sqrt[3]{27-x}+(\sqrt[3]{27-x})^2 &= 7 \\ \text{misalkan } \sqrt[3]{8+x}=a, 8+x=a^3, \sqrt[3]{27-x}=b \text{ dan } 27-x=b^3 \\ a^2-ab+b^2 &= 7 \\ a^3+b^3 &= 8+x+27-x \\ &= 35 \\ a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ 35 &= (a+b)(7) \\ a+b &= 5 \\ b &= 5-a \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ 5^3 &= 35+3ab(5) \\ 125 &= 35+15ab \\ 80 &= 15ab \\ ab &= 6 \\ a(5-a) &= 6 \\ 5a-a^2 &= 6 \\ a^2-5a+6 &= 6 \\ (a-2)(a-3) &= 6 \\ a=2 &\text{ atau } a=3 \\ a=2, b=3 \text{ dan } a=3,b=2 \\ 8+x &= a^3 \\ &= 2^3 \\ &= 8 \\ x &= 0 \\ 8+x &= a^3 \\ &= 3^3 \\ &= 27 \\ x &= 19 \\ \end{align} </math> </div></div> <ol start=15> <li>Berapa nilai x dari <math>3^x+5^x-9^x+15^x-25^x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3^x+5^x-9^x+15^x-25^x &= 1 \\ 3^x+5^x-(3^2)^x+(3 \cdot 5)^x-(5^2)^x &= 1 \\ 3^x+5^x-(3^x)^2+(3^x \cdot 5^x)-(5^x)^2 &= 1 \\ \text{misalkan } 3^x=a \text{ dan } 5^x=b \\ a+b-a^2+ab-b^2 &= 1 \\ a^2-ab+b^2-a-b+1 &= 0 \\ 2a^2-2ab+2b^2-2a-2b+2 &= 0 \\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1 &= 0 \\ (a-b)^2+(a-1)^2+(b-1)^2 &= 0 \\ a-b=0; a-1=0; b-1 &= 0 \\ a=b &= 1 \\ 3^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> <ol start=16> <li>Berapa nilai x dari <math>^6log x^2+^{6x}log \frac{6}{x}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ \text{misalkan } 6x=a \text{ maka } x=\frac{a}{6} \\ ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ ^6log (\frac{a}{6})^2+^{6 \frac{a}{6}}log \frac{6}{\frac{a}{6}} &= 1 \\ ^6log \frac{a^2}{6^2}+^alog \frac{6^2}{a} &= 1 \\ ^6log a^2-^6log 6^2+^alog 6^2-^alog a &= 1 \\ 2 ^6log a-2 ^6log 6+2 ^alog 6-^alog a &= 1 \\ 2 ^6log a-2+2 \frac{1}{^6log a}-1 &= 1 \\ 2 ^6log a+2 \frac{1}{^6log a}-4 &= 0 \\ 2 ^6log^2 a-4 ^6log a+2 &= 0 \\ ^6log^2 a-2 ^6log a+1 &= 0 \\ (^6log a-1)^2 &= 0 \\ ^6log a &= 1 \\ a &= 6 \\ x &= \frac{a}{6} \\ &= \frac{6}{6} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=17> <li>Berapa nilai x dari (x+500)³+x=20?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+500)^3+x &= 20 \\ \text{misalkan } a=x+500 \text{ maka } x=a-500 \\ a^3+a-500 &= 20 \\ a^3+a &= 520 \\ a(a^2+1) &= 8 \cdot 65 \\ a(a^2+1) &= 8(64+1) \\ a(a^2+1) &= 8(8^2+1) \\ a &= 8 \\ x &= 8-500 \\ &= -492 \\ \end{align} </math> </div></div> <ol start=18> <li>Berapa nilai x dari <math>\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2} &= 0 \\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}} &= \frac{1}{2} \\ \frac{x^n+4^n}{x^n+16^n} &= (\frac{1}{2})^n \\ \frac{x^n+4^n}{x^n+16^n} &= \frac{1}{2^n} \\ 2^n(x^n+4^n) &= x^n+16^n \\ 2^n(x^n+2^{2n}) &= x^n+2^{4n} \\ 2^n \cdot x^n+2^{3n} &= x^n+2^{4n} \\ 2^n \cdot x^n-x^n &= 2^{4n}-2^{3n} \\ x^n(2^n-1) &= 2^{3n}(2^n-1) \\ x^n &= 2^{3n} \\ x^n &= (2^3)^n \\ x^n &= 8^n \\ x &= 8 \\ \end{align} </math> </div></div> <ol start=19> <li>Berapa hasil dari <math>\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ x &= \frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5 \cdot 6}+\sqrt{5 \cdot 5}+\sqrt{6 \cdot 4}+\sqrt{5 \cdot 4}}{\sqrt{5 \cdot 4}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5} \cdot \sqrt{6}+\sqrt{5} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{4}}{2 \cdot \sqrt{5}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{6} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{5}+\sqrt{5} \cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}} \\ &= \frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ &= \frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ \frac{1}{x} &= \frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}} \\ &= \frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ &= \frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1} \\ &= \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5} \\ &= \sqrt{6}-\sqrt{4} \\ &= \sqrt{6}-2 \\ x &= \frac{1}{\sqrt{6}-2} \\ &= \frac{\sqrt{6}+2}{6-4} \\ &= \frac{\sqrt{6}+2}{2} \\ &= 1+\frac{\sqrt{6}}{2} \\ \end{align} </math> </div></div> <ol start=20> <li>Berapa hasil dari <math>(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5 \\ \text{misalkan } x=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ x &= \frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ &= \frac{\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}} \\ &= \frac{\sqrt{3}+1}{4} \\ 4x &= \sqrt{3}+1 \\ 4x-1 &= \sqrt{3} \\ (4x-1)^2 &= 3 \\ 16x^2-8x+1 &= 3 \\ 16x^2 &= 8x+2 \\ 8x^2 &= 4x+1 \\ x^2 &= \frac{4x+1}{8} \\ *cara 1 \\ x^3 &= x \cdot x^2 \\ &= x(\frac{4x+1}{8}) \\ &= \frac{4x^2+x}{8} \\ &= \frac{4x^2}{8}+\frac{x}{8} \\ &= \frac{4(\frac{4x+1}{8})}{8}+\frac{x}{8} \\ &= \frac{16x+4}{64}+\frac{x}{8} \\ &= \frac{4x+1}{16}+\frac{x}{8} \\ &= \frac{4x+1+2x}{16} \\ &= \frac{6x+1}{16} \\ x^5 &= x^2 \cdot x^3 \\ &= (\frac{4x+1}{8})(\frac{6x+1}{16}) \\ &= \frac{24x^2+10x+1}{128} \\ &= \frac{24x^2}{128}+\frac{10x+1}{128} \\ &= \frac{24(\frac{4x+1}{8})}{128}+\frac{10x+1}{128} \\ &= \frac{96x+24}{1024}+\frac{10x+1}{128} \\ &= \frac{96x+24+80x+8}{1024} \\ &= \frac{176x+32}{1024} \\ &= \frac{176x}{1024}+\frac{32}{1024} \\ &= \frac{176}{1024}(\frac{\sqrt{3}+1}{4})+\frac{32}{1024} \\ &= \frac{44(\sqrt{3}+1)}{1024}+\frac{32}{1024} \\ &= \frac{44\sqrt{3}+44}{1024}+\frac{32}{1024} \\ &= \frac{76+44\sqrt{3}}{1024} \\ &= \frac{19+11\sqrt{3}}{256} \\ *cara 2 \\ x^4 &= (x^2)^2 \\ &= (\frac{4x+1}{8})^2 \\ &= \frac{16x^2+8x+1}{64} \\ &= \frac{16x^2}{64}+\frac{8x}{64}+\frac{1}{64} \\ &= \frac{x^2}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{\frac{4x+1}{8}}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{4x}{32}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{8}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{4}+\frac{3}{64} \\ x^5 &= x \cdot x^4 \\ &= (\frac{\sqrt{3}+1}{4})(\frac{x}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\frac{\sqrt{3}+1}{4}}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\sqrt{3}+1}{16}+\frac{3}{64}) \\ &= \frac{(\sqrt{3}+1)^2}{64}+(\frac{\sqrt{3}+1}{4})\frac{3}{64} \\ &= \frac{3+2\sqrt{3}+1}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{4+2\sqrt{3}}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{16+8\sqrt{3}}{256}+\frac{3\sqrt{3}+3}{256} \\ &= \frac{19+11\sqrt{3}}{256} \\ \end{align} </math> </div></div> <ol start=21> <li>Berapa hasil dari <math>\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4} &= \frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots) \\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots &= \frac{1}{1-\frac{1}{4}} \\ &= \frac{4}{3} \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{4}{3}) \\ &= \frac{1}{4}+\frac{16}{3} \\ &= \frac{67}{12} \\ x &= \frac{67}{9} \\ \end{align} </math> </div></div> <ol start=22> <li>Berapa nilai y-x jika <math>\frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} = \frac{x}{y}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} &= \frac{x}{y} \\ \frac{\frac{106 \times 107}{2}}{\frac{106}{2}(4+109)} &= \frac{x}{y} \\ \frac{53 \times 107}{53 \times 113} &= \frac{x}{y} \\ y-x &= 113-107 = 6 \\ \end{align} </math> </div></div> <ol start=23> <li>Berapa angka satuan dari hasil 17²⁰²⁴?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan angka satuannya} \\ 17^1 &= 7 \\ 17^2 &= 9 \\ 17^3 &= 3 \\ 17^4 &= 1 \\ 17^5 &= 7 \\ 17^6 &= 9 \\ 17^7 &= 3 \\ 17^8 &= 1 \\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1} \end{align} </math> </div></div> <ol start=24> <li>Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ 1! + 2! + 3! + 4! + \dots + 2024! &= 1 + (1x2) + (1x2x3) + (1x2x3x4) + \dots + 2024! \\ &= 1 + 2 + 6 + 24 + 120 + 720 + \dots + 2024! \\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka } 1+2+6+24 = 33 \text{ jadi angka satuannya adalah } 3 \end{align} </math> </div></div> <ol start=25> <li>Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ \frac{1! + 2! + 3! + 4! + \dots + 2024!}{12} &= \frac{1 + 1x2 + 1x2x3 + 1x2x3x4 + \dots + 2024!}{12} \\ &= \frac{1 + 2 + 6 + 24 + \dots + 2024!}{12} \\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi } 1+2+6 = 9 \end{align} </math> </div></div> <ol start=26> <li>Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{penjumlahan} \\ 1+1+1+1+ \dots \text{ (sebanyak 88 buah) }+x+y &= A \\ 88+x+y &= A \\ \text{perkalian} \\ 1 \times 1 \times 1 \times \dots \text{ (sebanyak 88 buah) }\times x \times y &= A \\ x \times y &= A \\ 88+x+y &= xy \\ xy-y &= 88+x \\ y(x-1) &= 88+x \\ y &= \frac{88+x}{x-1} \\ \text{uji selidiki untuk x=2} \\ y &= \frac{88+2}{2-1} \\ &= 90 \\ \text{buktikan} \\ 88+x+y &= xy \\ 88+2+90 &= 2(90) \\ 180 &= 180 \\ \text{terbukti} \\ \text{nilai A adalah } 180 \\ \end{align} </math> </div></div> <ol start=27> <li>Berapakah nilai x, y dan z dari <math>x+y-z=1, x^2+y^2-z^2=-5 \text{ dan } x^3+y^3-z^3=-53</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+y-z &= 1 \\ x+y &= z+1 \\ x^2+2xy+y^2 &= z^2+2z+1 \\ x^2+y^2-z^2 &= 2z+1-2xy \\ -5 &= 2z+1-2xy \\ 2xy &= 2z+6 \\ xy &= z+3 \\ x^2+y^2-z^2 &= -5 \\ x^2+y^2 &= z^2-5 \\ x^3+y^3-z^3 &= -53 \\ (x+y)(x^2-xy+y^2)-z^3+53 &= 0 \\ (x+y)(x^2+y^2-xy)-z^3+53 &= 0 \\ (z+1)(z^2-5-(z+3))-z^3+53 &= 0 \\ (z+1)(z^2-z-8)-z^3+53 &= 0 \\ z^3-z^2-8z+z^2-z-8-z^3+53 &= 0 \\ -9z+45 &= 0 \\ -9z &= -45 \\ z &= 5 \\ x+y &= 5+1 \\ x+y &= 6 \\ x &= 6-y \\ xy &= 5+3 \\ xy &= 8 \\ (6-y)y &= 8 \\ 6y-y^2 &= 8 \\ y^2-6y+8 &= 0 \\ (y-4)(y-2) &= 0 \\ y=4 \text{ atau } y=2 \\ \text{jika } y=4 \\ x+y &= z+1 \\ x+4 &= 5+1 \\ x &= 2 \\ \text{jika } y=2 \\ x+y &= z+1 \\ x+2 &= 5+1 \\ x &= 4 \\ \end{align} </math> </div></div> <ol start=28> <li>Berapakah nilai titik koordinat (x,y) dari <math>\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}</math> dan <math>\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x+y}+\sqrt{x-y} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+y}-\sqrt{x-y} &= \sqrt{\frac{52y}{3x}} \\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y}) &= \sqrt{\frac{432x}{13y}} \cdot \sqrt{\frac{52y}{3x}} \\ x+y-x+y &= \sqrt{\frac{432x \cdot 52y}{13y \cdot 3x}} \\ 2y &= \sqrt{144 \cdot 4} \\ 2y &= \sqrt{576} \\ 2y &= 24 \\ y &= 12 \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13(12)}} \\ x+12+x-12+2 \cdot \sqrt{x+12} \cdot \sqrt{x-12} &= \frac{36x}{13} \\ 2x+2 \sqrt{x^2-144} &= \frac{36x}{13} \\ 2(x+\sqrt{x^2-144}) &= \frac{36x}{13} \\ x+\sqrt{x^2-144} &= \frac{18x}{13} \\ \sqrt{x^2-144} &= \frac{5x}{13} \\ x^2-144 &= \frac{25x^2}{169} \\ \frac{144x^2}{169}-144 &= 0 \\ \frac{x^2}{169}-1 &= 0 \\ x^2-169 &= 0 \\ (x-13)(x+13) &= 0 \\ x_1=13 &\text{ atau } x_2=-13 \text{ (TM) karena } x>y \\ \end{align} </math> jadi titik koordinat (13,12) </div></div> <ol start=29> <li>Berapakah nilai dari <math>x^2-7x</math> jika <math>(x-2)^2+\frac{1}{(x-2)^2} = 11</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x-2)^2+\frac{1}{(x-2)^2} &= 11 \\ (x-2)^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2)^2} &= 11-2 \\ (x-2-\frac{1}{x-2})^2 &= 9 \\ x-2-\frac{1}{x-2} &= 3 \\ (x-2)^2-1 &= 3(x-2) \\ x^2-4x+4-1 &= 3x-6 \\ x^2-7x &= -9 \\ \end{align} </math> </div></div> <ol start=30> <li>Berapakah nilai dari <math>\frac{(x+y)^2(x+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)}</math> jika xy+yz+xz=1?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy+yz+xz &= 1 \\ x^2+xy+yz+xz &= x^2+1 \\ x(x+y)+z(x+y) &= x^2+1 \\ (x+y)(x+z) &= x^2+1 \\ \text{dengan pola yang sama } \\ (y+x)(y+z) &= y^2+1 \\ (x+z)(y+z) &= z^2+1 \\ \frac{(x+y)^2(y+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)} &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)} \\ &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)^2(y+z)^2(x+z)^2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=31> <li>Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} w+5 &= w+x+y+z+5 \\ x+4 &= w+x+y+z+5 \\ y+3 &= w+x+y+z+5 \\ z+2 &= w+x+y+z+5 \\ \text{jumlahkan keempat persamaan } \\ w+x+y+z+14 &= 4(w+x+y+z+5) \\ w+x+y+z+14 &= 4(w+x+y+z)+20 \\ 3(w+x+y+z) &= -6 \\ w+x+y+z &= -2 \\ \end{align} </math> </div></div> <ol start=32> <li>Berapakah nilai dari <math>\frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2}</math> jika <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3</math> dan x+y+z=xyz?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2} &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2} \\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) \\ 3^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz}) \\ 9 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz}) \\ &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2 \\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} &= 7 \\ \end{align} </math> </div></div> <ol start=33> <li>Berapakah nilai dari <math>\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}</math> jika <math>x^2+y^2+z^2 = -2(ab+bc+ac)</math>?>/li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+z^2 &= -2(xy+yz+xz) \\ x^2+y^2+z^2+2(xy+yz+xz) &= 0 \\ (x+y+z)^2 &= 0 \\ x+y+z &= 0 \\ x+y &= -z \\ x+z &= -y \\ y+z &= -x \\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z} &= \frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x} \\ &= -2-(-5)-(-7) \\ &= 10 \\ \end{align} </math> </div></div> <ol start=34> <li>Berapakah nilai dari <math>\frac{20xyz}{xy+yz+xz}</math> jika <math>16^x = 256^y = 625^z = 40</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 16^x = 256^y = 625^z &= 40 \\ 2^{4x} = 4^{4y} = 5^{4z} &= 40 \\ 2^{4x} &= 40 \\ 2 &= 40^{\frac{1}{4x}} \\ 4^{4y} &= 40 \\ 4 &= 40^{\frac{1}{4y}} \\ 5^{4z} &= 40 \\ 5 &= 40^{\frac{1}{4z}} \\ 2 \cdot 4 \cdot 5 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z}} \\ 1 &= \frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z} \\ 4 &= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{20xyz}{xy+yz+xz} &= 20 \cdot \frac{xyz}{xy+yz+xz} \\ &= 20 \cdot (\frac{xy+yz+xz}{xyz})^{-1} \\ &= 20 \cdot (\frac{1}{z} + \frac{1}{x} + \frac{1}{y})^{-1} \\ &= 20 \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{-1} \\ &= 20 \cdot (4)^{-1} \\ &= 20 \cdot \frac{1}{4} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=35> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+3x^2+1}</math> jika <math>6x^2+25x+6=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 6x^2+25x+6 &= 0 \\ 6x+25+\frac{6}{x} &= 0 \\ 6(x+\frac{1}{x}) &= -25 \\ x+\frac{1}{x} &= \frac{-25}{6} \\ (c+\frac{1}{x})^2 &= (\frac{-25}{6})^2 \\ x^2+2+\frac{1}{x^2} &= \frac{625}{36} \\ x^2+\frac{1}{x^2} &= \frac{625}{36}-2 \\ x^2+\frac{1}{x^2} &= \frac{553}{36} \\ \frac{x^2}{x^4+3x^2+1} &= \frac{1}{x^2+3+\frac{1}{x^2}} \\ &= \frac{1}{a^2+\frac{1}{x^2}+3} \\ &= \frac{1}{\frac{553}{36}+3} \\ &= \frac{1}{\frac{661}{36}} \\ &= \frac{36}{661} \\ \end{align} </math> </div></div> <ol start=36> <li>Berapakah nilai dari <math>\frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5} &= \frac{(9+2\sqrt{20})^{1013}}{((2)^3+3(2)^2(\sqrt{5})+3(2)(\sqrt{5})^2+(\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{((2+\sqrt{5})^2)^{1013}}{((2+\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{(2+\sqrt{5})^{2026}}{(2+\sqrt{5})^{2025}}+6-\sqrt{5} \\ &= 2+\sqrt{5}+6-\sqrt{5} \\ &= 8 \\ \end{align} </math> </div></div> <ol start=37> <li>Berapakah nilai dari <math>27x^3+\frac{8}{x^3}</math> jika <math>3x+\frac{2}{x}=6</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3x+\frac{2}{x} &= 6 \\ (3x+\frac{2}{x})^3 &= 6^3 \\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3} &= 216 \\ 27x^3+18(6)+\frac{8}{x^3} &= 216 \\ 27x^3+108+\frac{8}{x^3} &= 216 \\ 27x^3+\frac{8}{x^3} &= 108 \\ \end{align} </math> </div></div> <ol start=38> <li>Berapakah nilai dari <math>x^6+\frac{8}{x^3}</math> jika <math>x^3+\frac{1}{x^3}=8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3+\frac{1}{x^3} &= 8 \\ x^3 &= 8-\frac{1}{x^3} \\ x^6 &= 8x^3-1 \\ x^6+\frac{8}{x^3} &= 8x^3-1+\frac{8}{x^3} \\ &= 8x^3+\frac{8}{x^3}-1 \\ &= 8(x^3+\frac{1}{x^3})-1 \\ &= 8(8)-1 \\ &= 63 \\ \end{align} </math> </div></div> <ol start=39> <li>Berapakah nilai dari <math>4x+\frac{25}{x}</math> jika <math>2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2\sqrt{x}+\frac{5}{\sqrt{x}} &= 4x-\frac{25}{x} \\ 2\sqrt{x}+\frac{5}{\sqrt{x}} &= (2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}}) \\ 1 &= 2\sqrt{x}-\frac{5}{\sqrt{x}} \\ 1^2 &= (2\sqrt{x}-\frac{5}{\sqrt{x}})^2 \\ 1 &= 4x-20+\frac{25}{x} \\ 4x+\frac{25}{x} &= 21 \\ \end{align} </math> </div></div> <ol start=40> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2-x+1}{x^2+x+1} &= \frac{5}{6} \\ \frac{x^2+1-x}{x^2+1+x} &= \frac{5}{6} \\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} &= \frac{5}{6} \\ \text{ misalkan } x+\frac{1}{x} &= y \\ \frac{y-1}{y+1} &= \frac{5}{6} \\ 6(y-1) &= 5(y+1) \\ 6y-6 &= 5y+5 \\ y &= 11 \\ x+\frac{1}{x} &= 11 \\ \end{align} </math> </div></div> <ol start=41> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt{x}+x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x}+x &= 1 \\ x-1 &= -\sqrt{x} \\ (x-1)^2 &= (-\sqrt{x})^2 \\ x^2-2x+1 &= x \\ x^2-3x+1 &= 0 \\ x-3+\frac{1}{x} &= 0 \\ x+\frac{1}{x} &= 3 \\ \end{align} </math> </div></div> <ol start=42> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt[3]{x}-\sqrt[3]{x-36}=3</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x}-\sqrt[3]{x-36} &= 3 \\ (\sqrt[3]{x}-\sqrt[3]{x-36})^3 &= 3^3 \\ x-(x-36)-3 \sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36}) &= 27 \\ 36-3 \sqrt[3]{x(x-36)}3 &= 27 \\ -9 \sqrt[3]{x(x-36)} &= -9 \\ \sqrt[3]{x(x-36)} &= 1 \\ x(x-36) &= 1 \\ x^2-36x-1 &= 0 \\ x-36-\frac{1}{x} &= 0 \\ x-\frac{1}{x} &= 36 \\ \end{align} </math> </div></div> <ol start=43> <li>Berapakah nilai dari <math>x+\frac{16}{x}</math> jika <math>x-3\sqrt{x}=4</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x-3\sqrt{x} &= 4 \\ x-4 &= 3\sqrt{x} \\ x^2-8x+16 &= 9x \\ x^2-17x+16 &= 0 \\ x-17+\frac{16}{x} &= 0 \\ x+\frac{16}{x} &= 17 \\ \end{align} </math> </div></div> <ol start=44> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+4}</math> jika <math>x^2-7x+2=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2-7x+2 &= 0 \\ x^2+2 &= 7x \\ x+\frac{2}{x} &= 7 \\ x^2+4+\frac{4}{x^2} &= 49 \\ x^2+\frac{4}{x^2} &= 45 \\ \frac{x^4+4}{x^2} &= 45 \\ \frac{x^2}{x^4+4} &= \frac{1}{45} \\ \end{align} </math> </div></div> <ol start=45> <li>Berapakah nilai dari <math>x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}</math> jika <math>x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}} &= 25 \\ x^{\frac{1}{2}}+x^{-\frac{1}{2}} &= 23 \\ x+2+x^{-1} &= 529 \\ x+x^{-1} &= 527 \\ x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+x^{-\frac{3}{4}} &= 110 \\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1} &= x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}} \\ &= 527+110 \\ &= 637 \\ \end{align} </math> </div></div> <ol start=46> <li>Berapakah nilai dari <math>\sqrt{8x^6+x^5+x^4+5x^3+1}</math> jika <math>\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5} &= 0 \\ \frac{x^2+x+1}{x^5} &= 0 \\ x^2+x+1 &= 0 \\ x^2+x+1 &= 0 \\ (x-1)(x^2+x+1) &= 0(x-1) \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ \sqrt{8x^6+x^5+x^4+5x^3+1} &= \sqrt{(2x^3)^2+x^3x^2+x^3x+5x^3+1} \\ &= \sqrt{(2(1))^2+(1)x^2+(1)x+5(1)+1} \\ &= \sqrt{(2)^2+x^2+x+5+1} \\ &= \sqrt{4+x^2+x+1+5} \\ &= \sqrt{4+0+5} \\ &= \sqrt{9} \\ &= 3 \\ \end{align} </math> </div></div> <ol start=47> <li>Berapakah nilai dari <math>f(1)+f(2)+f(3)+ \dots + f(99)</math> jika <math>f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{1}{\sqrt{x+1}+\sqrt{x}} \\ &= \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} \\ &= \sqrt{x+1}-\sqrt{x} \\ f(1)+f(2)+f(3)+ \dots + f(98)+f(99) &= \sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+ \cdot + \sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99} \\ &= \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+ \cdot + \sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ &= \sqrt{100}-\sqrt{1} \\ &= 10-1 \\ &= 9 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots + \frac{2024}{2025})</math> jika <math>h(x)=\frac{3}{3+9^x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} h(x) &= \frac{3}{3+9^x} \\ h(1-x) &= \frac{3}{3+9^{1-x}} \\ &= \frac{3}{3+\frac{9}{9^x}} \\ &= \frac{9^x}{3+9^x} \\ h(x)+h(1-x) &= \frac{3}{3+9^x}+\frac{9^x}{3+9^x} \\ &= \frac{3+9^x}{3+9^x} \\ &= 1 \\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots +(1-\frac{2}{2025})+(1-\frac{1}{2025})) \\ & 5(1+1+1+ \dots +1+1) \text{ sebanyak 1012 kali } \\ & 5(1012) \\ & 5060 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>\frac{7^{2025} - 7^{2023} + 432}{7^{2024} + 7^{2023} + 72}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72} &= \frac{7^{2023}7^{2}-7^{2023} + 48 \times 9}{7^{2023}7^1+7^{2023}+8 \times 9} \\ &= \frac{7^{2023}(7^{2}-1)+48 \times 9}{7^{2023}(7^1+1)+8 \times 9} \\ &= \frac{7^{2023}(49-1)+48 \times 9}{7^{2023}(7+1) + 8 \times 9} \\ &= \frac{7^{2023} \times 48+48 \times 9}{7^{2023} \times 8+8 \times 9} \\ &= \frac{48(7^{2023}+9)}{8(7^{2023}+9)} \\ &= \frac{48}{8} \\ &= 6 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>tan (x+\frac{\pi}{4})</math> jika <math>\frac{1}{cos x}-tan x = \frac{4}{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{cos x}-tan x &= \frac{4}{5} \\ sec x-tan x &= \frac{4}{5} \\ sec^2 x-tan^2 x &= 1 \\ (sec x+tan x)(sec x-tan x) &= 1 \\ (sec x+tan x)\frac{4}{5} &= 1 \\ sec x+tan x &= \frac{5}{4} \\ \text{kedua persamaan dengan cara metode eliminasi } \\ 2 tan x &= \frac{5}{4}-\frac{4}{5} \\ 2 tan x &= \frac{9}{20} \\ tan x &= \frac{9}{40} \\ tan (x+\frac{\pi}{4}) &= \frac{tan x+tan \frac{\pi}{4}}{1-tan x \cdot tan \frac{\pi}{4}} \\ &= \frac{\frac{9}{40}+1}{1-\frac{9}{40} \cdot 1} \\ &= \frac{\frac{49}{40}}{\frac{31}{40}} \\ &= \frac{49}{31} \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>sin^3 x+csc^3 x</math> jika <math>sin x-csc x = 8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ Dengan menggunakan rumus: } (a-b)^3 &= a^3-b^3-3ab(a-b) \\ (sin x-csc x)^3 &= sin^3 x-csc^3 x-3sin x csc x(sin x-csc x) \\ 8^3 &= sin^3 x-csc^3 x-3sin x (\frac{1}{sin x})(8) \\ 512 &= sin^3 x-csc^3 x-24 \\ sin^3 x-csc^3 x &= 512+24 \\ sin^3 x-csc^3 x &= 536 \\ \end{align} </math> </div></div> <ol start=52> <li>Berapakah nilai dari <math>(sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2</math> jika <math>\frac{1}{sin x}+\frac{1}{cos x} = 10</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{sin x}+\frac{1}{cos x} &= 10 \\ \frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} &= 100 \\ (sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2 &= sin^2 x+\frac{2sin x}{cos x}+\frac{1}{cos^2 x}+cos^2 x+\frac{2cos x}{sin x}+\frac{1}{sin^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2(sin^2 x+cos^2 x)}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+100 \\ &= 101 \\ \end{align} </math> </div></div> <ol start=53> <li>Berapakah nilai dari (x-1)⁶ jika <math>x=\frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin 80^\circ &= cos 10^\circ \\ sin 80^\circ-cos 10^\circ &= 0 \\ x &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+2 sin 20^\circ cos 20^\circ}{cos 10^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+4 sin 10^\circ cos 10^\circ cos 20^\circ}{cos 10^\circ} \\ &= 4 cos 55^\circ cos 25^\circ+4 sin 10^\circ cos 20^\circ \\ &= 2(2 cos 55^\circ cos 25^\circ+2 sin 10^\circ cos 20^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ+sin (-10)^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ-sin 10^\circ) \\ &= 2(cos 80^\circ-sin 10^\circ+cos 30^\circ+sin 30^\circ) \\ &= 2(cos 80^\circ-sin (90^\circ-80^\circ)+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(cos 80^\circ-cos 80^\circ+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= \sqrt{3}+1 \\ x-1 &= \sqrt{3} \\ (x-1)^6 &= (\sqrt{3})^6 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=54> <li>Berapakah nilai dari x jika <math>x=\frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ} \\ 2x sin 20^\circ-x sin 40 ^\circ &= x sin 20^\circ-x^2 sin 10^\circ \\ x^2 sin 10^\circ+x sin 20^\circ-x sin 40 ^\circ &= 0 \\ x(x sin 10^\circ+sin 20^\circ-sin 40 ^\circ) &= 0 \\ x = 0 &\text{ atau } x sin 10^\circ+sin 20^\circ-sin 40 ^\circ = 0 \\ x sin 10^\circ+sin 20^\circ-sin 40 ^\circ &= 0 \\ x sin 10^\circ &= sin 40 ^\circ-sin 20^\circ \\ x &= \frac{sin 40 ^\circ-sin 20^\circ}{sin 10^\circ} \\ &= \frac{2 cos 30 ^\circ sin 10^\circ}{sin 10^\circ} \\ &= 2 cos 30 ^\circ \\ &= \frac{2 \sqrt{3}}{2} \\ &= \sqrt{3} \\ \end{align} </math> </div></div> <ol start=55> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x^2}{x^2-16y^2} = \frac{625}{49}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2}{x^2-16y^2} &= \frac{625}{49} \\ \frac{x^2-16y^2}{x^2} &= \frac{49}{625} \text{ (terbalik posisinya)} \\ 1-\frac{16y^2}{x^2} &= \frac{49}{625} \\ \frac{16y^2}{x^2} &= 1 - \frac{49}{625} \\ (\frac{4y}{x})^2 &= \frac{576}{625} \\ (\frac{4y}{x})^2 &= (\frac{24}{25})^2 \\ \frac{4y}{x} &= \frac{24}{25} \\ \frac{y}{x} &= \frac{6}{25} \\ \frac{x}{y} &= \frac{25}{6} \\ \end{align} </math> </div></div> <ol start=56> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x}{y}+\frac{x+10y}{y+10x} = 2</math> serta bilangan real untuk x dan y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x}{y}+\frac{x+10y}{y+10x} &= 2 \\ \frac{x}{y}+\frac{\frac{x}{y}+10}{1+10\frac{x}{y}} &= 2 \\ \text{misalkan } \frac{x}{y} = a \\ a+\frac{a+10}{1+10a} &= 2 \\ a(1+10a)+a+10 &= 2(1+10a) \\ 10a^2+a+a+10 &= 2+20a \\ 10a^2-18a+8 &= 0 \\ 5a^2-9a+4 &= 0 \\ (5a-4)(a-1) &= 0 \\ a = \frac{4}{5} &\text{ atau } a = 1 \\ \text{jadi } \frac{x}{y} = {\frac{4}{5}, 1} \\ \end{align} </math> </div></div> <ol start=57> <li>Berapakah nilai dari xy jika <math>x^4+y^4+x^2y^2=15 \text{ dan } x^2+y^2+xy=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+xy &= 5 \\ x^2+y^2 &= 5-xy \\ x^4+y^4+x^2y^2 &= 15 \\ (x^2)^2+(y^2)^2+2x^2y^2-x^2y^2 &= 15 \\ (x^2+y^2)^2-x^2y^2 &= 15 \\ (5-xy)^2-x^2y^2 &= 15 \\ 25-10xy+x^2y^2-x^2y^2 &= 15 \\ 25-10xy &= 15 \\ 10xy &= 10 \\ xy &= 1 \\ \end{align} </math> </div></div> <ol start=58> <li>Berapakah nilai dari x jika <math>4^x = 63(4^3+1)(4^6+1)(4^{12}+1)+1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 4^x &= 63(4^3+1)(4^6+1)(4^{12}+1)+1 \\ 4^x-1 &= 63(4^3+1)(4^6+1)(4^{12}+1) \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{4^3-1} \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{63} \\ &= (4^3+1)(4^6+1)(4^{12}+1)(4^3-1) \\ &= (4^3-1)(4^3+1)(4^6+1)(4^{12}+1) \\ &= (4^6-1)(4^6+1)(4^{12}+1) \\ &= (4^{12}-1)(4^{12}+1) \\ &= 4^{24}-1 \\ 4^x &= 4^{24} \\ x &= 24 \\ \end{align} </math> </div></div> <ol start=59> <li>Berapakah nilai dari <math>\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}</math> jika <math>x=\sqrt{9+4\sqrt{5}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \sqrt{9+4\sqrt{5}} \\ x &= 2+\sqrt{5} \\ x^2 &= 9+4\sqrt{5} \\ x^2-4x &= 9+4\sqrt{5}-4(2+\sqrt{5}) \\ x^2-4x &= 1 \\ x^2 &= 4x+1 \\ x^3 &= x \cdot x^2 \\ &= x(4x+1) \\ &= 4x^2+x \\ &= 4(4x+1)+x \\ &= 16x+4+x \\ &= 17x+4 \\ x^4 &= x \cdot x^3 \\ &= x(17x+4) \\ &= 17x^2+4x \\ &= 17(4x+1)+4x \\ &= 68x+17+4x \\ &= 72x+17 \\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1} &= \frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1} \\ &= \frac{72x+17-85x-20+8x+2+5x+3}{2} \\ &= \frac{2}{2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=60> <li>Berapakah nilai dari <math>\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}</math> jika 2x-1=<math>\sqrt{61}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \frac{x^3+1}{x^5-x^4-x^3+x^2} = p \\ p &= \frac{x^3+1}{x^5-x^4-x^3+x^2} \\ &= \frac{x^3+1}{x^5-x^4-(x^3-x^2)} \\ &= \frac{x^3+1}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)} \\ &= \frac{x^2-x+1}{x^2(x-1)^2} \\ &= \frac{x^2-x+1}{(x(x-1))^2} \\ &= \frac{x(x-1)+1}{(x(x-1))^2} \\ 2x-1 &= \sqrt{61} \\ x &= \frac{\sqrt{61}+1}{2} \\ x-1 &= \frac{\sqrt{61}-1}{2} \\ x(x-1) &= (\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2}) \\ &= \frac{61-1}{4} \\ &= \frac{60}{4} \\ &= 15 \\ p &= \frac{x(x-1)+1}{(x(x-1))^2} \\ &= \frac{15+1}{15^2} \\ &= \frac{16}{15^2} \\ \sqrt{p} &= \sqrt{\frac{16}{15^2}} \\ &= \frac{4}{15} \\ \end{align} </math> </div></div> <ol start=61> <li>Berapakah nilai dari <math>(\frac{x-3}{x})^{25}</math> jika <math>x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\sqrt[5]{8}+\sqrt[5]{2} &= 1+\sqrt[5]{16}+\sqrt[5]{4} \\ x+(\sqrt[5]{2})^3+\sqrt[5]{2} &= 1+(\sqrt[5]{2})^4+(\sqrt[5]{2})^2 \\ x &= (\sqrt[5]{2})^4-(\sqrt[5]{2})^3+(\sqrt[5]{2})^2-\sqrt[5]{2}+1 \\ \text{misalkan } \sqrt[5]{2} = p \\ x &= p^4-p^3+p^2-p+1 \\ x &= \frac{p^5+1}{p+1} \\ (\frac{x-3}{x})^{25} &= (1-\frac{3}{x})^{25} \\ &= (1-\frac{3}{\frac{p^5+1}{p+1}})^{25} \\ &= (1-\frac{3(p+1)}{p^5+1})^{25} \\ &= (1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2})^5+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{2+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-3\sqrt[5]{2}-3)}{3})^{25} \\ &= (-\sqrt[5]{2})^{25} \\ &= (-2)^5 \\ &= -32 \\ \end{align} </math> </div></div> <ol start=62> <li>Berapakah nilai dari <math>x^{50}+x^{49}+x^{48}+x^{47}+x^{46}</math> jika <math>x^2+x+1=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+x+1 &= 0 \\ x^2+x &= -1 \\ \frac{x^3-1}{x-1} &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46} &= x^{48}(x^2+x+1)+x^{45}(x^2+x) \\ &= x^{48}(0)+(x^3)^{15}(-1) \\ &= 0+(1)^{15}(-1) \\ &= -1 \\ \end{align} </math> </div></div> <ol start=63> <li>Berapakah 2²⁴ dari <math>8^7+8^6+8^5+8^4+8^3+8^2+8+1=A</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= A \\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1) &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8 &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= 8A+1 \\ 8^8+A &= 8A+1 \\ 8^8 &= 7A+1 \\ (2^3)^8 &= 7A+1 \\ 2^{24} &= 7A+1 \\ \end{align} </math> </div></div> <ol start=64> <li>Berapakah nilai dari <math>x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1</math> jika <math>x+\frac{1}{x}=\sqrt{3}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{x} &= \sqrt{3} \\ x^2+2+\frac{1}{x^2} &= 3 \\ x^2-1+\frac{1}{x^2} &= 0 \\ x^2(x^2-1+\frac{1}{x^2}) &= x^2(0) \\ x^4-x^2+1 &= 0 \\ (x^2+1)(x^4-x^2+1) &= (x^2+1)0 \\ x^6-x^4+x^2+x^4-x^2+1 &= 0 \\ x^6+1 &= 0 \\ x^6 &= -1 \\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1 &= {x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1 \\ &= (-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2-1+1 \\ &= -1+1-1+1-1+1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=65> <li>Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari <math>\frac{c-b}{a-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= ax^2+bx+c \\ f(2) &= a(2)^2+2b+c = 4 \\ &= 4a+2b+c = 4 \\ f(7) &= a(7)^2+7b+c = 49 \\ &= 49a+7b+c = 49 \\ 49a+7b+c &= 49 \\ 4a+2b+c &= 4 \\ 45a+5b &= 45 \text{ (f(7) dikurangi f(2)) } \\ 9a+b &= 9 \\ b &= -9a+9 \\ 4a+2b+c &= 4 \\ 4a+2(-9a+9)+c &= 4 \\ 4a-18a+18+c &= 4 \\ -14a+18+c &= 4 \\ c &= 14a-14 \\ \frac{c-b}{a-1} &= \frac{14a-14-(-9a+9)}{a-1} \\ &= \frac{14(a-1)+9(a-1)}{a-1} \\ &= \frac{(14+9)(a-1)}{a-1} \\ &= 23 \\ \end{align} </math> </div></div> <ol start=66> <li>Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+y)^3 &= x^3+y^3+3xy(x+y) \\ 11^3 &= 242+3xy(11) \text{ (dibagi 11)} \\ 11^2 &= 22+3xy \\ 121 &= 22+3xy \\ 99 &= 3xy \\ xy &= 33 \\ (x-y)^2 &= x^2+y^2-2xy \\ &= ((x+y)^2-2xy)-2xy \\ &= (x+y)^2-4xy \\ &= 11^2-4(33) \\ &= 121-132 \\ &= -11 \\ \end{align} </math> </div></div> <ol start=67> <li>Berapa f(1)+f(-1) jika <math>f(\frac{ax-b}{bx-a})</math>=x²-5x+6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ jika} f(1) = f(\frac{ax-b}{bx-a}) \\ 1 &= \frac{ax-b}{bx-a} \\ bx-a &= ax-b \\ (b-a)x &= -b+a \\ &= -(b-a) \\ &= -1 \\ f(1) &= x^2-5x+6 \\ &= (-1)^2-5(-1)+6 \\ &= 12 \\ \text{ jika} f(-1) = f(\frac{ax-b}{bx-a}) \\ -1 &= \frac{ax-b}{bx-a} \\ -(bx-a) &= ax-b \\ -bx+a &= ax-b \\ (-b-a)x &= -b-a \\ &= 1 \\ f(-1) &= x^2-5x+6 \\ &= (1)^2-5(1)+6 \\ &= 2 \\ f(1)+f(-1) &= 12+2 \\ &= 14 \\ \end{align} </math> </div></div> <ol start=68> <li>berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)-x &= f(x-1) \\ f(x)-f(x-1) &= x \\ x=1 ; f(1)-f(0) &= 1 \\ x=2 ; f(2)-f(1) &= 2 \\ x=3 ; f(3)-f(2) &= 3 \\ x=4 ; f(4)-f(3) &= 4 \\ \dots \\ x=200 ; f(200)-f(199) &= 200 \\ \text{ jumlahkan tersebut menjadi } \\ f(200)-f(0) &= 1+2+3+4+\dots+200 \\ &= \frac{200 \cdot 201}{2} \\ &= 20.100 \\ f(200)-1 &= 20.100 \\ &= 20.101 \\ \end{align} </math> </div></div> <ol start=69> <li>Misalkan f(x) adalah fungsi rekursif yang berlaku ∀x ∈ R sebagai berikut: : f(x)+f(15-x) = 2024 : f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)+f(15-x) &= 2024 \\ f(15+x) &= f(x)+2020 \\ *cara 1 \\ \text{ganti x dengan 15+x } \\ f(15+x)+f(-x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ *cara 2 \\ \text{ganti x dengan -x } \\ f(-x)+f(15+x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ \end{align} </math> </div></div> <ol start=70> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>2f(\frac{2002}{x}) + f(x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2f(\frac{2002}{x}) + f(x) &= 3x \\ \text{ganti x dengan 2 } \\ 2f(\frac{2002}{2}) + f(2) &= 3(2) \\ 2f(1001) + f(2) &= 6 \\ \text{ganti x dengan 1001 } \\ 2f(\frac{2002}{1001}) + f(1001) &= 3(1001) \\ 2f(2) + f(1001) &= 3003 \\ 2f(2) + f(1001) &= 3003 \\ f(1001) &= 3003 - 2f(2) \\ 2f(1001) + f(2) &= 6 \\ 2(3003 - 2f(2)) + f(2) &= 6 \\ 6006 - 4f(2) + f(2) &= 6 \\ 3f(2) &= 6000 \\ f(2) &= 2000 \\ \end{align} </math> </div></div> <ol start=71> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>f(\frac{1}{x}) + \frac{1}{x}f(-x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{1}{x})+\frac{1}{x}f(-x) &= 3x \\ \text{ganti x dengan 1/3 } \\ f(3)+3f(-\frac{1}{3}) &= 1 \\ \text{ganti x dengan -3 } \\ f(-\frac{1}{3}) - \frac{1}{3}f(3) &= -9 \\ \text{dikalikan 3 } \\ 3f(-\frac{1}{3})-f(3) &= -27 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ 2f(3) &= 28 \\ f(3) &= 14 \\ \end{align} </math> </div></div> <ol start=72> <li>Diketahui polinom <math>f(7^b-1)=7^{3b}-10</math>. tentukan nilai f(5)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(5) &= f(7^b-1) \\ 5 &= 7^b-1 \\ 7^b &= 6 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(6-1) &= 6^3-10 \\ f(5) &= 216-10 \\ &= 206 \\ *cara 2 \\ \text{misalkan } 7^b-1=a \text{ maka } 7^b=a+1 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(a) &= (a+1)^3-10 \\ f(5) &= (5+1)^3-10 \\ &= 6^3-10 \\ &= 216-10 \\ &= 206 \\ \end{align} </math> </div></div> <ol start=73> <li>Diketahui polinom <math>f(6^b-7)=6^{3b}-2 \cdot 6^{2b}-4</math>. tentukan nilai f(-2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(-2) &= f(6^b-7) \\ -2 &= 6^b-7 \\ 6^b &= 5 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(5-7) &= 5^3-2 \cdot 5^2-4 \\ f(-2) &= 125-50-4 \\ &= 71 \\ *cara 2 \\ \text{misalkan } 6^b-7=a \text{ maka } 6^b=a+7 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(a) &= (a+7)^3-2(a+7)^2-4 \\ f(-2) &= (-2+7)^3-2(-2+7)^2-4 \\ &= 5^3-2(5)^2-4 \\ &= 125-50-4 \\ &= 71 \\ \end{align} </math> </div></div> <ol start=74> <li>Jika <math>f(xy)=\frac{f(x)}{y}</math> dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(10) &= 7 \\ f(2 \cdot 5) &= 7 \\ f(xy) &= \frac{f(x)}{y} \\ f(2 \cdot 5) &= \frac{f(2)}{5} \\ 7 &= \frac{f(2)}{5} \\ f(2) &= 35 \\ \end{align} </math> </div></div> <ol start=75> <li>Jika <math>f(xy)=\frac{f(x+y)}{xy}</math> dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(15) &= 16 \\ f(3 \cdot 5) &= 16 \\ f(xy) &= \frac{f(x+y)}{xy} \\ f(3 \cdot 5) &= \frac{f(3+5)}{3 \cdot 5} \\ f(15) &= \frac{f(8)}{15} \\ 16 &= \frac{f(8)}{15} \\ f(8) &= 240 \\ \end{align} </math> </div></div> <ol start=76> <li>Jika <math>f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15</math> maka tentukan nilai f(16)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x+\frac{1}{x}+6) &= x^2+\frac{1}{x^2}+15 \\ &= (x+\frac{1}{x})^2-2+15 \\ &= (x+\frac{1}{x})^2+13 \\ \text{misalkan } x+\frac{1}{x} &= p \\ f(x+\frac{1}{x}+6) &= (x+\frac{1}{x})^2+13 \\ f(p+6) &= p^2+13 \\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 } \\ f(p+6) &= p^2+13 \\ f(10+6) &= 10^2+13 \\ f(16) &= 100+13 \\ &= 113 \\ \end{align} </math> </div></div> <ol start=77> <li>Tentukan nilai x jika <math>f(x)=\frac{4}{4-x}</math> dan <math>f(x \cdot f(x))^{\frac{f(4x)}{f(x)}}=256</math>!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{4}{4-x} \\ f(4x) &= \frac{4}{4-4x} \\ \frac{f(4x)}{f(x)} &= \frac{\frac{4}{4-4x}}{\frac{4}{4-x}} \\ &= \frac{4-x}{4-4x} \\ f(x \cdot f(x)) &= f(x(\frac{4}{4-x})) \\ &= f(\frac{4x}{4-x}) \\ &= \frac{4}{4-(\frac{4x}{4-x})} \\ &= \frac{4}{\frac{16-4x-4x}{4-x}} \\ &= \frac{4}{\frac{16-8x}{4-x}} \\ &= \frac{4(4-x)}{4(4-4x)} \\ &= \frac{4-x}{4-4x} \\ \text{misalkan } \frac{4-x}{4-4x} &= a \\ f(x \cdot f(x))^{\frac{f(4x)}{f(x)}} &= 256 \\ a^a &= 256 \\ a^a &= 4^4 \\ a &= 4 \\ \frac{4-x}{4-4x} &= 4 \\ 4-x &= 16-16x \\ 15x &= 12 \\ x &= \frac{4}{5} \\ \end{align} </math> </div></div> <ol start=78> <li>Fungsi <math>f(x) = \frac{kx}{2x+1} \text{dengan } x \neq -\frac{1}{2}</math>. Dengan f(f(x)) = x maka tentukan nilai k!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{kx}{2x+1} \\ f(f(x)) &= x \\ f(\frac{kx}{2x+1}) &= x \\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1} &= x \\ \frac{\frac{k^2x}{2x+1}}{\frac{2kx+2x+1}{2x+1}} &= x \\ \frac{k^2x}{2kx+2x+1} &= x \\ \frac{k^2}{2kx+2x+1} &= 1 \\ k^2 &= 2kx+2x+1 \\ k^2-2kx &= 2x+1 \\ k^2-2kx+x^2 &= x^2+2x+1 \\ (k-x)^2 &= (x+1)^2 \\ (k-x)^2-(x+1)^2 &= 0 \\ (k-x+x+1)(k-x-(x+1)) &= 0 \\ k=-1 &\text{ atau } k=2x+1 &\text{ (TM) } \\ \end{align} </math> </div></div> <ol start=79> <li>Jika n = 2023²+2024² maka berapa hasil dari <math>\sqrt{2n-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n &= 2023^2+2024^2 \\ &= 2023^2+(2023+1)^2 \\ \text{misalkan 2023 = p} \\ n &= p^2+(p+1)^2 \\ &= p^2+p^2+2p+1 \\ &= 2p^2+2p+1 \\ \sqrt{2n-1} &= \sqrt{2(2p^2+2p+1)-1} \\ &= \sqrt{4p^2+4p+2-1} \\ &= \sqrt{4p^2+4p+1} \\ &= \sqrt{(2p+1)^2} \\ &= 2p+1 \\ &= 2(2023)+1 \\ &= 4046+1 \\ &= 4047 \\ \end{align} </math> </div></div> <ol start=80> <li>Tentukan nilai dari a+b+c merupakan bilangan bulat positif jika ab = 2, bc = 3 dan ac = 6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab \cdot bc \cdot ac &= 2 \cdot 3 \cdot 6 \\ (abc)^2 &= 36 \\ abc &= \pm 6 \\ abc &= 6 \\ \frac{abc}{ab} &= c = \frac{6}{2} = 3 \\ \frac{abc}{bc} &= a = \frac{6}{3} = 2 \\ \frac{abc}{ac} &= b = \frac{6}{6} = 1 \\ a+b+c &= 6 \\ \end{align} </math> </div></div> <ol start=81> <li>Tentukan nilai dari (a-c)^b jika <math>\frac{ab}{a+b} = \frac{1}{3}</math>, <math>\frac{bc}{b+c} = \frac{1}{4}</math> dan <math>\frac{ac}{a+c} = \frac{1}{9}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{ab}{a+b} &= \frac{1}{3} \\ \frac{a+b}{ab} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{b} + \frac{1}{a} &= 3 \\ \frac{bc}{b+c} &= \frac{1}{4} \\ \frac{b+c}{bc} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{b} &= 4 \\ \frac{ac}{a+c} &= \frac{1}{9} \\ \frac{a+c}{ac} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{a} &= 9 \\ \text{Misalkan 1/a = x, 1/b = y dan 1/c = z} \\ x+y &= 3 \\ y+z &= 4 \\ x+z &= 9 \\ x+y &= 3 \\ y+z &= 4 \\ x-z &= -1 \\ x-z &= -1 \\ x+z &= 9 \\ 2x &= 8 \\ x &= 4 \\ x-z &= -1 \\ 4-z &= -1 \\ z &= 5 \\ x+y &= 3 \\ 4+y &= 3 \\ y &= -1 \\ \frac{1}{a} &= 4 \\ a &= \frac{1}{4} \\ \frac{1}{b} &= -1 \\ b &= -1 \\ \frac{1}{c} &= 5 \\ c &= \frac{1}{5} \\ (a-c)^b &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=82> <li>Tentukan nilai dari a, b dan c jika <math>\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}</math> dan a+2b+3c=28?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan k untuk semua ketiga persamaan tersebut } \\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5} &= k \\ a+b &= 2k \\ a+c &= 4k \\ b+c &= 5k \\ 2a+b+c &= 6k \\ 2a+5k &= 6k \\ k &= 2a \\ a &= \frac{k}{2} \\ b &= \frac{3k}{2} \\ c &= \frac{7k}{2} \\ a+2b+3c &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ a &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ b &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ c &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=83> <li>Tentukan nilai dari (b+c)^a jika <math>\frac{a+b+c}{2} = \sqrt{a-2}+\sqrt{b-1}+\sqrt{c}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{a+b+c}{2} &= \sqrt{a-2}+\sqrt{b-1}+\sqrt{c} \\ a+b+c &= 2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}) \\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c} &= 0 \\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1 &= 0 \\ (\sqrt{a-2}-1)^2+(\sqrt{b-1}-1)^2+(\sqrt{c}-1)^2 &= 0 \\ (\sqrt{a-2}-1)^2 &= 0 \\ \sqrt{a-2}-1 &= 0 \\ \sqrt{a-2} &= 1 \\ a-2 &= 1 \\ a &= 3 \\ (\sqrt{b-1}-1)^2 &= 0 \\ \sqrt{b-1}-1 &= 0 \\ \sqrt{b-1} &= 1 \\ b-1 &= 1 \\ b &= 1 \\ (\sqrt{c}-1)^2 &= 0 \\ \sqrt{c}-1 &= 0 \\ \sqrt{c} &= 1 \\ c &= 1 \\ (b+c)^a &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=84> <li>x dan y merupakan bilangan tak nol. Jika xy = <math>\frac{x}{y}</math> = x-y maka berapa nilai x+y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy &= \frac{x}{y} \\ y^2 &= 1 \\ y^2 - 1 &= 0 \\ (y-1)(y+1) &= 0 \\ y = 1 &\text{ atau } y = -1 \\ \frac{x}{y} &= x-y \\ x &= xy-y^2 \\ x-xy &= -y^2 \\ x(1-y) &= -y^2 \\ x &= \frac{-y^2}{1-y} \\ \text{cek y=1 } \\ x &= \frac{-1^2}{1-1} \\ \text{tidak memenuhi syarat } \\ \text{cek y=-1 } \\ x &= \frac{-(-1)^2}{1-(-1)} \\ &= \frac{-1}{2} \\ x+y &= -1-\frac{1}{2} \\ &= -\frac{3}{2} \\ \end{align} </math> </div></div> <ol start=86> <li>Berapa nilai x dari <math>(\frac{a}{b})^3+(\frac{b}{a})^3 = 2\sqrt{x}</math> jika <math>\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b} \\ \frac{a+b}{ab} &= \frac{1}{a+b} \\ (a+b)^2 &= ab \\ a^2+2ab+b^2 &= ab \\ a^2+b^2 &= -ab \\ \text{misalkan } \frac{a}{b}+\frac{b}{a} = n \\ \frac{a}{b}+\frac{b}{a} &= n \\ \frac{a^2+b^2}{ab} &= n \\ a^2+b^2 &= nab \\ n &= -1 \\ \frac{a}{b}+\frac{b}{a} &= n \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3(\frac{a}{b}+\frac{b}{a}) &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3n &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3 &= n^3-3n \\ &= (-1)^3-3(-1) \\ &= 2 \\ 2\sqrt{x} &= 2 \\ \sqrt{x} &= 1 \\ x &= 1 \\ \end{align} </math> </div></div> <ol start=87> <li>Berapa nilai m dari <math>x^2-mx-1=0</math> jika <math>\sqrt[3]{x_1}+\sqrt[3]{x_2}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x_1} &= a \\ x_1 &= a^3 \\ \sqrt[3]{x_2} &= b \\ x_2 &= b^3 \\ \sqrt[3]{x_1}+\sqrt[3]{x_2} &= 1 \\ a+b &= 1 \\ x^2-mx-1 &= 0 \\ x_1+x_2 &= m \\ x_1 \cdot x_2 &= -1 \\ x_1+x_2 &= m \\ a^3+b^3 &= m \\ x_1 \cdot x_2 &= -1 \\ a^3 \cdot b^3 &= -1 \\ (ab)^2 &= (-1)^3 \\ ab &= -1 \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ (1)^3 &= m+3(-1)(1) \\ 1 &= m-3 \\ m &= 4 \\ \end{align} </math> </div></div> <ol start=88> <li>Berapa nilai <math>\frac{x_1}{x_2}</math> dari <math>ax^2-18x-b=0</math> jika <math>ab=45</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab &= 45 \\ b &= \frac{45}{a} \\ ax^2-18x-b &= 0 \\ ax^2-18x-\frac{45}{a} &= 0 \\ a^2x^2-18ax-45 &= 0 \\ (ax-3)(ax-15) &= 0 \\ ax-3 &= 0 \\ x &= \frac{3}{a} \\ ax-15 &= 0 \\ x &= \frac{15}{a} \\ \frac{x_1}{x_2} &= \frac{\frac{3}{a}}{\frac{15}{a}} \\ &= \frac{3}{15} \\ &= \frac{1}{5} \\ \frac{x_1}{x_2} &= \frac{\frac{15}{a}}{\frac{3}{a}} \\ &= \frac{15}{3} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=89> <li>Jika <math>\frac{u_3}{u_1+u_2} = \frac{7}{8}</math> merupakan barisan aritmetika maka berapa dari <math>\frac{u_2+u_3}{u_1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{u_3}{u_1+u_2} &= \frac{7}{8} \\ \frac{a+2b}{a+a+b} &= \frac{7}{8} \\ \frac{a+2b}{2a+b} &= \frac{7}{8} \\ 8(a+2b) &= 7(2a+b) \\ 8a+16b &= 14a+7b \\ 9b &= 6a \\ b &= \frac{2a}{3} \\ \frac{u_2+u_3}{u_1} &= \frac{a+b+a+2b}{a} \\ &= \frac{2a+3b}{a} \\ &= \frac{2a+3(\frac{2a}{3})}{a} \\ &= \frac{2a+2a}{a} \\ &= \frac{4a}{a} \\ &= 4 \\ \end{align} </math> </div></div> <ol start=90> <li>Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7p+q}{2p+q} &= \frac{17p+q}{7p+q} \\ (7p+q)^2 &= (17p+q)(2p+q) \\ 49p^2+14pq+q^2 &= 34p^2+19pq+q^2 \\ 15p^2 &= 5pq \\ 3p &= q \\ \frac{7p+q}{2p+q} &= \frac{7p+3p}{2p+3p} \\ &= \frac{10p}{5p} \\ &= 2 \\ \end{align} </math> </div></div> <ol start=91> <li>Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{rataan geometris } \\ \sqrt{a \cdot b} &= b-24 \\ a \cdot b &= (b-24)^2 \\ \text{rataan aritmatik } \\ \frac{a+b}{2} &= a+15 \\ a+b &= 2(a+15) \\ a+b &= 2a+30 \\ a &= b-30 \\ a \cdot b &= (b-24)^2 \\ (b-30)b &= (b-24)^2 \\ b^2-30b &= b^2-48b+576 \\ 18b &= 576 \\ b &= 32 \\ a &= b-30 \\ &= 32-30 \\ &= 2 \\ a+b &= 32+2 \\ &= 34 \\ \end{align} </math> </div></div> <ol start=91> <li>Segitiga lancip ABC dengan <math>\frac{a^4+b^4+c^4+a^2b^2}{c^2(a^2+b^2)}=2</math>. tentukan nilai sudut C?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{syarat segitiga lancip semua sudut masing-masing kurang dari } 90^\circ \\ c^2 &= a^2+b^2-2ab cos C \\ cos C &= \frac{a^2+b^2-c^2}{2ab} \\ a^4+b^4+c^4+a^2b^2 &= 2c^2(a^2+b^2) \\ a^4+b^4+a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-2c^2(a^2+b^2)+(c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= (ab)^2 \\ a^2+b^2-c^2 &= \pm ab \\ cos C &= \pm \frac{ab}{2ab} \\ &= \pm \frac{1}{2} \\ &= \frac{1}{2} \text{ (karena sudut harus kurang dari } 90^\circ) \\ C &= 60^\circ \\ \end{align} </math> </div></div> <ol start=92> <li>Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan panjang DA dan EA } = x \text{ dan panjang AB } = y \\ \text{luas segitiga CAB } &= \frac{CA \cdot AB}{2} \\ 45 &= \frac{(x+9)(x+5)}{2} \\ 90 &= x^2+14x+45 \\ x^2+14x &= 45 \\ y^2 &= (x+9)^2+(x+5)^2 \\ &= x^2+18x+81+x^2+10x+25 \\ &= 2x^2+28x+106 \\ &= 2(x^2+14x)+106 \\ &= 2(45)+106 \\ &= 196 \\ y &= 14 \\ \end{align} </math> jadi panjang BC adalah 14 cm </div></div> <ol start=93> <li>Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kita cari ukuran GC yaitu } \\ \frac{GC}{AD} &= \frac{CE}{DE} \\ \frac{GC}{15} &= \frac{6}{18} \\ GC &= 5 \\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC } \\ &= \frac{1}{2} \cdot BC \cdot CF - \frac{1}{2} \cdot GC \cdot CE \\ &= \frac{1}{2} \cdot 15 \cdot 18 - \frac{1}{2} \cdot 5 \cdot 6 \\ &= 135 - 15 \\ &= 120 \\ \end{align} </math> jadi luas daerah BFEG adalah 120 cm² </div></div> <ol start=94> <li>Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ AI &= \frac{1}{5} AF \\ &= \frac{1}{5} 10 \\ &= 2 \\ IF &= AF-AI \\ &= 10-2 \\ &= 8 \\ \text{luas trapesium AFGD } &= \frac{(AD+EF) \cdot AF}{2} \\ &= \frac{(4+6)10}{2} \\ &= 50 \\ \text{luas segitiga AID } &= \frac{AI \cdot AF}{2} \\ &= \frac{(2)4}{2} \\ &= 4 \\ \text{luas segitiga IFG } &= \frac{IF \cdot FG}{2} \\ &= \frac{(8)6}{2} \\ &= 24 \\ \text{luas daerah segitiga IGD } &= \text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG } \\ &= 50-4-24 \\ &= 22 \\ \end{align} </math> jadi luas daerah segitiga IGD adalah 22 cm² </div></div> <ol start=96> <li>Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-<math>\pi</math>) cm³ maka berapa luas permukaan kerucut tersebut?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ \text{volume balok} \\ V_b &= x^2(12) \\ \text{volume kerucut} \\ V_b &= \frac{1}{3}\pi x^2(12) \\ &= 4\pi x^2 \\ V_{b-k} &= Vb-Vk \\ 100(3-\pi) &= 12x^2-4\pi x^2 \\ 100(3-\pi) &= 4x^2(3-\pi) \\ x^2 &= 25 \\ x &= 5 \\ s &= \sqrt{12^2+5^2} \\ &= \sqrt{144+25} \\ &= \sqrt{169} \\ &= 13 \\ \text{luas permukaan kerucut } &= \pi r(r+s) \\ &= \pi(5)(5+13) \\ &= 90\pi \\ \end{align} </math> jadi luas daerah permukaan kerucut adalah 90<math>\pi</math> cm² </div></div> <ol start=97> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 3a+1 \\ B &= 3b+2 \\ A(A+1)+3B \\ (3a+1)(3a+1+1)+3(3b+2) \\ (3a+1)(3a+2)+9b+6 \\ 9a^2+9a+2+9b+6 \\ 9a^2+9a+9b+8 \\ 9(a^2+a+b)+8 \\ \text{sisa pembagiannya adalah } 8 \\ \end{align} </math> </div></div> <ol start=98> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 9a+7 \\ B &= 9b+8 \\ A(A-5)+9B \\ (9a+7)(9a+7-5)+9(9b+8) \\ (9a+7)(9a+2)+81b+72 \\ 81a^2+81a+14+81b+72 \\ 81a^2+81a+81b+86 \\ 81a^2+81a+81b+81+5 \\ 81(a^2+a+b+1)+5 \\ \text{sisa pembagiannya adalah } 5 \\ \end{align} </math> </div></div> <ol start=99> <li>Jika <math>\begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix}</math> maka berapa hasil dari A²¹+A²⁵+A⁴⁶?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A^2 &= A \cdot A \\ &= \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \\ A^3 &= A^2 \cdot A \\ &= \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ &= - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= -I \\ A^{21}+A^{25}+A^{46} &= A^{21} \cdot (I+A^4+A^{25}) \\ &= A^{21} \cdot (I+A^3 \cdot A +A^{24} \cdot A) \\ &= (A^3)^7 \cdot (I+A^3 \cdot A +(A^3)^8 \cdot A) \\ &= (-I)^7 \cdot (I-I \cdot A +(-I)^8 \cdot A) \\ &= -I \cdot (I-A+A) \\ &= -I \cdot I \\ &= -I \\ &= -\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ \end{align} </math> </div></div> <ol start=100> <li>Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida?</li></ol> : bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15 : ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9). : ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8) : maka Ida menulis bilangan 8. <ol start=101> <li>Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9?</li></ol> : Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu. : untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja. : untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja. : Jadi banyaknya bilangan mungkin 2. <ol start=102> <li>Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli!</li></ol> ;cara 1 # 8¹ = 1 # 8² = 1 (8²=8¹x8¹ sama dengan 1x1) # 8³ = 1 (8³=8¹x8² sama dengan 1x1) # 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2) # 8⁵ = 1 # 8ⁿ = 1 (semua n untuk bilangan asli) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli ;cara 2 # 8ⁿ = b mod 7 # 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil) # (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya) # 8ⁿ = 1ⁿ mod 7 # 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli <ol start=103> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 5?</li></ol> ;cara 1 # 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 # 7¹ = 7 (sisa 1) # 7² = 49 (sisa 2) # 7³ = 343 (sisa 3) # 7⁴ = 2,401 (sisa 0) # 7⁵ = 16,807 # 7⁶ = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3 ;cara 2 :17¹ = 2 :17² = 4 :17³ = 3 :17⁴ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3 :17⁹⁹ = (17⁴)²⁴ x 17³ Untuk 17⁴ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 3. Jadi 17⁹⁹ dibagi 7 bersisa 3 ;cara 3 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 17⁴ ::17⁴ = 1 mod 5 ::(17⁴)²⁴ = 1²⁴ mod 5 ::17⁹⁶ = 1²⁴ mod 5 ::17⁹⁶ = 1 mod 5 ::17⁹⁶ x 17³ = 1 x 17³ mod 5 ::17⁹⁹ = 17³ mod 5 ::17⁹⁹ = 17 x 17 x 17 mod 5 ::17⁹⁹ = 2 x 2 x 2 mod 5 ::17⁹⁹ = 8 mod 5 ::17⁹⁹ = 3 mod 5 Jadi 17⁹⁹ dibagi 5 bersisa 3 <ol start=104> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 7?</li></ol> ;cara 1 :17¹ = 3 :17² = 2 :17³ = 6 :17⁴ = 4 :17⁵ = 5 :17⁶ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3 :17⁹⁹ = (17⁶)¹⁶ x 17³ Untuk 17⁶ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 6. Jadi 17⁹⁹ dibagi 7 bersisa 6 ;cara 2 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 17⁶ ::17⁶ = 1 mod 7 ::(17⁶)¹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1 mod 7 ::17⁹⁶ x 17³ = 1 x 17³ mod 7 ::17⁹⁹ = 17³ mod 7 ::17⁹⁹ = 17 x 17 x 17 mod 7 ::17⁹⁹ = 3 x 3 x 3 mod 7 ::17⁹⁹ = 27 mod 7 ::17⁹⁹ = 6 mod 7 Jadi 17⁹⁹ dibagi 7 bersisa 6 <ol start=105> <li>Berapa hasil sisa dari 41²⁰²⁴ dibagi 33?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 41^{2024} &= 41^{2024} \text{ mod } 33 \\ &= (33 \times 3 + 2)^{2024} \text{ mod } 33 \\ &= 2^{2024} \text{ mod } 33 \\ &= 2^{2020} 2^4 \text{ mod } 33 \\ &= (2^5)^{404} 2^4 \text{ mod } 33 \\ &= (33 - 1)^{404} 2^4 \text{ mod } 33 \\ &= (-1)^{404} 2^4 \text{ mod } 33 \\ &= 2^4 \text{ mod } 33 \\ &= 16 \text{ mod } 33 \\ \text{Jadi hasil sisa adalah } 16 \\ \end{align} </math> </div></div> <ol start=106> <li>Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 99^{99} &= (3^2 \times 11)^{99} \\ &= 3^{198} \times 11^{99} \\ 243^n &= (3^5)^n \\ &= 3^{5n} \\ \text{agar bisa membagi, maka} \\ 5n &= 198 \\ n &= 39.6 \\ \text{jadi bilangan n terbesar adalah } 39 \\ \end{align} </math> </div></div> <ol start=107> <li>Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸?</lu></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 88^{88} &= (8 \times 11)^{88} \\ &= 8^{88} \times 11^{88} \\ &= 8^{87} \times 8 \times 11^{88} \\ &= (8^3)^{29} \times 8 \times 11^{88} \\ &= 512^{29} \times 8 \times 11^{88} \\ 512^n &= 512^{29} \\ \text{jadi bilangan n terbesar adalah } 29 \\ \end{align} </math> </div></div> <ol start=108> <li>Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6!</li></ol> ; cara 1 : KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. : N dibagi 3 sisa 1 : N dibagi 5 sisa 2 : N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a : KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70 : KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21 : KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97 ; cara 2 : Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35 : kemudian ditambahkan sisa masing-masing sesuai dengan KPK. : KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, <b>97</b> : KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, <b>97</b> : KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, <b>97</b> Jadi bilangan bulat positif adalah 97 :: NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 <ol start=109> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |- | 4 || 0 || Semua isi ember B dibuang |- | 1 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. <ol start=110> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |} nah ada ember A berisi 4 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |- | 1 || 0 || Semua isi ember B dibuang |- | 0 || 1 || Tuangkan 1 l dari ember A ke B sehingga ember A kosong |- | 3 || 1 || Isikan 3 l ke ember A |- | 0 || 4 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |} nah ada ember B berisi 4 liter. # Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kemungkinan 1} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 +3 \cdot ((\frac{4}{3})^x)^2 &= 7 \cdot (\frac{4}{3})^x \\ \text{misalkan } (\frac{4}{3})^x = y \\ 4 +3y^2 &= 7y \\ 3y^2-7y+4 &= 0 \\ (3y-4)(y-1) &= 0 \\ y_1 = \frac{4}{3} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{4}{3} \\ (\frac{4}{3})^x &= \frac{4}{3} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{4}{3})^x &= 1 \\ x &= 0 \\ \text{kemungkinan 2} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot ((\frac{3}{4})^x)^2 +3 &= 7 \cdot (\frac{3}{4})^x \\ \text{misalkan } (\frac{3}{4})^x = y \\ 4y^2 +3 &= 7y \\ 4y^2-7y+3 &= 0 \\ (4y-3)(y-1) &= 0 \\ y_1 = \frac{3}{4} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{3}{4} \\ (\frac{3}{4})^x &= \frac{3}{4} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{3}{4})^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-6}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{2x-4}{3}+2x-6 &= \frac{3}{2x-4}+\frac{1}{2x-6} \\ \frac{2x-4}{3}+2x-6 &= \frac{1}{\frac{2x-4}{3}}+\frac{1}{2x-6} \\ \text{misalkan } \frac{2x-4}{3} = y \text{ dan } 2x-6 = z \\ y+z &= \frac{1}{y}+\frac{1}{z} \\ y+z &= \frac{y+z}{xy} \\ y+z-\frac{y+z}{xy} &= 0 \\ (y+z)(1-\frac{1}{xy}) &= 0 \\ *y+z=0 \\ y+z &= 0 \\ y &= -z \\ \frac{2x-4}{3} &= -(2x-6) \\ \frac{2(x-2)}{3} &= -2(x-3) \\ \frac{x-2}{3} &= -x+3 \\ x-2 &= -3x+9 \\ 4x &= 11 \\ x &= \frac{11}{4} \\ *1-\frac{1}{xy}=0 \\ 1-\frac{1}{xy} &= 0 \\ xy-1 &= 0 \\ xy &= 0 \\ (\frac{2x-4}{3})(2x-6) &= 0 \\ \frac{4x^2-20x+24}{3} &= 0 \\ 4x^2-20x+24 &= 0 \\ x^2-5x+6 &= 0 \\ (x-2)(x-3) &= 0 \\ x = 2 &\text{ (TM) atau } x = 3 \text{ (TM) } \\ \end{align} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 5ayuhp96j4sko38284wjw3jz7m3lyb1 117496 117494 2026-07-13T05:12:41Z Akuindo 8654 117496 wikitext text/x-wiki contoh soal <ol start=1> <li>Berapa hasil dari <math>\sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Misalkan 2020 = p} \\ \sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64} &= \sqrt{(2020-5) \cdot (2020-3) \cdot (2020+3) \cdot (2020+5) + 64} \\ &= \sqrt{(p-5) \cdot (p-3) \cdot (p+3) \cdot (p+5) + 64} \\ &= \sqrt{(p-5) \cdot (p+5) \cdot (p-3) \cdot (p+3) + 64} \\ &= \sqrt{(p^2-25) \cdot (p^2-9) + 64} \\ &= \sqrt{p^4-34p^2+ 225 + 64} \\ &= \sqrt{p^4-34p^2+ 289} \\ &= \sqrt{(p^2-17)^2} \\ &= p^2-17 \\ &= 2020^2-17 \\ &= (2000+20)^2-17 \\ &= 4.000.000+80.000+400-17 \\ &= 4.080.383 \\ \end{align} </math> </div></div> <ol start=2> <li>Berapa nilai x dari <math>\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} = \frac{9}{10}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} &= \frac{9}{10} \\ \text{misalkan untuk } \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} = p \\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} &= p \\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}} &= p^2 \\ x^2-x-p &= p^2 \\ x^2-2x+1+x-1 &= p^2+p \\ (x-1)^2+(x-1) &= p^2+p \\ x-1 &= p \\ \text{misalkan untuk } \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}} &= q^3 \\ x^2 q &= q^3 \\ x^2 &= q^2 \\ x &= q \\ \frac{x-1}{x} &= \frac{9}{10} \\ x &= 10 \\ \end{align} </math> </div></div> <ol start=3> <li>Berapa nilai x dari <math>(\frac{x}{x+10})^{x+10}=\frac{1}{1024}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{x+10}{x})^{-(x+10)} &= (1024)^{-1} \\ (\frac{x+10}{x})^{x+10} &= 1024 \\ (\frac{x+10}{x})^{x+10} &= 2^{10} \\ (\frac{x+10}{x})^{\frac{x+10}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (\frac{1}{2})^{-1} \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (1+(-\frac{1}{2}))^{(1+(-\frac{2}{1}))} \\ \frac{10}{x} &= -\frac{1}{2} \\ x &= -20 \\ \end{align} </math> </div></div> <ol start=4> <li>Berapa nilai x dari <math>x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{2}+\sqrt{x+\frac{1}{4}} &= (\sqrt{x+\frac{1}{4}})^2+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+(\frac{1}{2})^2 \\ &= (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 \\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}} &= 4 \\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2} &= 4 \\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 4 \\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 &= 4 \\ \sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 2 \\ \sqrt{x+\frac{1}{4}} &= \frac{3}{2} \\ x+\frac{1}{4} &= \frac{9}{4} \\ x &= 2 \\ \end{align} </math> </div></div> <ol start=5> <li>Berapa nilai x dari <math>\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^3}{\sqrt{8-x^2}}+x^2-8 &= 0 \\ \frac{x^3}{\sqrt{8-x^2}} &= 8-x^2 \\ x^3 &= (8-x^2)^{\frac{3}{2}} \\ x &= (8-x^2)^{\frac{1}{2}} \\ x^2 &= 8-x^2 \\ 2x^2-8 &= 0 \\ x^2-4 &= 0 \\ (x-2)(x+2) &= 0 \\ \text{membuktikan } \\ x=2 \text{ maka hasilnya 0 } \\ x=-2 \text{ maka hasilnya -8 } \\ \text{jadi } x=2 \\ \end{align} </math> </div></div> <ol start=6> <li>Berapa nilai x dari <math>\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} = 5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} &= 5 \\ \frac{x^{50}+x^{60}+x^{70}}{31}} &= 5^5 \\ x^{50}+x^{60}+x^{70} &= 5^5 \cdot 31 \\ x^{50}(1+x^{10}+x^{20}) &= 5^5 \cdot 31 \\ (x^{10}^5)(1+x^{10}+(x^{10}^2) &= 5^5 \cdot 31 \\ \text{ misalkan } x^{10} = a \\ a^5(1+a+a^2) &= 5^5 \cdot 31 \\ a &= 5 \\ x^{10} &= 5 \\ x &= ^5 log 10 \\ \end{align} </math> </div></div> <ol start=7> <li>Berapa nilai x dari <math>\sqrt{3x+5+\sqrt{4x+5}} = x</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{3x+5+\sqrt{4x+5}} &= x \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \text{misalkan } \sqrt{4x+5}=y \text{ dan } 4x+5=y^2 \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \sqrt{y^2+y-x} &= x \\ y^2+y &= x^2+x \\ y=x \\ 4x+5 &= y^2 \\ 4x+5 &= x^2 \\ x^2-4x-5 &= 0 \\ (x-5)(x+1) &= 0 \\ x=5 &\text{ atau } x=-1 \text{ (TM) } \\ \end{align} </math> </div></div> <ol start=8> <li>Berapa nilai x dari <math>\sqrt{1+\sqrt{1+x}} = \sqrt[3]{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{1+\sqrt{1+x}} &= \sqrt[3]{x} \\ \sqrt[3]{x} &= n \\ x &= n^3 \\ \sqrt{1+\sqrt{1+n^3}} &= n \\ 1+\sqrt{1+n^3} &= n^2 \\ \sqrt{1+n^3} &= n^2-1 \\ 1+n^3 &= n^4-2n^2+1 \\ n^4-n^3-2n^2 &= 0 \\ n^2(n^2-n-2) &= 0 \\ n^2(n-2)(n+1) &= 0 \\ n=0, n=2 \text{ atau } n=-1 \\ n &= 0 \\ x &= 0^3 \\ &= 0 \\ n &= 2 \\ x &= 2^3 \\ &= 8 \\ n &= -1 \\ x &= (-1)^3 \\ &= -1 \\ \text{yang paling mungkin untuk nilai x adalah } 8 \\ \end{align} </math> </div></div> <ol start=9> <li>Berapa nilai x dari <math>\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} &= \frac{\sqrt{1+x}}{\sqrt{x}} \\ \sqrt{x}(\sqrt{1+x}+\sqrt{x}) &= (\sqrt{1+x}-\sqrt{x})\sqrt{1+x} \\ \sqrt{x(1+x)}+x &= 1+x-\sqrt{x(1+x)} \\ 2\sqrt{x(1+x)} &= 1 \\ \sqrt{x(1+x)} &= \frac{1}{2} \\ x(1+x) &= \frac{1}{4} \\ x^2+x &= \frac{1}{4} \\ 4x^2+4x &= 1 \\ 4x^2+4x-1 &= 0 \\ x &= \frac{-4 \pm \sqrt{4^2-4(4)(-1)}}{2(4)} \\ &= \frac{-4 \pm \sqrt{32}}{8} \\ &= \frac{-4 \pm 4\sqrt{2}}{8} \\ &= \frac{-1 \pm \sqrt{2}}{2} \\ \text{karena akar x harus minimal nol jadi } x = \frac{-1+\sqrt{2}}{2} \\ \end{align} </math> </div></div> <ol start=10> <li>Berapa nilai x dari <math>\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x-\sqrt{x+1}}{x+\sqrt{x+1}} &= \frac{11}{19} \\ \text{misalkan } \sqrt{x+1}=y \text{ dan } x=y^2-1 \\ \frac{y^2-1-y}{y^2-1+y} &= \frac{11}{19} \\ 19(y^2-y-1) &= 11(y^2+y-1) \\ 19y^2-19y-19 &= 11y^2+11y-11 \\ 8y^2-30y-8 &= 0 \\ 4y^2-15y-4 &= 0 \\ (4y+1)(y-4) &= 0 \\ y=-\frac{1}{4} \text{ (TM) atau } & y=4 \\ x &= 4^2-1 \\ &= 15 \\ \end{align} </math> </div></div> <ol start=11> <li>Berapa nilai x dari <math>\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} &= 98 \\ \text{misalkan } \sqrt{x^2-1}=y \\ \frac{x+y}{x-y}+\frac{x-y}{x+y} &= 98 \\ \frac{(x+y)^2+(x-y)^2}{(x-y)(x+y)} &= 98 \\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} &= 98 \\ \frac{2(x^2+y^2)}{x^2-y^2} &= 98 \\ \frac{x^2+y^2}{x^2-y^2} &= 49 \\ x^2+y^2 &= 49(x^2-y^2) \\ x^2+y^2 &= 49x^2-49y^2 \\ 48x^2 &= 50y^2 \\ 24x^2 &= 25y^2 \\ 24x^2 &= 25(\sqrt{x^2-1})^2 \\ 24x^2 &= 25(x^2-1) \\ 24x^2 &= 25x^2-25 \\ x^2 &= 25 \\ x &= \pm 5 \\ \end{align} </math> </div></div> <ol start=12> <li>Berapa nilai x dari <math>\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \sqrt{x}=y \text{ dan } x=y^2 \\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} &= \frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\sqrt{\frac{y^2}{y^2+y}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\frac{y}{\sqrt{y^2+y}} \\ y^2+y-\sqrt{(y^2+y)(y^2-y)} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^4-y^2} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^2(y^2-1)} &= \frac{5}{4}y \\ y(y+1)-y\sqrt{y^2-1} &= \frac{5}{4}y \\ y+1-\sqrt{y^2-1} &= \frac{5}{4} \\ -\sqrt{y^2-1} &= \frac{1}{4}-y \\ y^2-1 &= (\frac{1}{4}-y)^2 \\ y^2-1 &= \frac{1}{16}-\frac{1}{2}y+y^2 \\ -1 &= \frac{1}{16}-\frac{1}{2}y \\ \frac{1}{2}y &= \frac{1}{16}+1 \\ \frac{1}{2}y &= \frac{17}{16} \\ y &= \frac{17}{8} \\ x &= (\frac{17}{8})^2 \\ &= \frac{289}{64} \\ \end{align} </math> </div></div> <ol start=13> <li>Berapa nilai x dari <math>\sqrt[4]{62+x}+\sqrt[4]{275-x}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ misalkan } \sqrt[4]{62+x}=a, 62+x=a^4, \sqrt[4]{275-x}=b \text{ dan } 275-x=b^4 \\ a+b &= 7 \\ (a+b)^2 &= 49 \\ a^2+b^2+2ab &= 49 \\ a^2+b^2 &= 49-2ab \\ a^4+b^4 &= 62+x+275-x \\ (a^2+b^2)^2-2(ab)^2 &= 337 \\ (49-2ab)^2-2(ab)^2 &= 337 \\ 2401-196ab+4(ab)^2-2(ab)^2 &= 337 \\ 2(ab)^2-196ab+2064 &= 0 \\ (ab)^2-98ab+1032 &= 0 \\ (ab-12)(ab-86) &= 0 \\ ab = 12 \text{ atau } & ab = 86 \text{ (TM) karena hasil kali maksimum yaitu 12 } \\ ab =12 \text{ dan } a+b=7 \\ a+b &= 7 \\ b &= 7-a \\ ab &= 12 \\ a(7-a) &= 12 \\ -a^2+7a &= 12 \\ a^2-7a+12 &= 0 \\ (a-3)(a-4) &= 0 \\ a=3 \text{ atau } & a=4 \\ a=3, b=4 \\ 62+x &= a^4 \\ 62+x &= (3)^4 \\ 62+x &= 81 \\ x &= 19 \\ a=4, b=3 \\ 62+x &= a^4 \\ 62+x &= (4)^4 \\ 62+x &= 256 \\ x &= 194 \\ \end{align} </math> </div></div> <ol start=14> <li>Berapa nilai x dari <math>\sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2} &= 7 \\ (\sqrt[3]{8+x})^2-\sqrt[3]{8+x} \sqrt[3]{27-x}+(\sqrt[3]{27-x})^2 &= 7 \\ \text{misalkan } \sqrt[3]{8+x}=a, 8+x=a^3, \sqrt[3]{27-x}=b \text{ dan } 27-x=b^3 \\ a^2-ab+b^2 &= 7 \\ a^3+b^3 &= 8+x+27-x \\ &= 35 \\ a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ 35 &= (a+b)(7) \\ a+b &= 5 \\ b &= 5-a \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ 5^3 &= 35+3ab(5) \\ 125 &= 35+15ab \\ 80 &= 15ab \\ ab &= 6 \\ a(5-a) &= 6 \\ 5a-a^2 &= 6 \\ a^2-5a+6 &= 6 \\ (a-2)(a-3) &= 6 \\ a=2 &\text{ atau } a=3 \\ a=2, b=3 \text{ dan } a=3,b=2 \\ 8+x &= a^3 \\ &= 2^3 \\ &= 8 \\ x &= 0 \\ 8+x &= a^3 \\ &= 3^3 \\ &= 27 \\ x &= 19 \\ \end{align} </math> </div></div> <ol start=15> <li>Berapa nilai x dari <math>3^x+5^x-9^x+15^x-25^x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3^x+5^x-9^x+15^x-25^x &= 1 \\ 3^x+5^x-(3^2)^x+(3 \cdot 5)^x-(5^2)^x &= 1 \\ 3^x+5^x-(3^x)^2+(3^x \cdot 5^x)-(5^x)^2 &= 1 \\ \text{misalkan } 3^x=a \text{ dan } 5^x=b \\ a+b-a^2+ab-b^2 &= 1 \\ a^2-ab+b^2-a-b+1 &= 0 \\ 2a^2-2ab+2b^2-2a-2b+2 &= 0 \\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1 &= 0 \\ (a-b)^2+(a-1)^2+(b-1)^2 &= 0 \\ a-b=0; a-1=0; b-1 &= 0 \\ a=b &= 1 \\ 3^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> <ol start=16> <li>Berapa nilai x dari <math>^6log x^2+^{6x}log \frac{6}{x}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ \text{misalkan } 6x=a \text{ maka } x=\frac{a}{6} \\ ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ ^6log (\frac{a}{6})^2+^{6 \frac{a}{6}}log \frac{6}{\frac{a}{6}} &= 1 \\ ^6log \frac{a^2}{6^2}+^alog \frac{6^2}{a} &= 1 \\ ^6log a^2-^6log 6^2+^alog 6^2-^alog a &= 1 \\ 2 ^6log a-2 ^6log 6+2 ^alog 6-^alog a &= 1 \\ 2 ^6log a-2+2 \frac{1}{^6log a}-1 &= 1 \\ 2 ^6log a+2 \frac{1}{^6log a}-4 &= 0 \\ 2 ^6log^2 a-4 ^6log a+2 &= 0 \\ ^6log^2 a-2 ^6log a+1 &= 0 \\ (^6log a-1)^2 &= 0 \\ ^6log a &= 1 \\ a &= 6 \\ x &= \frac{a}{6} \\ &= \frac{6}{6} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=17> <li>Berapa nilai x dari (x+500)³+x=20?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+500)^3+x &= 20 \\ \text{misalkan } a=x+500 \text{ maka } x=a-500 \\ a^3+a-500 &= 20 \\ a^3+a &= 520 \\ a(a^2+1) &= 8 \cdot 65 \\ a(a^2+1) &= 8(64+1) \\ a(a^2+1) &= 8(8^2+1) \\ a &= 8 \\ x &= 8-500 \\ &= -492 \\ \end{align} </math> </div></div> <ol start=18> <li>Berapa nilai x dari <math>\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2} &= 0 \\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}} &= \frac{1}{2} \\ \frac{x^n+4^n}{x^n+16^n} &= (\frac{1}{2})^n \\ \frac{x^n+4^n}{x^n+16^n} &= \frac{1}{2^n} \\ 2^n(x^n+4^n) &= x^n+16^n \\ 2^n(x^n+2^{2n}) &= x^n+2^{4n} \\ 2^n \cdot x^n+2^{3n} &= x^n+2^{4n} \\ 2^n \cdot x^n-x^n &= 2^{4n}-2^{3n} \\ x^n(2^n-1) &= 2^{3n}(2^n-1) \\ x^n &= 2^{3n} \\ x^n &= (2^3)^n \\ x^n &= 8^n \\ x &= 8 \\ \end{align} </math> </div></div> <ol start=19> <li>Berapa hasil dari <math>\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ x &= \frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5 \cdot 6}+\sqrt{5 \cdot 5}+\sqrt{6 \cdot 4}+\sqrt{5 \cdot 4}}{\sqrt{5 \cdot 4}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5} \cdot \sqrt{6}+\sqrt{5} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{4}}{2 \cdot \sqrt{5}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{6} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{5}+\sqrt{5} \cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}} \\ &= \frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ &= \frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ \frac{1}{x} &= \frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}} \\ &= \frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ &= \frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1} \\ &= \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5} \\ &= \sqrt{6}-\sqrt{4} \\ &= \sqrt{6}-2 \\ x &= \frac{1}{\sqrt{6}-2} \\ &= \frac{\sqrt{6}+2}{6-4} \\ &= \frac{\sqrt{6}+2}{2} \\ &= 1+\frac{\sqrt{6}}{2} \\ \end{align} </math> </div></div> <ol start=20> <li>Berapa hasil dari <math>(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5 \\ \text{misalkan } x=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ x &= \frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ &= \frac{\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}} \\ &= \frac{\sqrt{3}+1}{4} \\ 4x &= \sqrt{3}+1 \\ 4x-1 &= \sqrt{3} \\ (4x-1)^2 &= 3 \\ 16x^2-8x+1 &= 3 \\ 16x^2 &= 8x+2 \\ 8x^2 &= 4x+1 \\ x^2 &= \frac{4x+1}{8} \\ *cara 1 \\ x^3 &= x \cdot x^2 \\ &= x(\frac{4x+1}{8}) \\ &= \frac{4x^2+x}{8} \\ &= \frac{4x^2}{8}+\frac{x}{8} \\ &= \frac{4(\frac{4x+1}{8})}{8}+\frac{x}{8} \\ &= \frac{16x+4}{64}+\frac{x}{8} \\ &= \frac{4x+1}{16}+\frac{x}{8} \\ &= \frac{4x+1+2x}{16} \\ &= \frac{6x+1}{16} \\ x^5 &= x^2 \cdot x^3 \\ &= (\frac{4x+1}{8})(\frac{6x+1}{16}) \\ &= \frac{24x^2+10x+1}{128} \\ &= \frac{24x^2}{128}+\frac{10x+1}{128} \\ &= \frac{24(\frac{4x+1}{8})}{128}+\frac{10x+1}{128} \\ &= \frac{96x+24}{1024}+\frac{10x+1}{128} \\ &= \frac{96x+24+80x+8}{1024} \\ &= \frac{176x+32}{1024} \\ &= \frac{176x}{1024}+\frac{32}{1024} \\ &= \frac{176}{1024}(\frac{\sqrt{3}+1}{4})+\frac{32}{1024} \\ &= \frac{44(\sqrt{3}+1)}{1024}+\frac{32}{1024} \\ &= \frac{44\sqrt{3}+44}{1024}+\frac{32}{1024} \\ &= \frac{76+44\sqrt{3}}{1024} \\ &= \frac{19+11\sqrt{3}}{256} \\ *cara 2 \\ x^4 &= (x^2)^2 \\ &= (\frac{4x+1}{8})^2 \\ &= \frac{16x^2+8x+1}{64} \\ &= \frac{16x^2}{64}+\frac{8x}{64}+\frac{1}{64} \\ &= \frac{x^2}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{\frac{4x+1}{8}}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{4x}{32}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{8}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{4}+\frac{3}{64} \\ x^5 &= x \cdot x^4 \\ &= (\frac{\sqrt{3}+1}{4})(\frac{x}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\frac{\sqrt{3}+1}{4}}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\sqrt{3}+1}{16}+\frac{3}{64}) \\ &= \frac{(\sqrt{3}+1)^2}{64}+(\frac{\sqrt{3}+1}{4})\frac{3}{64} \\ &= \frac{3+2\sqrt{3}+1}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{4+2\sqrt{3}}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{16+8\sqrt{3}}{256}+\frac{3\sqrt{3}+3}{256} \\ &= \frac{19+11\sqrt{3}}{256} \\ \end{align} </math> </div></div> <ol start=21> <li>Berapa hasil dari <math>\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4} &= \frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots) \\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots &= \frac{1}{1-\frac{1}{4}} \\ &= \frac{4}{3} \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{4}{3}) \\ &= \frac{1}{4}+\frac{16}{3} \\ &= \frac{67}{12} \\ x &= \frac{67}{9} \\ \end{align} </math> </div></div> <ol start=22> <li>Berapa nilai y-x jika <math>\frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} = \frac{x}{y}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} &= \frac{x}{y} \\ \frac{\frac{106 \times 107}{2}}{\frac{106}{2}(4+109)} &= \frac{x}{y} \\ \frac{53 \times 107}{53 \times 113} &= \frac{x}{y} \\ y-x &= 113-107 = 6 \\ \end{align} </math> </div></div> <ol start=23> <li>Berapa angka satuan dari hasil 17²⁰²⁴?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan angka satuannya} \\ 17^1 &= 7 \\ 17^2 &= 9 \\ 17^3 &= 3 \\ 17^4 &= 1 \\ 17^5 &= 7 \\ 17^6 &= 9 \\ 17^7 &= 3 \\ 17^8 &= 1 \\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1} \end{align} </math> </div></div> <ol start=24> <li>Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ 1! + 2! + 3! + 4! + \dots + 2024! &= 1 + (1x2) + (1x2x3) + (1x2x3x4) + \dots + 2024! \\ &= 1 + 2 + 6 + 24 + 120 + 720 + \dots + 2024! \\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka } 1+2+6+24 = 33 \text{ jadi angka satuannya adalah } 3 \end{align} </math> </div></div> <ol start=25> <li>Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ \frac{1! + 2! + 3! + 4! + \dots + 2024!}{12} &= \frac{1 + 1x2 + 1x2x3 + 1x2x3x4 + \dots + 2024!}{12} \\ &= \frac{1 + 2 + 6 + 24 + \dots + 2024!}{12} \\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi } 1+2+6 = 9 \end{align} </math> </div></div> <ol start=26> <li>Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{penjumlahan} \\ 1+1+1+1+ \dots \text{ (sebanyak 88 buah) }+x+y &= A \\ 88+x+y &= A \\ \text{perkalian} \\ 1 \times 1 \times 1 \times \dots \text{ (sebanyak 88 buah) }\times x \times y &= A \\ x \times y &= A \\ 88+x+y &= xy \\ xy-y &= 88+x \\ y(x-1) &= 88+x \\ y &= \frac{88+x}{x-1} \\ \text{uji selidiki untuk x=2} \\ y &= \frac{88+2}{2-1} \\ &= 90 \\ \text{buktikan} \\ 88+x+y &= xy \\ 88+2+90 &= 2(90) \\ 180 &= 180 \\ \text{terbukti} \\ \text{nilai A adalah } 180 \\ \end{align} </math> </div></div> <ol start=27> <li>Berapakah nilai x, y dan z dari <math>x+y-z=1, x^2+y^2-z^2=-5 \text{ dan } x^3+y^3-z^3=-53</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+y-z &= 1 \\ x+y &= z+1 \\ x^2+2xy+y^2 &= z^2+2z+1 \\ x^2+y^2-z^2 &= 2z+1-2xy \\ -5 &= 2z+1-2xy \\ 2xy &= 2z+6 \\ xy &= z+3 \\ x^2+y^2-z^2 &= -5 \\ x^2+y^2 &= z^2-5 \\ x^3+y^3-z^3 &= -53 \\ (x+y)(x^2-xy+y^2)-z^3+53 &= 0 \\ (x+y)(x^2+y^2-xy)-z^3+53 &= 0 \\ (z+1)(z^2-5-(z+3))-z^3+53 &= 0 \\ (z+1)(z^2-z-8)-z^3+53 &= 0 \\ z^3-z^2-8z+z^2-z-8-z^3+53 &= 0 \\ -9z+45 &= 0 \\ -9z &= -45 \\ z &= 5 \\ x+y &= 5+1 \\ x+y &= 6 \\ x &= 6-y \\ xy &= 5+3 \\ xy &= 8 \\ (6-y)y &= 8 \\ 6y-y^2 &= 8 \\ y^2-6y+8 &= 0 \\ (y-4)(y-2) &= 0 \\ y=4 \text{ atau } y=2 \\ \text{jika } y=4 \\ x+y &= z+1 \\ x+4 &= 5+1 \\ x &= 2 \\ \text{jika } y=2 \\ x+y &= z+1 \\ x+2 &= 5+1 \\ x &= 4 \\ \end{align} </math> </div></div> <ol start=28> <li>Berapakah nilai titik koordinat (x,y) dari <math>\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}</math> dan <math>\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x+y}+\sqrt{x-y} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+y}-\sqrt{x-y} &= \sqrt{\frac{52y}{3x}} \\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y}) &= \sqrt{\frac{432x}{13y}} \cdot \sqrt{\frac{52y}{3x}} \\ x+y-x+y &= \sqrt{\frac{432x \cdot 52y}{13y \cdot 3x}} \\ 2y &= \sqrt{144 \cdot 4} \\ 2y &= \sqrt{576} \\ 2y &= 24 \\ y &= 12 \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13(12)}} \\ x+12+x-12+2 \cdot \sqrt{x+12} \cdot \sqrt{x-12} &= \frac{36x}{13} \\ 2x+2 \sqrt{x^2-144} &= \frac{36x}{13} \\ 2(x+\sqrt{x^2-144}) &= \frac{36x}{13} \\ x+\sqrt{x^2-144} &= \frac{18x}{13} \\ \sqrt{x^2-144} &= \frac{5x}{13} \\ x^2-144 &= \frac{25x^2}{169} \\ \frac{144x^2}{169}-144 &= 0 \\ \frac{x^2}{169}-1 &= 0 \\ x^2-169 &= 0 \\ (x-13)(x+13) &= 0 \\ x_1=13 &\text{ atau } x_2=-13 \text{ (TM) karena } x>y \\ \end{align} </math> jadi titik koordinat (13,12) </div></div> <ol start=29> <li>Berapakah nilai dari <math>x^2-7x</math> jika <math>(x-2)^2+\frac{1}{(x-2)^2} = 11</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x-2)^2+\frac{1}{(x-2)^2} &= 11 \\ (x-2)^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2)^2} &= 11-2 \\ (x-2-\frac{1}{x-2})^2 &= 9 \\ x-2-\frac{1}{x-2} &= 3 \\ (x-2)^2-1 &= 3(x-2) \\ x^2-4x+4-1 &= 3x-6 \\ x^2-7x &= -9 \\ \end{align} </math> </div></div> <ol start=30> <li>Berapakah nilai dari <math>\frac{(x+y)^2(x+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)}</math> jika xy+yz+xz=1?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy+yz+xz &= 1 \\ x^2+xy+yz+xz &= x^2+1 \\ x(x+y)+z(x+y) &= x^2+1 \\ (x+y)(x+z) &= x^2+1 \\ \text{dengan pola yang sama } \\ (y+x)(y+z) &= y^2+1 \\ (x+z)(y+z) &= z^2+1 \\ \frac{(x+y)^2(y+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)} &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)} \\ &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)^2(y+z)^2(x+z)^2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=31> <li>Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} w+5 &= w+x+y+z+5 \\ x+4 &= w+x+y+z+5 \\ y+3 &= w+x+y+z+5 \\ z+2 &= w+x+y+z+5 \\ \text{jumlahkan keempat persamaan } \\ w+x+y+z+14 &= 4(w+x+y+z+5) \\ w+x+y+z+14 &= 4(w+x+y+z)+20 \\ 3(w+x+y+z) &= -6 \\ w+x+y+z &= -2 \\ \end{align} </math> </div></div> <ol start=32> <li>Berapakah nilai dari <math>\frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2}</math> jika <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3</math> dan x+y+z=xyz?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2} &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2} \\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) \\ 3^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz}) \\ 9 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz}) \\ &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2 \\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} &= 7 \\ \end{align} </math> </div></div> <ol start=33> <li>Berapakah nilai dari <math>\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}</math> jika <math>x^2+y^2+z^2 = -2(ab+bc+ac)</math>?>/li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+z^2 &= -2(xy+yz+xz) \\ x^2+y^2+z^2+2(xy+yz+xz) &= 0 \\ (x+y+z)^2 &= 0 \\ x+y+z &= 0 \\ x+y &= -z \\ x+z &= -y \\ y+z &= -x \\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z} &= \frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x} \\ &= -2-(-5)-(-7) \\ &= 10 \\ \end{align} </math> </div></div> <ol start=34> <li>Berapakah nilai dari <math>\frac{20xyz}{xy+yz+xz}</math> jika <math>16^x = 256^y = 625^z = 40</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 16^x = 256^y = 625^z &= 40 \\ 2^{4x} = 4^{4y} = 5^{4z} &= 40 \\ 2^{4x} &= 40 \\ 2 &= 40^{\frac{1}{4x}} \\ 4^{4y} &= 40 \\ 4 &= 40^{\frac{1}{4y}} \\ 5^{4z} &= 40 \\ 5 &= 40^{\frac{1}{4z}} \\ 2 \cdot 4 \cdot 5 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z}} \\ 1 &= \frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z} \\ 4 &= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{20xyz}{xy+yz+xz} &= 20 \cdot \frac{xyz}{xy+yz+xz} \\ &= 20 \cdot (\frac{xy+yz+xz}{xyz})^{-1} \\ &= 20 \cdot (\frac{1}{z} + \frac{1}{x} + \frac{1}{y})^{-1} \\ &= 20 \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{-1} \\ &= 20 \cdot (4)^{-1} \\ &= 20 \cdot \frac{1}{4} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=35> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+3x^2+1}</math> jika <math>6x^2+25x+6=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 6x^2+25x+6 &= 0 \\ 6x+25+\frac{6}{x} &= 0 \\ 6(x+\frac{1}{x}) &= -25 \\ x+\frac{1}{x} &= \frac{-25}{6} \\ (c+\frac{1}{x})^2 &= (\frac{-25}{6})^2 \\ x^2+2+\frac{1}{x^2} &= \frac{625}{36} \\ x^2+\frac{1}{x^2} &= \frac{625}{36}-2 \\ x^2+\frac{1}{x^2} &= \frac{553}{36} \\ \frac{x^2}{x^4+3x^2+1} &= \frac{1}{x^2+3+\frac{1}{x^2}} \\ &= \frac{1}{a^2+\frac{1}{x^2}+3} \\ &= \frac{1}{\frac{553}{36}+3} \\ &= \frac{1}{\frac{661}{36}} \\ &= \frac{36}{661} \\ \end{align} </math> </div></div> <ol start=36> <li>Berapakah nilai dari <math>\frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5} &= \frac{(9+2\sqrt{20})^{1013}}{((2)^3+3(2)^2(\sqrt{5})+3(2)(\sqrt{5})^2+(\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{((2+\sqrt{5})^2)^{1013}}{((2+\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{(2+\sqrt{5})^{2026}}{(2+\sqrt{5})^{2025}}+6-\sqrt{5} \\ &= 2+\sqrt{5}+6-\sqrt{5} \\ &= 8 \\ \end{align} </math> </div></div> <ol start=37> <li>Berapakah nilai dari <math>27x^3+\frac{8}{x^3}</math> jika <math>3x+\frac{2}{x}=6</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3x+\frac{2}{x} &= 6 \\ (3x+\frac{2}{x})^3 &= 6^3 \\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3} &= 216 \\ 27x^3+18(6)+\frac{8}{x^3} &= 216 \\ 27x^3+108+\frac{8}{x^3} &= 216 \\ 27x^3+\frac{8}{x^3} &= 108 \\ \end{align} </math> </div></div> <ol start=38> <li>Berapakah nilai dari <math>x^6+\frac{8}{x^3}</math> jika <math>x^3+\frac{1}{x^3}=8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3+\frac{1}{x^3} &= 8 \\ x^3 &= 8-\frac{1}{x^3} \\ x^6 &= 8x^3-1 \\ x^6+\frac{8}{x^3} &= 8x^3-1+\frac{8}{x^3} \\ &= 8x^3+\frac{8}{x^3}-1 \\ &= 8(x^3+\frac{1}{x^3})-1 \\ &= 8(8)-1 \\ &= 63 \\ \end{align} </math> </div></div> <ol start=39> <li>Berapakah nilai dari <math>4x+\frac{25}{x}</math> jika <math>2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2\sqrt{x}+\frac{5}{\sqrt{x}} &= 4x-\frac{25}{x} \\ 2\sqrt{x}+\frac{5}{\sqrt{x}} &= (2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}}) \\ 1 &= 2\sqrt{x}-\frac{5}{\sqrt{x}} \\ 1^2 &= (2\sqrt{x}-\frac{5}{\sqrt{x}})^2 \\ 1 &= 4x-20+\frac{25}{x} \\ 4x+\frac{25}{x} &= 21 \\ \end{align} </math> </div></div> <ol start=40> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2-x+1}{x^2+x+1} &= \frac{5}{6} \\ \frac{x^2+1-x}{x^2+1+x} &= \frac{5}{6} \\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} &= \frac{5}{6} \\ \text{ misalkan } x+\frac{1}{x} &= y \\ \frac{y-1}{y+1} &= \frac{5}{6} \\ 6(y-1) &= 5(y+1) \\ 6y-6 &= 5y+5 \\ y &= 11 \\ x+\frac{1}{x} &= 11 \\ \end{align} </math> </div></div> <ol start=41> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt{x}+x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x}+x &= 1 \\ x-1 &= -\sqrt{x} \\ (x-1)^2 &= (-\sqrt{x})^2 \\ x^2-2x+1 &= x \\ x^2-3x+1 &= 0 \\ x-3+\frac{1}{x} &= 0 \\ x+\frac{1}{x} &= 3 \\ \end{align} </math> </div></div> <ol start=42> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt[3]{x}-\sqrt[3]{x-36}=3</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x}-\sqrt[3]{x-36} &= 3 \\ (\sqrt[3]{x}-\sqrt[3]{x-36})^3 &= 3^3 \\ x-(x-36)-3 \sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36}) &= 27 \\ 36-3 \sqrt[3]{x(x-36)}3 &= 27 \\ -9 \sqrt[3]{x(x-36)} &= -9 \\ \sqrt[3]{x(x-36)} &= 1 \\ x(x-36) &= 1 \\ x^2-36x-1 &= 0 \\ x-36-\frac{1}{x} &= 0 \\ x-\frac{1}{x} &= 36 \\ \end{align} </math> </div></div> <ol start=43> <li>Berapakah nilai dari <math>x+\frac{16}{x}</math> jika <math>x-3\sqrt{x}=4</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x-3\sqrt{x} &= 4 \\ x-4 &= 3\sqrt{x} \\ x^2-8x+16 &= 9x \\ x^2-17x+16 &= 0 \\ x-17+\frac{16}{x} &= 0 \\ x+\frac{16}{x} &= 17 \\ \end{align} </math> </div></div> <ol start=44> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+4}</math> jika <math>x^2-7x+2=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2-7x+2 &= 0 \\ x^2+2 &= 7x \\ x+\frac{2}{x} &= 7 \\ x^2+4+\frac{4}{x^2} &= 49 \\ x^2+\frac{4}{x^2} &= 45 \\ \frac{x^4+4}{x^2} &= 45 \\ \frac{x^2}{x^4+4} &= \frac{1}{45} \\ \end{align} </math> </div></div> <ol start=45> <li>Berapakah nilai dari <math>x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}</math> jika <math>x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}} &= 25 \\ x^{\frac{1}{2}}+x^{-\frac{1}{2}} &= 23 \\ x+2+x^{-1} &= 529 \\ x+x^{-1} &= 527 \\ x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+x^{-\frac{3}{4}} &= 110 \\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1} &= x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}} \\ &= 527+110 \\ &= 637 \\ \end{align} </math> </div></div> <ol start=46> <li>Berapakah nilai dari <math>\sqrt{8x^6+x^5+x^4+5x^3+1}</math> jika <math>\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5} &= 0 \\ \frac{x^2+x+1}{x^5} &= 0 \\ x^2+x+1 &= 0 \\ x^2+x+1 &= 0 \\ (x-1)(x^2+x+1) &= 0(x-1) \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ \sqrt{8x^6+x^5+x^4+5x^3+1} &= \sqrt{(2x^3)^2+x^3x^2+x^3x+5x^3+1} \\ &= \sqrt{(2(1))^2+(1)x^2+(1)x+5(1)+1} \\ &= \sqrt{(2)^2+x^2+x+5+1} \\ &= \sqrt{4+x^2+x+1+5} \\ &= \sqrt{4+0+5} \\ &= \sqrt{9} \\ &= 3 \\ \end{align} </math> </div></div> <ol start=47> <li>Berapakah nilai dari <math>f(1)+f(2)+f(3)+ \dots + f(99)</math> jika <math>f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{1}{\sqrt{x+1}+\sqrt{x}} \\ &= \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} \\ &= \sqrt{x+1}-\sqrt{x} \\ f(1)+f(2)+f(3)+ \dots + f(98)+f(99) &= \sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+ \cdot + \sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99} \\ &= \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+ \cdot + \sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ &= \sqrt{100}-\sqrt{1} \\ &= 10-1 \\ &= 9 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots + \frac{2024}{2025})</math> jika <math>h(x)=\frac{3}{3+9^x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} h(x) &= \frac{3}{3+9^x} \\ h(1-x) &= \frac{3}{3+9^{1-x}} \\ &= \frac{3}{3+\frac{9}{9^x}} \\ &= \frac{9^x}{3+9^x} \\ h(x)+h(1-x) &= \frac{3}{3+9^x}+\frac{9^x}{3+9^x} \\ &= \frac{3+9^x}{3+9^x} \\ &= 1 \\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots +(1-\frac{2}{2025})+(1-\frac{1}{2025})) \\ & 5(1+1+1+ \dots +1+1) \text{ sebanyak 1012 kali } \\ & 5(1012) \\ & 5060 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>\frac{7^{2025} - 7^{2023} + 432}{7^{2024} + 7^{2023} + 72}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72} &= \frac{7^{2023}7^{2}-7^{2023} + 48 \times 9}{7^{2023}7^1+7^{2023}+8 \times 9} \\ &= \frac{7^{2023}(7^{2}-1)+48 \times 9}{7^{2023}(7^1+1)+8 \times 9} \\ &= \frac{7^{2023}(49-1)+48 \times 9}{7^{2023}(7+1) + 8 \times 9} \\ &= \frac{7^{2023} \times 48+48 \times 9}{7^{2023} \times 8+8 \times 9} \\ &= \frac{48(7^{2023}+9)}{8(7^{2023}+9)} \\ &= \frac{48}{8} \\ &= 6 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>tan (x+\frac{\pi}{4})</math> jika <math>\frac{1}{cos x}-tan x = \frac{4}{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{cos x}-tan x &= \frac{4}{5} \\ sec x-tan x &= \frac{4}{5} \\ sec^2 x-tan^2 x &= 1 \\ (sec x+tan x)(sec x-tan x) &= 1 \\ (sec x+tan x)\frac{4}{5} &= 1 \\ sec x+tan x &= \frac{5}{4} \\ \text{kedua persamaan dengan cara metode eliminasi } \\ 2 tan x &= \frac{5}{4}-\frac{4}{5} \\ 2 tan x &= \frac{9}{20} \\ tan x &= \frac{9}{40} \\ tan (x+\frac{\pi}{4}) &= \frac{tan x+tan \frac{\pi}{4}}{1-tan x \cdot tan \frac{\pi}{4}} \\ &= \frac{\frac{9}{40}+1}{1-\frac{9}{40} \cdot 1} \\ &= \frac{\frac{49}{40}}{\frac{31}{40}} \\ &= \frac{49}{31} \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>sin^3 x+csc^3 x</math> jika <math>sin x-csc x = 8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ Dengan menggunakan rumus: } (a-b)^3 &= a^3-b^3-3ab(a-b) \\ (sin x-csc x)^3 &= sin^3 x-csc^3 x-3sin x csc x(sin x-csc x) \\ 8^3 &= sin^3 x-csc^3 x-3sin x (\frac{1}{sin x})(8) \\ 512 &= sin^3 x-csc^3 x-24 \\ sin^3 x-csc^3 x &= 512+24 \\ sin^3 x-csc^3 x &= 536 \\ \end{align} </math> </div></div> <ol start=52> <li>Berapakah nilai dari <math>(sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2</math> jika <math>\frac{1}{sin x}+\frac{1}{cos x} = 10</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{sin x}+\frac{1}{cos x} &= 10 \\ \frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} &= 100 \\ (sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2 &= sin^2 x+\frac{2sin x}{cos x}+\frac{1}{cos^2 x}+cos^2 x+\frac{2cos x}{sin x}+\frac{1}{sin^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2(sin^2 x+cos^2 x)}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+100 \\ &= 101 \\ \end{align} </math> </div></div> <ol start=53> <li>Berapakah nilai dari (x-1)⁶ jika <math>x=\frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin 80^\circ &= cos 10^\circ \\ sin 80^\circ-cos 10^\circ &= 0 \\ x &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+2 sin 20^\circ cos 20^\circ}{cos 10^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+4 sin 10^\circ cos 10^\circ cos 20^\circ}{cos 10^\circ} \\ &= 4 cos 55^\circ cos 25^\circ+4 sin 10^\circ cos 20^\circ \\ &= 2(2 cos 55^\circ cos 25^\circ+2 sin 10^\circ cos 20^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ+sin (-10)^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ-sin 10^\circ) \\ &= 2(cos 80^\circ-sin 10^\circ+cos 30^\circ+sin 30^\circ) \\ &= 2(cos 80^\circ-sin (90^\circ-80^\circ)+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(cos 80^\circ-cos 80^\circ+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= \sqrt{3}+1 \\ x-1 &= \sqrt{3} \\ (x-1)^6 &= (\sqrt{3})^6 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=54> <li>Berapakah nilai dari x jika <math>x=\frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ} \\ 2x sin 20^\circ-x sin 40 ^\circ &= x sin 20^\circ-x^2 sin 10^\circ \\ x^2 sin 10^\circ+x sin 20^\circ-x sin 40 ^\circ &= 0 \\ x(x sin 10^\circ+sin 20^\circ-sin 40 ^\circ) &= 0 \\ x = 0 &\text{ atau } x sin 10^\circ+sin 20^\circ-sin 40 ^\circ = 0 \\ x sin 10^\circ+sin 20^\circ-sin 40 ^\circ &= 0 \\ x sin 10^\circ &= sin 40 ^\circ-sin 20^\circ \\ x &= \frac{sin 40 ^\circ-sin 20^\circ}{sin 10^\circ} \\ &= \frac{2 cos 30 ^\circ sin 10^\circ}{sin 10^\circ} \\ &= 2 cos 30 ^\circ \\ &= \frac{2 \sqrt{3}}{2} \\ &= \sqrt{3} \\ \end{align} </math> </div></div> <ol start=55> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x^2}{x^2-16y^2} = \frac{625}{49}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2}{x^2-16y^2} &= \frac{625}{49} \\ \frac{x^2-16y^2}{x^2} &= \frac{49}{625} \text{ (terbalik posisinya)} \\ 1-\frac{16y^2}{x^2} &= \frac{49}{625} \\ \frac{16y^2}{x^2} &= 1 - \frac{49}{625} \\ (\frac{4y}{x})^2 &= \frac{576}{625} \\ (\frac{4y}{x})^2 &= (\frac{24}{25})^2 \\ \frac{4y}{x} &= \frac{24}{25} \\ \frac{y}{x} &= \frac{6}{25} \\ \frac{x}{y} &= \frac{25}{6} \\ \end{align} </math> </div></div> <ol start=56> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x}{y}+\frac{x+10y}{y+10x} = 2</math> serta bilangan real untuk x dan y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x}{y}+\frac{x+10y}{y+10x} &= 2 \\ \frac{x}{y}+\frac{\frac{x}{y}+10}{1+10\frac{x}{y}} &= 2 \\ \text{misalkan } \frac{x}{y} = a \\ a+\frac{a+10}{1+10a} &= 2 \\ a(1+10a)+a+10 &= 2(1+10a) \\ 10a^2+a+a+10 &= 2+20a \\ 10a^2-18a+8 &= 0 \\ 5a^2-9a+4 &= 0 \\ (5a-4)(a-1) &= 0 \\ a = \frac{4}{5} &\text{ atau } a = 1 \\ \text{jadi } \frac{x}{y} = {\frac{4}{5}, 1} \\ \end{align} </math> </div></div> <ol start=57> <li>Berapakah nilai dari xy jika <math>x^4+y^4+x^2y^2=15 \text{ dan } x^2+y^2+xy=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+xy &= 5 \\ x^2+y^2 &= 5-xy \\ x^4+y^4+x^2y^2 &= 15 \\ (x^2)^2+(y^2)^2+2x^2y^2-x^2y^2 &= 15 \\ (x^2+y^2)^2-x^2y^2 &= 15 \\ (5-xy)^2-x^2y^2 &= 15 \\ 25-10xy+x^2y^2-x^2y^2 &= 15 \\ 25-10xy &= 15 \\ 10xy &= 10 \\ xy &= 1 \\ \end{align} </math> </div></div> <ol start=58> <li>Berapakah nilai dari x jika <math>4^x = 63(4^3+1)(4^6+1)(4^{12}+1)+1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 4^x &= 63(4^3+1)(4^6+1)(4^{12}+1)+1 \\ 4^x-1 &= 63(4^3+1)(4^6+1)(4^{12}+1) \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{4^3-1} \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{63} \\ &= (4^3+1)(4^6+1)(4^{12}+1)(4^3-1) \\ &= (4^3-1)(4^3+1)(4^6+1)(4^{12}+1) \\ &= (4^6-1)(4^6+1)(4^{12}+1) \\ &= (4^{12}-1)(4^{12}+1) \\ &= 4^{24}-1 \\ 4^x &= 4^{24} \\ x &= 24 \\ \end{align} </math> </div></div> <ol start=59> <li>Berapakah nilai dari <math>\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}</math> jika <math>x=\sqrt{9+4\sqrt{5}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \sqrt{9+4\sqrt{5}} \\ x &= 2+\sqrt{5} \\ x^2 &= 9+4\sqrt{5} \\ x^2-4x &= 9+4\sqrt{5}-4(2+\sqrt{5}) \\ x^2-4x &= 1 \\ x^2 &= 4x+1 \\ x^3 &= x \cdot x^2 \\ &= x(4x+1) \\ &= 4x^2+x \\ &= 4(4x+1)+x \\ &= 16x+4+x \\ &= 17x+4 \\ x^4 &= x \cdot x^3 \\ &= x(17x+4) \\ &= 17x^2+4x \\ &= 17(4x+1)+4x \\ &= 68x+17+4x \\ &= 72x+17 \\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1} &= \frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1} \\ &= \frac{72x+17-85x-20+8x+2+5x+3}{2} \\ &= \frac{2}{2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=60> <li>Berapakah nilai dari <math>\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}</math> jika 2x-1=<math>\sqrt{61}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \frac{x^3+1}{x^5-x^4-x^3+x^2} = p \\ p &= \frac{x^3+1}{x^5-x^4-x^3+x^2} \\ &= \frac{x^3+1}{x^5-x^4-(x^3-x^2)} \\ &= \frac{x^3+1}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)} \\ &= \frac{x^2-x+1}{x^2(x-1)^2} \\ &= \frac{x^2-x+1}{(x(x-1))^2} \\ &= \frac{x(x-1)+1}{(x(x-1))^2} \\ 2x-1 &= \sqrt{61} \\ x &= \frac{\sqrt{61}+1}{2} \\ x-1 &= \frac{\sqrt{61}-1}{2} \\ x(x-1) &= (\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2}) \\ &= \frac{61-1}{4} \\ &= \frac{60}{4} \\ &= 15 \\ p &= \frac{x(x-1)+1}{(x(x-1))^2} \\ &= \frac{15+1}{15^2} \\ &= \frac{16}{15^2} \\ \sqrt{p} &= \sqrt{\frac{16}{15^2}} \\ &= \frac{4}{15} \\ \end{align} </math> </div></div> <ol start=61> <li>Berapakah nilai dari <math>(\frac{x-3}{x})^{25}</math> jika <math>x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\sqrt[5]{8}+\sqrt[5]{2} &= 1+\sqrt[5]{16}+\sqrt[5]{4} \\ x+(\sqrt[5]{2})^3+\sqrt[5]{2} &= 1+(\sqrt[5]{2})^4+(\sqrt[5]{2})^2 \\ x &= (\sqrt[5]{2})^4-(\sqrt[5]{2})^3+(\sqrt[5]{2})^2-\sqrt[5]{2}+1 \\ \text{misalkan } \sqrt[5]{2} = p \\ x &= p^4-p^3+p^2-p+1 \\ x &= \frac{p^5+1}{p+1} \\ (\frac{x-3}{x})^{25} &= (1-\frac{3}{x})^{25} \\ &= (1-\frac{3}{\frac{p^5+1}{p+1}})^{25} \\ &= (1-\frac{3(p+1)}{p^5+1})^{25} \\ &= (1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2})^5+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{2+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-3\sqrt[5]{2}-3)}{3})^{25} \\ &= (-\sqrt[5]{2})^{25} \\ &= (-2)^5 \\ &= -32 \\ \end{align} </math> </div></div> <ol start=62> <li>Berapakah nilai dari <math>x^{50}+x^{49}+x^{48}+x^{47}+x^{46}</math> jika <math>x^2+x+1=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+x+1 &= 0 \\ x^2+x &= -1 \\ \frac{x^3-1}{x-1} &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46} &= x^{48}(x^2+x+1)+x^{45}(x^2+x) \\ &= x^{48}(0)+(x^3)^{15}(-1) \\ &= 0+(1)^{15}(-1) \\ &= -1 \\ \end{align} </math> </div></div> <ol start=63> <li>Berapakah 2²⁴ dari <math>8^7+8^6+8^5+8^4+8^3+8^2+8+1=A</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= A \\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1) &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8 &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= 8A+1 \\ 8^8+A &= 8A+1 \\ 8^8 &= 7A+1 \\ (2^3)^8 &= 7A+1 \\ 2^{24} &= 7A+1 \\ \end{align} </math> </div></div> <ol start=64> <li>Berapakah nilai dari <math>x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1</math> jika <math>x+\frac{1}{x}=\sqrt{3}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{x} &= \sqrt{3} \\ x^2+2+\frac{1}{x^2} &= 3 \\ x^2-1+\frac{1}{x^2} &= 0 \\ x^2(x^2-1+\frac{1}{x^2}) &= x^2(0) \\ x^4-x^2+1 &= 0 \\ (x^2+1)(x^4-x^2+1) &= (x^2+1)0 \\ x^6-x^4+x^2+x^4-x^2+1 &= 0 \\ x^6+1 &= 0 \\ x^6 &= -1 \\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1 &= {x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1 \\ &= (-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2-1+1 \\ &= -1+1-1+1-1+1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=65> <li>Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari <math>\frac{c-b}{a-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= ax^2+bx+c \\ f(2) &= a(2)^2+2b+c = 4 \\ &= 4a+2b+c = 4 \\ f(7) &= a(7)^2+7b+c = 49 \\ &= 49a+7b+c = 49 \\ 49a+7b+c &= 49 \\ 4a+2b+c &= 4 \\ 45a+5b &= 45 \text{ (f(7) dikurangi f(2)) } \\ 9a+b &= 9 \\ b &= -9a+9 \\ 4a+2b+c &= 4 \\ 4a+2(-9a+9)+c &= 4 \\ 4a-18a+18+c &= 4 \\ -14a+18+c &= 4 \\ c &= 14a-14 \\ \frac{c-b}{a-1} &= \frac{14a-14-(-9a+9)}{a-1} \\ &= \frac{14(a-1)+9(a-1)}{a-1} \\ &= \frac{(14+9)(a-1)}{a-1} \\ &= 23 \\ \end{align} </math> </div></div> <ol start=66> <li>Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+y)^3 &= x^3+y^3+3xy(x+y) \\ 11^3 &= 242+3xy(11) \text{ (dibagi 11)} \\ 11^2 &= 22+3xy \\ 121 &= 22+3xy \\ 99 &= 3xy \\ xy &= 33 \\ (x-y)^2 &= x^2+y^2-2xy \\ &= ((x+y)^2-2xy)-2xy \\ &= (x+y)^2-4xy \\ &= 11^2-4(33) \\ &= 121-132 \\ &= -11 \\ \end{align} </math> </div></div> <ol start=67> <li>Berapa f(1)+f(-1) jika <math>f(\frac{ax-b}{bx-a})</math>=x²-5x+6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ jika} f(1) = f(\frac{ax-b}{bx-a}) \\ 1 &= \frac{ax-b}{bx-a} \\ bx-a &= ax-b \\ (b-a)x &= -b+a \\ &= -(b-a) \\ &= -1 \\ f(1) &= x^2-5x+6 \\ &= (-1)^2-5(-1)+6 \\ &= 12 \\ \text{ jika} f(-1) = f(\frac{ax-b}{bx-a}) \\ -1 &= \frac{ax-b}{bx-a} \\ -(bx-a) &= ax-b \\ -bx+a &= ax-b \\ (-b-a)x &= -b-a \\ &= 1 \\ f(-1) &= x^2-5x+6 \\ &= (1)^2-5(1)+6 \\ &= 2 \\ f(1)+f(-1) &= 12+2 \\ &= 14 \\ \end{align} </math> </div></div> <ol start=68> <li>berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)-x &= f(x-1) \\ f(x)-f(x-1) &= x \\ x=1 ; f(1)-f(0) &= 1 \\ x=2 ; f(2)-f(1) &= 2 \\ x=3 ; f(3)-f(2) &= 3 \\ x=4 ; f(4)-f(3) &= 4 \\ \dots \\ x=200 ; f(200)-f(199) &= 200 \\ \text{ jumlahkan tersebut menjadi } \\ f(200)-f(0) &= 1+2+3+4+\dots+200 \\ &= \frac{200 \cdot 201}{2} \\ &= 20.100 \\ f(200)-1 &= 20.100 \\ &= 20.101 \\ \end{align} </math> </div></div> <ol start=69> <li>Misalkan f(x) adalah fungsi rekursif yang berlaku ∀x ∈ R sebagai berikut: : f(x)+f(15-x) = 2024 : f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)+f(15-x) &= 2024 \\ f(15+x) &= f(x)+2020 \\ *cara 1 \\ \text{ganti x dengan 15+x } \\ f(15+x)+f(-x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ *cara 2 \\ \text{ganti x dengan -x } \\ f(-x)+f(15+x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ \end{align} </math> </div></div> <ol start=70> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>2f(\frac{2002}{x}) + f(x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2f(\frac{2002}{x}) + f(x) &= 3x \\ \text{ganti x dengan 2 } \\ 2f(\frac{2002}{2}) + f(2) &= 3(2) \\ 2f(1001) + f(2) &= 6 \\ \text{ganti x dengan 1001 } \\ 2f(\frac{2002}{1001}) + f(1001) &= 3(1001) \\ 2f(2) + f(1001) &= 3003 \\ 2f(2) + f(1001) &= 3003 \\ f(1001) &= 3003 - 2f(2) \\ 2f(1001) + f(2) &= 6 \\ 2(3003 - 2f(2)) + f(2) &= 6 \\ 6006 - 4f(2) + f(2) &= 6 \\ 3f(2) &= 6000 \\ f(2) &= 2000 \\ \end{align} </math> </div></div> <ol start=71> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>f(\frac{1}{x}) + \frac{1}{x}f(-x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{1}{x})+\frac{1}{x}f(-x) &= 3x \\ \text{ganti x dengan 1/3 } \\ f(3)+3f(-\frac{1}{3}) &= 1 \\ \text{ganti x dengan -3 } \\ f(-\frac{1}{3}) - \frac{1}{3}f(3) &= -9 \\ \text{dikalikan 3 } \\ 3f(-\frac{1}{3})-f(3) &= -27 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ 2f(3) &= 28 \\ f(3) &= 14 \\ \end{align} </math> </div></div> <ol start=72> <li>Diketahui polinom <math>f(7^b-1)=7^{3b}-10</math>. tentukan nilai f(5)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(5) &= f(7^b-1) \\ 5 &= 7^b-1 \\ 7^b &= 6 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(6-1) &= 6^3-10 \\ f(5) &= 216-10 \\ &= 206 \\ *cara 2 \\ \text{misalkan } 7^b-1=a \text{ maka } 7^b=a+1 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(a) &= (a+1)^3-10 \\ f(5) &= (5+1)^3-10 \\ &= 6^3-10 \\ &= 216-10 \\ &= 206 \\ \end{align} </math> </div></div> <ol start=73> <li>Diketahui polinom <math>f(6^b-7)=6^{3b}-2 \cdot 6^{2b}-4</math>. tentukan nilai f(-2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(-2) &= f(6^b-7) \\ -2 &= 6^b-7 \\ 6^b &= 5 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(5-7) &= 5^3-2 \cdot 5^2-4 \\ f(-2) &= 125-50-4 \\ &= 71 \\ *cara 2 \\ \text{misalkan } 6^b-7=a \text{ maka } 6^b=a+7 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(a) &= (a+7)^3-2(a+7)^2-4 \\ f(-2) &= (-2+7)^3-2(-2+7)^2-4 \\ &= 5^3-2(5)^2-4 \\ &= 125-50-4 \\ &= 71 \\ \end{align} </math> </div></div> <ol start=74> <li>Jika <math>f(xy)=\frac{f(x)}{y}</math> dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(10) &= 7 \\ f(2 \cdot 5) &= 7 \\ f(xy) &= \frac{f(x)}{y} \\ f(2 \cdot 5) &= \frac{f(2)}{5} \\ 7 &= \frac{f(2)}{5} \\ f(2) &= 35 \\ \end{align} </math> </div></div> <ol start=75> <li>Jika <math>f(xy)=\frac{f(x+y)}{xy}</math> dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(15) &= 16 \\ f(3 \cdot 5) &= 16 \\ f(xy) &= \frac{f(x+y)}{xy} \\ f(3 \cdot 5) &= \frac{f(3+5)}{3 \cdot 5} \\ f(15) &= \frac{f(8)}{15} \\ 16 &= \frac{f(8)}{15} \\ f(8) &= 240 \\ \end{align} </math> </div></div> <ol start=76> <li>Jika <math>f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15</math> maka tentukan nilai f(16)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x+\frac{1}{x}+6) &= x^2+\frac{1}{x^2}+15 \\ &= (x+\frac{1}{x})^2-2+15 \\ &= (x+\frac{1}{x})^2+13 \\ \text{misalkan } x+\frac{1}{x} &= p \\ f(x+\frac{1}{x}+6) &= (x+\frac{1}{x})^2+13 \\ f(p+6) &= p^2+13 \\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 } \\ f(p+6) &= p^2+13 \\ f(10+6) &= 10^2+13 \\ f(16) &= 100+13 \\ &= 113 \\ \end{align} </math> </div></div> <ol start=77> <li>Tentukan nilai x jika <math>f(x)=\frac{4}{4-x}</math> dan <math>f(x \cdot f(x))^{\frac{f(4x)}{f(x)}}=256</math>!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{4}{4-x} \\ f(4x) &= \frac{4}{4-4x} \\ \frac{f(4x)}{f(x)} &= \frac{\frac{4}{4-4x}}{\frac{4}{4-x}} \\ &= \frac{4-x}{4-4x} \\ f(x \cdot f(x)) &= f(x(\frac{4}{4-x})) \\ &= f(\frac{4x}{4-x}) \\ &= \frac{4}{4-(\frac{4x}{4-x})} \\ &= \frac{4}{\frac{16-4x-4x}{4-x}} \\ &= \frac{4}{\frac{16-8x}{4-x}} \\ &= \frac{4(4-x)}{4(4-4x)} \\ &= \frac{4-x}{4-4x} \\ \text{misalkan } \frac{4-x}{4-4x} &= a \\ f(x \cdot f(x))^{\frac{f(4x)}{f(x)}} &= 256 \\ a^a &= 256 \\ a^a &= 4^4 \\ a &= 4 \\ \frac{4-x}{4-4x} &= 4 \\ 4-x &= 16-16x \\ 15x &= 12 \\ x &= \frac{4}{5} \\ \end{align} </math> </div></div> <ol start=78> <li>Fungsi <math>f(x) = \frac{kx}{2x+1} \text{dengan } x \neq -\frac{1}{2}</math>. Dengan f(f(x)) = x maka tentukan nilai k!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{kx}{2x+1} \\ f(f(x)) &= x \\ f(\frac{kx}{2x+1}) &= x \\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1} &= x \\ \frac{\frac{k^2x}{2x+1}}{\frac{2kx+2x+1}{2x+1}} &= x \\ \frac{k^2x}{2kx+2x+1} &= x \\ \frac{k^2}{2kx+2x+1} &= 1 \\ k^2 &= 2kx+2x+1 \\ k^2-2kx &= 2x+1 \\ k^2-2kx+x^2 &= x^2+2x+1 \\ (k-x)^2 &= (x+1)^2 \\ (k-x)^2-(x+1)^2 &= 0 \\ (k-x+x+1)(k-x-(x+1)) &= 0 \\ k=-1 &\text{ atau } k=2x+1 &\text{ (TM) } \\ \end{align} </math> </div></div> <ol start=79> <li>Jika n = 2023²+2024² maka berapa hasil dari <math>\sqrt{2n-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n &= 2023^2+2024^2 \\ &= 2023^2+(2023+1)^2 \\ \text{misalkan 2023 = p} \\ n &= p^2+(p+1)^2 \\ &= p^2+p^2+2p+1 \\ &= 2p^2+2p+1 \\ \sqrt{2n-1} &= \sqrt{2(2p^2+2p+1)-1} \\ &= \sqrt{4p^2+4p+2-1} \\ &= \sqrt{4p^2+4p+1} \\ &= \sqrt{(2p+1)^2} \\ &= 2p+1 \\ &= 2(2023)+1 \\ &= 4046+1 \\ &= 4047 \\ \end{align} </math> </div></div> <ol start=80> <li>Tentukan nilai dari a+b+c merupakan bilangan bulat positif jika ab = 2, bc = 3 dan ac = 6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab \cdot bc \cdot ac &= 2 \cdot 3 \cdot 6 \\ (abc)^2 &= 36 \\ abc &= \pm 6 \\ abc &= 6 \\ \frac{abc}{ab} &= c = \frac{6}{2} = 3 \\ \frac{abc}{bc} &= a = \frac{6}{3} = 2 \\ \frac{abc}{ac} &= b = \frac{6}{6} = 1 \\ a+b+c &= 6 \\ \end{align} </math> </div></div> <ol start=81> <li>Tentukan nilai dari (a-c)^b jika <math>\frac{ab}{a+b} = \frac{1}{3}</math>, <math>\frac{bc}{b+c} = \frac{1}{4}</math> dan <math>\frac{ac}{a+c} = \frac{1}{9}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{ab}{a+b} &= \frac{1}{3} \\ \frac{a+b}{ab} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{b} + \frac{1}{a} &= 3 \\ \frac{bc}{b+c} &= \frac{1}{4} \\ \frac{b+c}{bc} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{b} &= 4 \\ \frac{ac}{a+c} &= \frac{1}{9} \\ \frac{a+c}{ac} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{a} &= 9 \\ \text{Misalkan 1/a = x, 1/b = y dan 1/c = z} \\ x+y &= 3 \\ y+z &= 4 \\ x+z &= 9 \\ x+y &= 3 \\ y+z &= 4 \\ x-z &= -1 \\ x-z &= -1 \\ x+z &= 9 \\ 2x &= 8 \\ x &= 4 \\ x-z &= -1 \\ 4-z &= -1 \\ z &= 5 \\ x+y &= 3 \\ 4+y &= 3 \\ y &= -1 \\ \frac{1}{a} &= 4 \\ a &= \frac{1}{4} \\ \frac{1}{b} &= -1 \\ b &= -1 \\ \frac{1}{c} &= 5 \\ c &= \frac{1}{5} \\ (a-c)^b &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=82> <li>Tentukan nilai dari a, b dan c jika <math>\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}</math> dan a+2b+3c=28?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan k untuk semua ketiga persamaan tersebut } \\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5} &= k \\ a+b &= 2k \\ a+c &= 4k \\ b+c &= 5k \\ 2a+b+c &= 6k \\ 2a+5k &= 6k \\ k &= 2a \\ a &= \frac{k}{2} \\ b &= \frac{3k}{2} \\ c &= \frac{7k}{2} \\ a+2b+3c &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ a &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ b &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ c &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=83> <li>Tentukan nilai dari (b+c)^a jika <math>\frac{a+b+c}{2} = \sqrt{a-2}+\sqrt{b-1}+\sqrt{c}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{a+b+c}{2} &= \sqrt{a-2}+\sqrt{b-1}+\sqrt{c} \\ a+b+c &= 2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}) \\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c} &= 0 \\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1 &= 0 \\ (\sqrt{a-2}-1)^2+(\sqrt{b-1}-1)^2+(\sqrt{c}-1)^2 &= 0 \\ (\sqrt{a-2}-1)^2 &= 0 \\ \sqrt{a-2}-1 &= 0 \\ \sqrt{a-2} &= 1 \\ a-2 &= 1 \\ a &= 3 \\ (\sqrt{b-1}-1)^2 &= 0 \\ \sqrt{b-1}-1 &= 0 \\ \sqrt{b-1} &= 1 \\ b-1 &= 1 \\ b &= 1 \\ (\sqrt{c}-1)^2 &= 0 \\ \sqrt{c}-1 &= 0 \\ \sqrt{c} &= 1 \\ c &= 1 \\ (b+c)^a &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=84> <li>x dan y merupakan bilangan tak nol. Jika xy = <math>\frac{x}{y}</math> = x-y maka berapa nilai x+y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy &= \frac{x}{y} \\ y^2 &= 1 \\ y^2 - 1 &= 0 \\ (y-1)(y+1) &= 0 \\ y = 1 &\text{ atau } y = -1 \\ \frac{x}{y} &= x-y \\ x &= xy-y^2 \\ x-xy &= -y^2 \\ x(1-y) &= -y^2 \\ x &= \frac{-y^2}{1-y} \\ \text{cek y=1 } \\ x &= \frac{-1^2}{1-1} \\ \text{tidak memenuhi syarat } \\ \text{cek y=-1 } \\ x &= \frac{-(-1)^2}{1-(-1)} \\ &= \frac{-1}{2} \\ x+y &= -1-\frac{1}{2} \\ &= -\frac{3}{2} \\ \end{align} </math> </div></div> <ol start=86> <li>Berapa nilai x dari <math>(\frac{a}{b})^3+(\frac{b}{a})^3 = 2\sqrt{x}</math> jika <math>\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b} \\ \frac{a+b}{ab} &= \frac{1}{a+b} \\ (a+b)^2 &= ab \\ a^2+2ab+b^2 &= ab \\ a^2+b^2 &= -ab \\ \text{misalkan } \frac{a}{b}+\frac{b}{a} = n \\ \frac{a}{b}+\frac{b}{a} &= n \\ \frac{a^2+b^2}{ab} &= n \\ a^2+b^2 &= nab \\ n &= -1 \\ \frac{a}{b}+\frac{b}{a} &= n \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3(\frac{a}{b}+\frac{b}{a}) &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3n &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3 &= n^3-3n \\ &= (-1)^3-3(-1) \\ &= 2 \\ 2\sqrt{x} &= 2 \\ \sqrt{x} &= 1 \\ x &= 1 \\ \end{align} </math> </div></div> <ol start=87> <li>Berapa nilai m dari <math>x^2-mx-1=0</math> jika <math>\sqrt[3]{x_1}+\sqrt[3]{x_2}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x_1} &= a \\ x_1 &= a^3 \\ \sqrt[3]{x_2} &= b \\ x_2 &= b^3 \\ \sqrt[3]{x_1}+\sqrt[3]{x_2} &= 1 \\ a+b &= 1 \\ x^2-mx-1 &= 0 \\ x_1+x_2 &= m \\ x_1 \cdot x_2 &= -1 \\ x_1+x_2 &= m \\ a^3+b^3 &= m \\ x_1 \cdot x_2 &= -1 \\ a^3 \cdot b^3 &= -1 \\ (ab)^2 &= (-1)^3 \\ ab &= -1 \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ (1)^3 &= m+3(-1)(1) \\ 1 &= m-3 \\ m &= 4 \\ \end{align} </math> </div></div> <ol start=88> <li>Berapa nilai <math>\frac{x_1}{x_2}</math> dari <math>ax^2-18x-b=0</math> jika <math>ab=45</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab &= 45 \\ b &= \frac{45}{a} \\ ax^2-18x-b &= 0 \\ ax^2-18x-\frac{45}{a} &= 0 \\ a^2x^2-18ax-45 &= 0 \\ (ax-3)(ax-15) &= 0 \\ ax-3 &= 0 \\ x &= \frac{3}{a} \\ ax-15 &= 0 \\ x &= \frac{15}{a} \\ \frac{x_1}{x_2} &= \frac{\frac{3}{a}}{\frac{15}{a}} \\ &= \frac{3}{15} \\ &= \frac{1}{5} \\ \frac{x_1}{x_2} &= \frac{\frac{15}{a}}{\frac{3}{a}} \\ &= \frac{15}{3} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=89> <li>Jika <math>\frac{u_3}{u_1+u_2} = \frac{7}{8}</math> merupakan barisan aritmetika maka berapa dari <math>\frac{u_2+u_3}{u_1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{u_3}{u_1+u_2} &= \frac{7}{8} \\ \frac{a+2b}{a+a+b} &= \frac{7}{8} \\ \frac{a+2b}{2a+b} &= \frac{7}{8} \\ 8(a+2b) &= 7(2a+b) \\ 8a+16b &= 14a+7b \\ 9b &= 6a \\ b &= \frac{2a}{3} \\ \frac{u_2+u_3}{u_1} &= \frac{a+b+a+2b}{a} \\ &= \frac{2a+3b}{a} \\ &= \frac{2a+3(\frac{2a}{3})}{a} \\ &= \frac{2a+2a}{a} \\ &= \frac{4a}{a} \\ &= 4 \\ \end{align} </math> </div></div> <ol start=90> <li>Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7p+q}{2p+q} &= \frac{17p+q}{7p+q} \\ (7p+q)^2 &= (17p+q)(2p+q) \\ 49p^2+14pq+q^2 &= 34p^2+19pq+q^2 \\ 15p^2 &= 5pq \\ 3p &= q \\ \frac{7p+q}{2p+q} &= \frac{7p+3p}{2p+3p} \\ &= \frac{10p}{5p} \\ &= 2 \\ \end{align} </math> </div></div> <ol start=91> <li>Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{rataan geometris } \\ \sqrt{a \cdot b} &= b-24 \\ a \cdot b &= (b-24)^2 \\ \text{rataan aritmatik } \\ \frac{a+b}{2} &= a+15 \\ a+b &= 2(a+15) \\ a+b &= 2a+30 \\ a &= b-30 \\ a \cdot b &= (b-24)^2 \\ (b-30)b &= (b-24)^2 \\ b^2-30b &= b^2-48b+576 \\ 18b &= 576 \\ b &= 32 \\ a &= b-30 \\ &= 32-30 \\ &= 2 \\ a+b &= 32+2 \\ &= 34 \\ \end{align} </math> </div></div> <ol start=91> <li>Segitiga lancip ABC dengan <math>\frac{a^4+b^4+c^4+a^2b^2}{c^2(a^2+b^2)}=2</math>. tentukan nilai sudut C?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{syarat segitiga lancip semua sudut masing-masing kurang dari } 90^\circ \\ c^2 &= a^2+b^2-2ab cos C \\ cos C &= \frac{a^2+b^2-c^2}{2ab} \\ a^4+b^4+c^4+a^2b^2 &= 2c^2(a^2+b^2) \\ a^4+b^4+a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-2c^2(a^2+b^2)+(c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= (ab)^2 \\ a^2+b^2-c^2 &= \pm ab \\ cos C &= \pm \frac{ab}{2ab} \\ &= \pm \frac{1}{2} \\ &= \frac{1}{2} \text{ (karena sudut harus kurang dari } 90^\circ) \\ C &= 60^\circ \\ \end{align} </math> </div></div> <ol start=92> <li>Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan panjang DA dan EA } = x \text{ dan panjang AB } = y \\ \text{luas segitiga CAB } &= \frac{CA \cdot AB}{2} \\ 45 &= \frac{(x+9)(x+5)}{2} \\ 90 &= x^2+14x+45 \\ x^2+14x &= 45 \\ y^2 &= (x+9)^2+(x+5)^2 \\ &= x^2+18x+81+x^2+10x+25 \\ &= 2x^2+28x+106 \\ &= 2(x^2+14x)+106 \\ &= 2(45)+106 \\ &= 196 \\ y &= 14 \\ \end{align} </math> jadi panjang BC adalah 14 cm </div></div> <ol start=93> <li>Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kita cari ukuran GC yaitu } \\ \frac{GC}{AD} &= \frac{CE}{DE} \\ \frac{GC}{15} &= \frac{6}{18} \\ GC &= 5 \\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC } \\ &= \frac{1}{2} \cdot BC \cdot CF - \frac{1}{2} \cdot GC \cdot CE \\ &= \frac{1}{2} \cdot 15 \cdot 18 - \frac{1}{2} \cdot 5 \cdot 6 \\ &= 135 - 15 \\ &= 120 \\ \end{align} </math> jadi luas daerah BFEG adalah 120 cm² </div></div> <ol start=94> <li>Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ AI &= \frac{1}{5} AF \\ &= \frac{1}{5} 10 \\ &= 2 \\ IF &= AF-AI \\ &= 10-2 \\ &= 8 \\ \text{luas trapesium AFGD } &= \frac{(AD+EF) \cdot AF}{2} \\ &= \frac{(4+6)10}{2} \\ &= 50 \\ \text{luas segitiga AID } &= \frac{AI \cdot AF}{2} \\ &= \frac{(2)4}{2} \\ &= 4 \\ \text{luas segitiga IFG } &= \frac{IF \cdot FG}{2} \\ &= \frac{(8)6}{2} \\ &= 24 \\ \text{luas daerah segitiga IGD } &= \text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG } \\ &= 50-4-24 \\ &= 22 \\ \end{align} </math> jadi luas daerah segitiga IGD adalah 22 cm² </div></div> <ol start=96> <li>Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-<math>\pi</math>) cm³ maka berapa luas permukaan kerucut tersebut?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ \text{volume balok} \\ V_b &= x^2(12) \\ \text{volume kerucut} \\ V_b &= \frac{1}{3}\pi x^2(12) \\ &= 4\pi x^2 \\ V_{b-k} &= Vb-Vk \\ 100(3-\pi) &= 12x^2-4\pi x^2 \\ 100(3-\pi) &= 4x^2(3-\pi) \\ x^2 &= 25 \\ x &= 5 \\ s &= \sqrt{12^2+5^2} \\ &= \sqrt{144+25} \\ &= \sqrt{169} \\ &= 13 \\ \text{luas permukaan kerucut } &= \pi r(r+s) \\ &= \pi(5)(5+13) \\ &= 90\pi \\ \end{align} </math> jadi luas daerah permukaan kerucut adalah 90<math>\pi</math> cm² </div></div> <ol start=97> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 3a+1 \\ B &= 3b+2 \\ A(A+1)+3B \\ (3a+1)(3a+1+1)+3(3b+2) \\ (3a+1)(3a+2)+9b+6 \\ 9a^2+9a+2+9b+6 \\ 9a^2+9a+9b+8 \\ 9(a^2+a+b)+8 \\ \text{sisa pembagiannya adalah } 8 \\ \end{align} </math> </div></div> <ol start=98> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 9a+7 \\ B &= 9b+8 \\ A(A-5)+9B \\ (9a+7)(9a+7-5)+9(9b+8) \\ (9a+7)(9a+2)+81b+72 \\ 81a^2+81a+14+81b+72 \\ 81a^2+81a+81b+86 \\ 81a^2+81a+81b+81+5 \\ 81(a^2+a+b+1)+5 \\ \text{sisa pembagiannya adalah } 5 \\ \end{align} </math> </div></div> <ol start=99> <li>Jika <math>\begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix}</math> maka berapa hasil dari A²¹+A²⁵+A⁴⁶?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A^2 &= A \cdot A \\ &= \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \\ A^3 &= A^2 \cdot A \\ &= \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ &= - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= -I \\ A^{21}+A^{25}+A^{46} &= A^{21} \cdot (I+A^4+A^{25}) \\ &= A^{21} \cdot (I+A^3 \cdot A +A^{24} \cdot A) \\ &= (A^3)^7 \cdot (I+A^3 \cdot A +(A^3)^8 \cdot A) \\ &= (-I)^7 \cdot (I-I \cdot A +(-I)^8 \cdot A) \\ &= -I \cdot (I-A+A) \\ &= -I \cdot I \\ &= -I \\ &= -\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ \end{align} </math> </div></div> <ol start=100> <li>Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida?</li></ol> : bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15 : ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9). : ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8) : maka Ida menulis bilangan 8. <ol start=101> <li>Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9?</li></ol> : Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu. : untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja. : untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja. : Jadi banyaknya bilangan mungkin 2. <ol start=102> <li>Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli!</li></ol> ;cara 1 # 8¹ = 1 # 8² = 1 (8²=8¹x8¹ sama dengan 1x1) # 8³ = 1 (8³=8¹x8² sama dengan 1x1) # 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2) # 8⁵ = 1 # 8ⁿ = 1 (semua n untuk bilangan asli) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli ;cara 2 # 8ⁿ = b mod 7 # 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil) # (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya) # 8ⁿ = 1ⁿ mod 7 # 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli <ol start=103> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 5?</li></ol> ;cara 1 # 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 # 7¹ = 7 (sisa 1) # 7² = 49 (sisa 2) # 7³ = 343 (sisa 3) # 7⁴ = 2,401 (sisa 0) # 7⁵ = 16,807 # 7⁶ = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3 ;cara 2 :17¹ = 2 :17² = 4 :17³ = 3 :17⁴ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3 :17⁹⁹ = (17⁴)²⁴ x 17³ Untuk 17⁴ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 3. Jadi 17⁹⁹ dibagi 7 bersisa 3 ;cara 3 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 17⁴ ::17⁴ = 1 mod 5 ::(17⁴)²⁴ = 1²⁴ mod 5 ::17⁹⁶ = 1²⁴ mod 5 ::17⁹⁶ = 1 mod 5 ::17⁹⁶ x 17³ = 1 x 17³ mod 5 ::17⁹⁹ = 17³ mod 5 ::17⁹⁹ = 17 x 17 x 17 mod 5 ::17⁹⁹ = 2 x 2 x 2 mod 5 ::17⁹⁹ = 8 mod 5 ::17⁹⁹ = 3 mod 5 Jadi 17⁹⁹ dibagi 5 bersisa 3 <ol start=104> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 7?</li></ol> ;cara 1 :17¹ = 3 :17² = 2 :17³ = 6 :17⁴ = 4 :17⁵ = 5 :17⁶ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3 :17⁹⁹ = (17⁶)¹⁶ x 17³ Untuk 17⁶ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 6. Jadi 17⁹⁹ dibagi 7 bersisa 6 ;cara 2 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 17⁶ ::17⁶ = 1 mod 7 ::(17⁶)¹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1 mod 7 ::17⁹⁶ x 17³ = 1 x 17³ mod 7 ::17⁹⁹ = 17³ mod 7 ::17⁹⁹ = 17 x 17 x 17 mod 7 ::17⁹⁹ = 3 x 3 x 3 mod 7 ::17⁹⁹ = 27 mod 7 ::17⁹⁹ = 6 mod 7 Jadi 17⁹⁹ dibagi 7 bersisa 6 <ol start=105> <li>Berapa hasil sisa dari 41²⁰²⁴ dibagi 33?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 41^{2024} &= 41^{2024} \text{ mod } 33 \\ &= (33 \times 3 + 2)^{2024} \text{ mod } 33 \\ &= 2^{2024} \text{ mod } 33 \\ &= 2^{2020} 2^4 \text{ mod } 33 \\ &= (2^5)^{404} 2^4 \text{ mod } 33 \\ &= (33 - 1)^{404} 2^4 \text{ mod } 33 \\ &= (-1)^{404} 2^4 \text{ mod } 33 \\ &= 2^4 \text{ mod } 33 \\ &= 16 \text{ mod } 33 \\ \text{Jadi hasil sisa adalah } 16 \\ \end{align} </math> </div></div> <ol start=106> <li>Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 99^{99} &= (3^2 \times 11)^{99} \\ &= 3^{198} \times 11^{99} \\ 243^n &= (3^5)^n \\ &= 3^{5n} \\ \text{agar bisa membagi, maka} \\ 5n &= 198 \\ n &= 39.6 \\ \text{jadi bilangan n terbesar adalah } 39 \\ \end{align} </math> </div></div> <ol start=107> <li>Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸?</lu></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 88^{88} &= (8 \times 11)^{88} \\ &= 8^{88} \times 11^{88} \\ &= 8^{87} \times 8 \times 11^{88} \\ &= (8^3)^{29} \times 8 \times 11^{88} \\ &= 512^{29} \times 8 \times 11^{88} \\ 512^n &= 512^{29} \\ \text{jadi bilangan n terbesar adalah } 29 \\ \end{align} </math> </div></div> <ol start=108> <li>Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6!</li></ol> ; cara 1 : KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. : N dibagi 3 sisa 1 : N dibagi 5 sisa 2 : N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a : KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70 : KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21 : KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97 ; cara 2 : Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35 : kemudian ditambahkan sisa masing-masing sesuai dengan KPK. : KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, <b>97</b> : KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, <b>97</b> : KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, <b>97</b> Jadi bilangan bulat positif adalah 97 :: NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 <ol start=109> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |- | 4 || 0 || Semua isi ember B dibuang |- | 1 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. <ol start=110> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |} nah ada ember A berisi 4 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |- | 1 || 0 || Semua isi ember B dibuang |- | 0 || 1 || Tuangkan 1 l dari ember A ke B sehingga ember A kosong |- | 3 || 1 || Isikan 3 l ke ember A |- | 0 || 4 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |} nah ada ember B berisi 4 liter. # Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kemungkinan 1} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 +3 \cdot ((\frac{4}{3})^x)^2 &= 7 \cdot (\frac{4}{3})^x \\ \text{misalkan } (\frac{4}{3})^x = y \\ 4 +3y^2 &= 7y \\ 3y^2-7y+4 &= 0 \\ (3y-4)(y-1) &= 0 \\ y_1 = \frac{4}{3} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{4}{3} \\ (\frac{4}{3})^x &= \frac{4}{3} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{4}{3})^x &= 1 \\ x &= 0 \\ \text{kemungkinan 2} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot ((\frac{3}{4})^x)^2 +3 &= 7 \cdot (\frac{3}{4})^x \\ \text{misalkan } (\frac{3}{4})^x = y \\ 4y^2 +3 &= 7y \\ 4y^2-7y+3 &= 0 \\ (4y-3)(y-1) &= 0 \\ y_1 = \frac{3}{4} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{3}{4} \\ (\frac{3}{4})^x &= \frac{3}{4} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{3}{4})^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-6}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{2x-4}{3}+2x-6 &= \frac{3}{2x-4}+\frac{1}{2x-6} \\ \frac{2x-4}{3}+2x-6 &= \frac{1}{\frac{2x-4}{3}}+\frac{1}{2x-6} \\ \text{misalkan } \frac{2x-4}{3} = y \text{ dan } 2x-6 = z \\ y+z &= \frac{1}{y}+\frac{1}{z} \\ y+z &= \frac{y+z}{yz} \\ y+z-\frac{y+z}{yz} &= 0 \\ (y+z)(1-\frac{1}{yz}) &= 0 \\ *y+z=0 \\ y+z &= 0 \\ y &= -z \\ \frac{2x-4}{3} &= -(2x-6) \\ \frac{2(x-2)}{3} &= -2(x-3) \\ \frac{x-2}{3} &= -x+3 \\ x-2 &= -3x+9 \\ 4x &= 11 \\ x &= \frac{11}{4} \\ *1-\frac{1}{yz}=0 \\ 1-\frac{1}{yz} &= 0 \\ yz-1 &= 0 \\ yz &= 0 \\ (\frac{2x-4}{3})(2x-6) &= 0 \\ \frac{4x^2-20x+24}{3} &= 0 \\ 4x^2-20x+24 &= 0 \\ x^2-5x+6 &= 0 \\ (x-2)(x-3) &= 0 \\ x = 2 &\text{ (TM) atau } x = 3 \text{ (TM) } \\ \end{align} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 8gij6pn1j99j9do98uw47yhx114dn8y 117497 117496 2026-07-13T05:15:53Z Akuindo 8654 117497 wikitext text/x-wiki contoh soal <ol start=1> <li>Berapa hasil dari <math>\sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Misalkan 2020 = p} \\ \sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64} &= \sqrt{(2020-5) \cdot (2020-3) \cdot (2020+3) \cdot (2020+5) + 64} \\ &= \sqrt{(p-5) \cdot (p-3) \cdot (p+3) \cdot (p+5) + 64} \\ &= \sqrt{(p-5) \cdot (p+5) \cdot (p-3) \cdot (p+3) + 64} \\ &= \sqrt{(p^2-25) \cdot (p^2-9) + 64} \\ &= \sqrt{p^4-34p^2+ 225 + 64} \\ &= \sqrt{p^4-34p^2+ 289} \\ &= \sqrt{(p^2-17)^2} \\ &= p^2-17 \\ &= 2020^2-17 \\ &= (2000+20)^2-17 \\ &= 4.000.000+80.000+400-17 \\ &= 4.080.383 \\ \end{align} </math> </div></div> <ol start=2> <li>Berapa nilai x dari <math>\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} = \frac{9}{10}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} &= \frac{9}{10} \\ \text{misalkan untuk } \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} = p \\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} &= p \\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}} &= p^2 \\ x^2-x-p &= p^2 \\ x^2-2x+1+x-1 &= p^2+p \\ (x-1)^2+(x-1) &= p^2+p \\ x-1 &= p \\ \text{misalkan untuk } \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}} &= q^3 \\ x^2 q &= q^3 \\ x^2 &= q^2 \\ x &= q \\ \frac{x-1}{x} &= \frac{9}{10} \\ x &= 10 \\ \end{align} </math> </div></div> <ol start=3> <li>Berapa nilai x dari <math>(\frac{x}{x+10})^{x+10}=\frac{1}{1024}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{x+10}{x})^{-(x+10)} &= (1024)^{-1} \\ (\frac{x+10}{x})^{x+10} &= 1024 \\ (\frac{x+10}{x})^{x+10} &= 2^{10} \\ (\frac{x+10}{x})^{\frac{x+10}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (\frac{1}{2})^{-1} \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (1+(-\frac{1}{2}))^{(1+(-\frac{2}{1}))} \\ \frac{10}{x} &= -\frac{1}{2} \\ x &= -20 \\ \end{align} </math> </div></div> <ol start=4> <li>Berapa nilai x dari <math>x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{2}+\sqrt{x+\frac{1}{4}} &= (\sqrt{x+\frac{1}{4}})^2+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+(\frac{1}{2})^2 \\ &= (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 \\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}} &= 4 \\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2} &= 4 \\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 4 \\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 &= 4 \\ \sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 2 \\ \sqrt{x+\frac{1}{4}} &= \frac{3}{2} \\ x+\frac{1}{4} &= \frac{9}{4} \\ x &= 2 \\ \end{align} </math> </div></div> <ol start=5> <li>Berapa nilai x dari <math>\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^3}{\sqrt{8-x^2}}+x^2-8 &= 0 \\ \frac{x^3}{\sqrt{8-x^2}} &= 8-x^2 \\ x^3 &= (8-x^2)^{\frac{3}{2}} \\ x &= (8-x^2)^{\frac{1}{2}} \\ x^2 &= 8-x^2 \\ 2x^2-8 &= 0 \\ x^2-4 &= 0 \\ (x-2)(x+2) &= 0 \\ \text{membuktikan } \\ x=2 \text{ maka hasilnya 0 } \\ x=-2 \text{ maka hasilnya -8 } \\ \text{jadi } x=2 \\ \end{align} </math> </div></div> <ol start=6> <li>Berapa nilai x dari <math>\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} = 5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} &= 5 \\ \frac{x^{50}+x^{60}+x^{70}}{31}} &= 5^5 \\ x^{50}+x^{60}+x^{70} &= 5^5 \cdot 31 \\ x^{50}(1+x^{10}+x^{20}) &= 5^5 \cdot 31 \\ (x^{10}^5)(1+x^{10}+(x^{10}^2) &= 5^5 \cdot 31 \\ \text{ misalkan } x^{10} = a \\ a^5(1+a+a^2) &= 5^5 \cdot 31 \\ a &= 5 \\ x^{10} &= 5 \\ x &= ^5 log 10 \\ \end{align} </math> </div></div> <ol start=7> <li>Berapa nilai x dari <math>\sqrt{3x+5+\sqrt{4x+5}} = x</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{3x+5+\sqrt{4x+5}} &= x \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \text{misalkan } \sqrt{4x+5}=y \text{ dan } 4x+5=y^2 \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \sqrt{y^2+y-x} &= x \\ y^2+y &= x^2+x \\ y=x \\ 4x+5 &= y^2 \\ 4x+5 &= x^2 \\ x^2-4x-5 &= 0 \\ (x-5)(x+1) &= 0 \\ x=5 &\text{ atau } x=-1 \text{ (TM) } \\ \end{align} </math> </div></div> <ol start=8> <li>Berapa nilai x dari <math>\sqrt{1+\sqrt{1+x}} = \sqrt[3]{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{1+\sqrt{1+x}} &= \sqrt[3]{x} \\ \sqrt[3]{x} &= n \\ x &= n^3 \\ \sqrt{1+\sqrt{1+n^3}} &= n \\ 1+\sqrt{1+n^3} &= n^2 \\ \sqrt{1+n^3} &= n^2-1 \\ 1+n^3 &= n^4-2n^2+1 \\ n^4-n^3-2n^2 &= 0 \\ n^2(n^2-n-2) &= 0 \\ n^2(n-2)(n+1) &= 0 \\ n=0, n=2 \text{ atau } n=-1 \\ n &= 0 \\ x &= 0^3 \\ &= 0 \\ n &= 2 \\ x &= 2^3 \\ &= 8 \\ n &= -1 \\ x &= (-1)^3 \\ &= -1 \\ \text{yang paling mungkin untuk nilai x adalah } 8 \\ \end{align} </math> </div></div> <ol start=9> <li>Berapa nilai x dari <math>\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} &= \frac{\sqrt{1+x}}{\sqrt{x}} \\ \sqrt{x}(\sqrt{1+x}+\sqrt{x}) &= (\sqrt{1+x}-\sqrt{x})\sqrt{1+x} \\ \sqrt{x(1+x)}+x &= 1+x-\sqrt{x(1+x)} \\ 2\sqrt{x(1+x)} &= 1 \\ \sqrt{x(1+x)} &= \frac{1}{2} \\ x(1+x) &= \frac{1}{4} \\ x^2+x &= \frac{1}{4} \\ 4x^2+4x &= 1 \\ 4x^2+4x-1 &= 0 \\ x &= \frac{-4 \pm \sqrt{4^2-4(4)(-1)}}{2(4)} \\ &= \frac{-4 \pm \sqrt{32}}{8} \\ &= \frac{-4 \pm 4\sqrt{2}}{8} \\ &= \frac{-1 \pm \sqrt{2}}{2} \\ \text{karena akar x harus minimal nol jadi } x = \frac{-1+\sqrt{2}}{2} \\ \end{align} </math> </div></div> <ol start=10> <li>Berapa nilai x dari <math>\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x-\sqrt{x+1}}{x+\sqrt{x+1}} &= \frac{11}{19} \\ \text{misalkan } \sqrt{x+1}=y \text{ dan } x=y^2-1 \\ \frac{y^2-1-y}{y^2-1+y} &= \frac{11}{19} \\ 19(y^2-y-1) &= 11(y^2+y-1) \\ 19y^2-19y-19 &= 11y^2+11y-11 \\ 8y^2-30y-8 &= 0 \\ 4y^2-15y-4 &= 0 \\ (4y+1)(y-4) &= 0 \\ y=-\frac{1}{4} \text{ (TM) atau } & y=4 \\ x &= 4^2-1 \\ &= 15 \\ \end{align} </math> </div></div> <ol start=11> <li>Berapa nilai x dari <math>\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} &= 98 \\ \text{misalkan } \sqrt{x^2-1}=y \\ \frac{x+y}{x-y}+\frac{x-y}{x+y} &= 98 \\ \frac{(x+y)^2+(x-y)^2}{(x-y)(x+y)} &= 98 \\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} &= 98 \\ \frac{2(x^2+y^2)}{x^2-y^2} &= 98 \\ \frac{x^2+y^2}{x^2-y^2} &= 49 \\ x^2+y^2 &= 49(x^2-y^2) \\ x^2+y^2 &= 49x^2-49y^2 \\ 48x^2 &= 50y^2 \\ 24x^2 &= 25y^2 \\ 24x^2 &= 25(\sqrt{x^2-1})^2 \\ 24x^2 &= 25(x^2-1) \\ 24x^2 &= 25x^2-25 \\ x^2 &= 25 \\ x &= \pm 5 \\ \end{align} </math> </div></div> <ol start=12> <li>Berapa nilai x dari <math>\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \sqrt{x}=y \text{ dan } x=y^2 \\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} &= \frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\sqrt{\frac{y^2}{y^2+y}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\frac{y}{\sqrt{y^2+y}} \\ y^2+y-\sqrt{(y^2+y)(y^2-y)} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^4-y^2} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^2(y^2-1)} &= \frac{5}{4}y \\ y(y+1)-y\sqrt{y^2-1} &= \frac{5}{4}y \\ y+1-\sqrt{y^2-1} &= \frac{5}{4} \\ -\sqrt{y^2-1} &= \frac{1}{4}-y \\ y^2-1 &= (\frac{1}{4}-y)^2 \\ y^2-1 &= \frac{1}{16}-\frac{1}{2}y+y^2 \\ -1 &= \frac{1}{16}-\frac{1}{2}y \\ \frac{1}{2}y &= \frac{1}{16}+1 \\ \frac{1}{2}y &= \frac{17}{16} \\ y &= \frac{17}{8} \\ x &= (\frac{17}{8})^2 \\ &= \frac{289}{64} \\ \end{align} </math> </div></div> <ol start=13> <li>Berapa nilai x dari <math>\sqrt[4]{62+x}+\sqrt[4]{275-x}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ misalkan } \sqrt[4]{62+x}=a, 62+x=a^4, \sqrt[4]{275-x}=b \text{ dan } 275-x=b^4 \\ a+b &= 7 \\ (a+b)^2 &= 49 \\ a^2+b^2+2ab &= 49 \\ a^2+b^2 &= 49-2ab \\ a^4+b^4 &= 62+x+275-x \\ (a^2+b^2)^2-2(ab)^2 &= 337 \\ (49-2ab)^2-2(ab)^2 &= 337 \\ 2401-196ab+4(ab)^2-2(ab)^2 &= 337 \\ 2(ab)^2-196ab+2064 &= 0 \\ (ab)^2-98ab+1032 &= 0 \\ (ab-12)(ab-86) &= 0 \\ ab = 12 \text{ atau } & ab = 86 \text{ (TM) karena hasil kali maksimum yaitu 12 } \\ ab =12 \text{ dan } a+b=7 \\ a+b &= 7 \\ b &= 7-a \\ ab &= 12 \\ a(7-a) &= 12 \\ -a^2+7a &= 12 \\ a^2-7a+12 &= 0 \\ (a-3)(a-4) &= 0 \\ a=3 \text{ atau } & a=4 \\ a=3, b=4 \\ 62+x &= a^4 \\ 62+x &= (3)^4 \\ 62+x &= 81 \\ x &= 19 \\ a=4, b=3 \\ 62+x &= a^4 \\ 62+x &= (4)^4 \\ 62+x &= 256 \\ x &= 194 \\ \end{align} </math> </div></div> <ol start=14> <li>Berapa nilai x dari <math>\sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2} &= 7 \\ (\sqrt[3]{8+x})^2-\sqrt[3]{8+x} \sqrt[3]{27-x}+(\sqrt[3]{27-x})^2 &= 7 \\ \text{misalkan } \sqrt[3]{8+x}=a, 8+x=a^3, \sqrt[3]{27-x}=b \text{ dan } 27-x=b^3 \\ a^2-ab+b^2 &= 7 \\ a^3+b^3 &= 8+x+27-x \\ &= 35 \\ a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ 35 &= (a+b)(7) \\ a+b &= 5 \\ b &= 5-a \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ 5^3 &= 35+3ab(5) \\ 125 &= 35+15ab \\ 80 &= 15ab \\ ab &= 6 \\ a(5-a) &= 6 \\ 5a-a^2 &= 6 \\ a^2-5a+6 &= 6 \\ (a-2)(a-3) &= 6 \\ a=2 &\text{ atau } a=3 \\ a=2, b=3 \text{ dan } a=3,b=2 \\ 8+x &= a^3 \\ &= 2^3 \\ &= 8 \\ x &= 0 \\ 8+x &= a^3 \\ &= 3^3 \\ &= 27 \\ x &= 19 \\ \end{align} </math> </div></div> <ol start=15> <li>Berapa nilai x dari <math>3^x+5^x-9^x+15^x-25^x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3^x+5^x-9^x+15^x-25^x &= 1 \\ 3^x+5^x-(3^2)^x+(3 \cdot 5)^x-(5^2)^x &= 1 \\ 3^x+5^x-(3^x)^2+(3^x \cdot 5^x)-(5^x)^2 &= 1 \\ \text{misalkan } 3^x=a \text{ dan } 5^x=b \\ a+b-a^2+ab-b^2 &= 1 \\ a^2-ab+b^2-a-b+1 &= 0 \\ 2a^2-2ab+2b^2-2a-2b+2 &= 0 \\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1 &= 0 \\ (a-b)^2+(a-1)^2+(b-1)^2 &= 0 \\ a-b=0; a-1=0; b-1 &= 0 \\ a=b &= 1 \\ 3^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> <ol start=16> <li>Berapa nilai x dari <math>^6log x^2+^{6x}log \frac{6}{x}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ \text{misalkan } 6x=a \text{ maka } x=\frac{a}{6} \\ ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ ^6log (\frac{a}{6})^2+^{6 \frac{a}{6}}log \frac{6}{\frac{a}{6}} &= 1 \\ ^6log \frac{a^2}{6^2}+^alog \frac{6^2}{a} &= 1 \\ ^6log a^2-^6log 6^2+^alog 6^2-^alog a &= 1 \\ 2 ^6log a-2 ^6log 6+2 ^alog 6-^alog a &= 1 \\ 2 ^6log a-2+2 \frac{1}{^6log a}-1 &= 1 \\ 2 ^6log a+2 \frac{1}{^6log a}-4 &= 0 \\ 2 ^6log^2 a-4 ^6log a+2 &= 0 \\ ^6log^2 a-2 ^6log a+1 &= 0 \\ (^6log a-1)^2 &= 0 \\ ^6log a &= 1 \\ a &= 6 \\ x &= \frac{a}{6} \\ &= \frac{6}{6} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=17> <li>Berapa nilai x dari (x+500)³+x=20?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+500)^3+x &= 20 \\ \text{misalkan } a=x+500 \text{ maka } x=a-500 \\ a^3+a-500 &= 20 \\ a^3+a &= 520 \\ a(a^2+1) &= 8 \cdot 65 \\ a(a^2+1) &= 8(64+1) \\ a(a^2+1) &= 8(8^2+1) \\ a &= 8 \\ x &= 8-500 \\ &= -492 \\ \end{align} </math> </div></div> <ol start=18> <li>Berapa nilai x dari <math>\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2} &= 0 \\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}} &= \frac{1}{2} \\ \frac{x^n+4^n}{x^n+16^n} &= (\frac{1}{2})^n \\ \frac{x^n+4^n}{x^n+16^n} &= \frac{1}{2^n} \\ 2^n(x^n+4^n) &= x^n+16^n \\ 2^n(x^n+2^{2n}) &= x^n+2^{4n} \\ 2^n \cdot x^n+2^{3n} &= x^n+2^{4n} \\ 2^n \cdot x^n-x^n &= 2^{4n}-2^{3n} \\ x^n(2^n-1) &= 2^{3n}(2^n-1) \\ x^n &= 2^{3n} \\ x^n &= (2^3)^n \\ x^n &= 8^n \\ x &= 8 \\ \end{align} </math> </div></div> <ol start=19> <li>Berapa hasil dari <math>\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ x &= \frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5 \cdot 6}+\sqrt{5 \cdot 5}+\sqrt{6 \cdot 4}+\sqrt{5 \cdot 4}}{\sqrt{5 \cdot 4}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5} \cdot \sqrt{6}+\sqrt{5} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{4}}{2 \cdot \sqrt{5}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{6} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{5}+\sqrt{5} \cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}} \\ &= \frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ &= \frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ \frac{1}{x} &= \frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}} \\ &= \frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ &= \frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1} \\ &= \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5} \\ &= \sqrt{6}-\sqrt{4} \\ &= \sqrt{6}-2 \\ x &= \frac{1}{\sqrt{6}-2} \\ &= \frac{\sqrt{6}+2}{6-4} \\ &= \frac{\sqrt{6}+2}{2} \\ &= 1+\frac{\sqrt{6}}{2} \\ \end{align} </math> </div></div> <ol start=20> <li>Berapa hasil dari <math>(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5 \\ \text{misalkan } x=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ x &= \frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ &= \frac{\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}} \\ &= \frac{\sqrt{3}+1}{4} \\ 4x &= \sqrt{3}+1 \\ 4x-1 &= \sqrt{3} \\ (4x-1)^2 &= 3 \\ 16x^2-8x+1 &= 3 \\ 16x^2 &= 8x+2 \\ 8x^2 &= 4x+1 \\ x^2 &= \frac{4x+1}{8} \\ *cara 1 \\ x^3 &= x \cdot x^2 \\ &= x(\frac{4x+1}{8}) \\ &= \frac{4x^2+x}{8} \\ &= \frac{4x^2}{8}+\frac{x}{8} \\ &= \frac{4(\frac{4x+1}{8})}{8}+\frac{x}{8} \\ &= \frac{16x+4}{64}+\frac{x}{8} \\ &= \frac{4x+1}{16}+\frac{x}{8} \\ &= \frac{4x+1+2x}{16} \\ &= \frac{6x+1}{16} \\ x^5 &= x^2 \cdot x^3 \\ &= (\frac{4x+1}{8})(\frac{6x+1}{16}) \\ &= \frac{24x^2+10x+1}{128} \\ &= \frac{24x^2}{128}+\frac{10x+1}{128} \\ &= \frac{24(\frac{4x+1}{8})}{128}+\frac{10x+1}{128} \\ &= \frac{96x+24}{1024}+\frac{10x+1}{128} \\ &= \frac{96x+24+80x+8}{1024} \\ &= \frac{176x+32}{1024} \\ &= \frac{176x}{1024}+\frac{32}{1024} \\ &= \frac{176}{1024}(\frac{\sqrt{3}+1}{4})+\frac{32}{1024} \\ &= \frac{44(\sqrt{3}+1)}{1024}+\frac{32}{1024} \\ &= \frac{44\sqrt{3}+44}{1024}+\frac{32}{1024} \\ &= \frac{76+44\sqrt{3}}{1024} \\ &= \frac{19+11\sqrt{3}}{256} \\ *cara 2 \\ x^4 &= (x^2)^2 \\ &= (\frac{4x+1}{8})^2 \\ &= \frac{16x^2+8x+1}{64} \\ &= \frac{16x^2}{64}+\frac{8x}{64}+\frac{1}{64} \\ &= \frac{x^2}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{\frac{4x+1}{8}}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{4x}{32}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{8}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{4}+\frac{3}{64} \\ x^5 &= x \cdot x^4 \\ &= (\frac{\sqrt{3}+1}{4})(\frac{x}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\frac{\sqrt{3}+1}{4}}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\sqrt{3}+1}{16}+\frac{3}{64}) \\ &= \frac{(\sqrt{3}+1)^2}{64}+(\frac{\sqrt{3}+1}{4})\frac{3}{64} \\ &= \frac{3+2\sqrt{3}+1}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{4+2\sqrt{3}}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{16+8\sqrt{3}}{256}+\frac{3\sqrt{3}+3}{256} \\ &= \frac{19+11\sqrt{3}}{256} \\ \end{align} </math> </div></div> <ol start=21> <li>Berapa hasil dari <math>\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4} &= \frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots) \\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots &= \frac{1}{1-\frac{1}{4}} \\ &= \frac{4}{3} \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{4}{3}) \\ &= \frac{1}{4}+\frac{16}{3} \\ &= \frac{67}{12} \\ x &= \frac{67}{9} \\ \end{align} </math> </div></div> <ol start=22> <li>Berapa nilai y-x jika <math>\frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} = \frac{x}{y}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} &= \frac{x}{y} \\ \frac{\frac{106 \times 107}{2}}{\frac{106}{2}(4+109)} &= \frac{x}{y} \\ \frac{53 \times 107}{53 \times 113} &= \frac{x}{y} \\ y-x &= 113-107 = 6 \\ \end{align} </math> </div></div> <ol start=23> <li>Berapa angka satuan dari hasil 17²⁰²⁴?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan angka satuannya} \\ 17^1 &= 7 \\ 17^2 &= 9 \\ 17^3 &= 3 \\ 17^4 &= 1 \\ 17^5 &= 7 \\ 17^6 &= 9 \\ 17^7 &= 3 \\ 17^8 &= 1 \\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1} \end{align} </math> </div></div> <ol start=24> <li>Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ 1! + 2! + 3! + 4! + \dots + 2024! &= 1 + (1x2) + (1x2x3) + (1x2x3x4) + \dots + 2024! \\ &= 1 + 2 + 6 + 24 + 120 + 720 + \dots + 2024! \\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka } 1+2+6+24 = 33 \text{ jadi angka satuannya adalah } 3 \end{align} </math> </div></div> <ol start=25> <li>Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ \frac{1! + 2! + 3! + 4! + \dots + 2024!}{12} &= \frac{1 + 1x2 + 1x2x3 + 1x2x3x4 + \dots + 2024!}{12} \\ &= \frac{1 + 2 + 6 + 24 + \dots + 2024!}{12} \\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi } 1+2+6 = 9 \end{align} </math> </div></div> <ol start=26> <li>Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{penjumlahan} \\ 1+1+1+1+ \dots \text{ (sebanyak 88 buah) }+x+y &= A \\ 88+x+y &= A \\ \text{perkalian} \\ 1 \times 1 \times 1 \times \dots \text{ (sebanyak 88 buah) }\times x \times y &= A \\ x \times y &= A \\ 88+x+y &= xy \\ xy-y &= 88+x \\ y(x-1) &= 88+x \\ y &= \frac{88+x}{x-1} \\ \text{uji selidiki untuk x=2} \\ y &= \frac{88+2}{2-1} \\ &= 90 \\ \text{buktikan} \\ 88+x+y &= xy \\ 88+2+90 &= 2(90) \\ 180 &= 180 \\ \text{terbukti} \\ \text{nilai A adalah } 180 \\ \end{align} </math> </div></div> <ol start=27> <li>Berapakah nilai x, y dan z dari <math>x+y-z=1, x^2+y^2-z^2=-5 \text{ dan } x^3+y^3-z^3=-53</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+y-z &= 1 \\ x+y &= z+1 \\ x^2+2xy+y^2 &= z^2+2z+1 \\ x^2+y^2-z^2 &= 2z+1-2xy \\ -5 &= 2z+1-2xy \\ 2xy &= 2z+6 \\ xy &= z+3 \\ x^2+y^2-z^2 &= -5 \\ x^2+y^2 &= z^2-5 \\ x^3+y^3-z^3 &= -53 \\ (x+y)(x^2-xy+y^2)-z^3+53 &= 0 \\ (x+y)(x^2+y^2-xy)-z^3+53 &= 0 \\ (z+1)(z^2-5-(z+3))-z^3+53 &= 0 \\ (z+1)(z^2-z-8)-z^3+53 &= 0 \\ z^3-z^2-8z+z^2-z-8-z^3+53 &= 0 \\ -9z+45 &= 0 \\ -9z &= -45 \\ z &= 5 \\ x+y &= 5+1 \\ x+y &= 6 \\ x &= 6-y \\ xy &= 5+3 \\ xy &= 8 \\ (6-y)y &= 8 \\ 6y-y^2 &= 8 \\ y^2-6y+8 &= 0 \\ (y-4)(y-2) &= 0 \\ y=4 \text{ atau } y=2 \\ \text{jika } y=4 \\ x+y &= z+1 \\ x+4 &= 5+1 \\ x &= 2 \\ \text{jika } y=2 \\ x+y &= z+1 \\ x+2 &= 5+1 \\ x &= 4 \\ \end{align} </math> </div></div> <ol start=28> <li>Berapakah nilai titik koordinat (x,y) dari <math>\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}</math> dan <math>\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x+y}+\sqrt{x-y} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+y}-\sqrt{x-y} &= \sqrt{\frac{52y}{3x}} \\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y}) &= \sqrt{\frac{432x}{13y}} \cdot \sqrt{\frac{52y}{3x}} \\ x+y-x+y &= \sqrt{\frac{432x \cdot 52y}{13y \cdot 3x}} \\ 2y &= \sqrt{144 \cdot 4} \\ 2y &= \sqrt{576} \\ 2y &= 24 \\ y &= 12 \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13(12)}} \\ x+12+x-12+2 \cdot \sqrt{x+12} \cdot \sqrt{x-12} &= \frac{36x}{13} \\ 2x+2 \sqrt{x^2-144} &= \frac{36x}{13} \\ 2(x+\sqrt{x^2-144}) &= \frac{36x}{13} \\ x+\sqrt{x^2-144} &= \frac{18x}{13} \\ \sqrt{x^2-144} &= \frac{5x}{13} \\ x^2-144 &= \frac{25x^2}{169} \\ \frac{144x^2}{169}-144 &= 0 \\ \frac{x^2}{169}-1 &= 0 \\ x^2-169 &= 0 \\ (x-13)(x+13) &= 0 \\ x_1=13 &\text{ atau } x_2=-13 \text{ (TM) karena } x>y \\ \end{align} </math> jadi titik koordinat (13,12) </div></div> <ol start=29> <li>Berapakah nilai dari <math>x^2-7x</math> jika <math>(x-2)^2+\frac{1}{(x-2)^2} = 11</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x-2)^2+\frac{1}{(x-2)^2} &= 11 \\ (x-2)^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2)^2} &= 11-2 \\ (x-2-\frac{1}{x-2})^2 &= 9 \\ x-2-\frac{1}{x-2} &= 3 \\ (x-2)^2-1 &= 3(x-2) \\ x^2-4x+4-1 &= 3x-6 \\ x^2-7x &= -9 \\ \end{align} </math> </div></div> <ol start=30> <li>Berapakah nilai dari <math>\frac{(x+y)^2(x+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)}</math> jika xy+yz+xz=1?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy+yz+xz &= 1 \\ x^2+xy+yz+xz &= x^2+1 \\ x(x+y)+z(x+y) &= x^2+1 \\ (x+y)(x+z) &= x^2+1 \\ \text{dengan pola yang sama } \\ (y+x)(y+z) &= y^2+1 \\ (x+z)(y+z) &= z^2+1 \\ \frac{(x+y)^2(y+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)} &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)} \\ &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)^2(y+z)^2(x+z)^2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=31> <li>Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} w+5 &= w+x+y+z+5 \\ x+4 &= w+x+y+z+5 \\ y+3 &= w+x+y+z+5 \\ z+2 &= w+x+y+z+5 \\ \text{jumlahkan keempat persamaan } \\ w+x+y+z+14 &= 4(w+x+y+z+5) \\ w+x+y+z+14 &= 4(w+x+y+z)+20 \\ 3(w+x+y+z) &= -6 \\ w+x+y+z &= -2 \\ \end{align} </math> </div></div> <ol start=32> <li>Berapakah nilai dari <math>\frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2}</math> jika <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3</math> dan x+y+z=xyz?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2} &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2} \\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) \\ 3^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz}) \\ 9 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz}) \\ &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2 \\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} &= 7 \\ \end{align} </math> </div></div> <ol start=33> <li>Berapakah nilai dari <math>\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}</math> jika <math>x^2+y^2+z^2 = -2(ab+bc+ac)</math>?>/li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+z^2 &= -2(xy+yz+xz) \\ x^2+y^2+z^2+2(xy+yz+xz) &= 0 \\ (x+y+z)^2 &= 0 \\ x+y+z &= 0 \\ x+y &= -z \\ x+z &= -y \\ y+z &= -x \\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z} &= \frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x} \\ &= -2-(-5)-(-7) \\ &= 10 \\ \end{align} </math> </div></div> <ol start=34> <li>Berapakah nilai dari <math>\frac{20xyz}{xy+yz+xz}</math> jika <math>16^x = 256^y = 625^z = 40</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 16^x = 256^y = 625^z &= 40 \\ 2^{4x} = 4^{4y} = 5^{4z} &= 40 \\ 2^{4x} &= 40 \\ 2 &= 40^{\frac{1}{4x}} \\ 4^{4y} &= 40 \\ 4 &= 40^{\frac{1}{4y}} \\ 5^{4z} &= 40 \\ 5 &= 40^{\frac{1}{4z}} \\ 2 \cdot 4 \cdot 5 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z}} \\ 1 &= \frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z} \\ 4 &= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{20xyz}{xy+yz+xz} &= 20 \cdot \frac{xyz}{xy+yz+xz} \\ &= 20 \cdot (\frac{xy+yz+xz}{xyz})^{-1} \\ &= 20 \cdot (\frac{1}{z} + \frac{1}{x} + \frac{1}{y})^{-1} \\ &= 20 \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{-1} \\ &= 20 \cdot (4)^{-1} \\ &= 20 \cdot \frac{1}{4} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=35> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+3x^2+1}</math> jika <math>6x^2+25x+6=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 6x^2+25x+6 &= 0 \\ 6x+25+\frac{6}{x} &= 0 \\ 6(x+\frac{1}{x}) &= -25 \\ x+\frac{1}{x} &= \frac{-25}{6} \\ (c+\frac{1}{x})^2 &= (\frac{-25}{6})^2 \\ x^2+2+\frac{1}{x^2} &= \frac{625}{36} \\ x^2+\frac{1}{x^2} &= \frac{625}{36}-2 \\ x^2+\frac{1}{x^2} &= \frac{553}{36} \\ \frac{x^2}{x^4+3x^2+1} &= \frac{1}{x^2+3+\frac{1}{x^2}} \\ &= \frac{1}{a^2+\frac{1}{x^2}+3} \\ &= \frac{1}{\frac{553}{36}+3} \\ &= \frac{1}{\frac{661}{36}} \\ &= \frac{36}{661} \\ \end{align} </math> </div></div> <ol start=36> <li>Berapakah nilai dari <math>\frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5} &= \frac{(9+2\sqrt{20})^{1013}}{((2)^3+3(2)^2(\sqrt{5})+3(2)(\sqrt{5})^2+(\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{((2+\sqrt{5})^2)^{1013}}{((2+\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{(2+\sqrt{5})^{2026}}{(2+\sqrt{5})^{2025}}+6-\sqrt{5} \\ &= 2+\sqrt{5}+6-\sqrt{5} \\ &= 8 \\ \end{align} </math> </div></div> <ol start=37> <li>Berapakah nilai dari <math>27x^3+\frac{8}{x^3}</math> jika <math>3x+\frac{2}{x}=6</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3x+\frac{2}{x} &= 6 \\ (3x+\frac{2}{x})^3 &= 6^3 \\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3} &= 216 \\ 27x^3+18(6)+\frac{8}{x^3} &= 216 \\ 27x^3+108+\frac{8}{x^3} &= 216 \\ 27x^3+\frac{8}{x^3} &= 108 \\ \end{align} </math> </div></div> <ol start=38> <li>Berapakah nilai dari <math>x^6+\frac{8}{x^3}</math> jika <math>x^3+\frac{1}{x^3}=8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3+\frac{1}{x^3} &= 8 \\ x^3 &= 8-\frac{1}{x^3} \\ x^6 &= 8x^3-1 \\ x^6+\frac{8}{x^3} &= 8x^3-1+\frac{8}{x^3} \\ &= 8x^3+\frac{8}{x^3}-1 \\ &= 8(x^3+\frac{1}{x^3})-1 \\ &= 8(8)-1 \\ &= 63 \\ \end{align} </math> </div></div> <ol start=39> <li>Berapakah nilai dari <math>4x+\frac{25}{x}</math> jika <math>2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2\sqrt{x}+\frac{5}{\sqrt{x}} &= 4x-\frac{25}{x} \\ 2\sqrt{x}+\frac{5}{\sqrt{x}} &= (2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}}) \\ 1 &= 2\sqrt{x}-\frac{5}{\sqrt{x}} \\ 1^2 &= (2\sqrt{x}-\frac{5}{\sqrt{x}})^2 \\ 1 &= 4x-20+\frac{25}{x} \\ 4x+\frac{25}{x} &= 21 \\ \end{align} </math> </div></div> <ol start=40> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2-x+1}{x^2+x+1} &= \frac{5}{6} \\ \frac{x^2+1-x}{x^2+1+x} &= \frac{5}{6} \\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} &= \frac{5}{6} \\ \text{ misalkan } x+\frac{1}{x} &= y \\ \frac{y-1}{y+1} &= \frac{5}{6} \\ 6(y-1) &= 5(y+1) \\ 6y-6 &= 5y+5 \\ y &= 11 \\ x+\frac{1}{x} &= 11 \\ \end{align} </math> </div></div> <ol start=41> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt{x}+x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x}+x &= 1 \\ x-1 &= -\sqrt{x} \\ (x-1)^2 &= (-\sqrt{x})^2 \\ x^2-2x+1 &= x \\ x^2-3x+1 &= 0 \\ x-3+\frac{1}{x} &= 0 \\ x+\frac{1}{x} &= 3 \\ \end{align} </math> </div></div> <ol start=42> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt[3]{x}-\sqrt[3]{x-36}=3</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x}-\sqrt[3]{x-36} &= 3 \\ (\sqrt[3]{x}-\sqrt[3]{x-36})^3 &= 3^3 \\ x-(x-36)-3 \sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36}) &= 27 \\ 36-3 \sqrt[3]{x(x-36)}3 &= 27 \\ -9 \sqrt[3]{x(x-36)} &= -9 \\ \sqrt[3]{x(x-36)} &= 1 \\ x(x-36) &= 1 \\ x^2-36x-1 &= 0 \\ x-36-\frac{1}{x} &= 0 \\ x-\frac{1}{x} &= 36 \\ \end{align} </math> </div></div> <ol start=43> <li>Berapakah nilai dari <math>x+\frac{16}{x}</math> jika <math>x-3\sqrt{x}=4</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x-3\sqrt{x} &= 4 \\ x-4 &= 3\sqrt{x} \\ x^2-8x+16 &= 9x \\ x^2-17x+16 &= 0 \\ x-17+\frac{16}{x} &= 0 \\ x+\frac{16}{x} &= 17 \\ \end{align} </math> </div></div> <ol start=44> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+4}</math> jika <math>x^2-7x+2=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2-7x+2 &= 0 \\ x^2+2 &= 7x \\ x+\frac{2}{x} &= 7 \\ x^2+4+\frac{4}{x^2} &= 49 \\ x^2+\frac{4}{x^2} &= 45 \\ \frac{x^4+4}{x^2} &= 45 \\ \frac{x^2}{x^4+4} &= \frac{1}{45} \\ \end{align} </math> </div></div> <ol start=45> <li>Berapakah nilai dari <math>x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}</math> jika <math>x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}} &= 25 \\ x^{\frac{1}{2}}+x^{-\frac{1}{2}} &= 23 \\ x+2+x^{-1} &= 529 \\ x+x^{-1} &= 527 \\ x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+x^{-\frac{3}{4}} &= 110 \\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1} &= x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}} \\ &= 527+110 \\ &= 637 \\ \end{align} </math> </div></div> <ol start=46> <li>Berapakah nilai dari <math>\sqrt{8x^6+x^5+x^4+5x^3+1}</math> jika <math>\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5} &= 0 \\ \frac{x^2+x+1}{x^5} &= 0 \\ x^2+x+1 &= 0 \\ x^2+x+1 &= 0 \\ (x-1)(x^2+x+1) &= 0(x-1) \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ \sqrt{8x^6+x^5+x^4+5x^3+1} &= \sqrt{(2x^3)^2+x^3x^2+x^3x+5x^3+1} \\ &= \sqrt{(2(1))^2+(1)x^2+(1)x+5(1)+1} \\ &= \sqrt{(2)^2+x^2+x+5+1} \\ &= \sqrt{4+x^2+x+1+5} \\ &= \sqrt{4+0+5} \\ &= \sqrt{9} \\ &= 3 \\ \end{align} </math> </div></div> <ol start=47> <li>Berapakah nilai dari <math>f(1)+f(2)+f(3)+ \dots + f(99)</math> jika <math>f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{1}{\sqrt{x+1}+\sqrt{x}} \\ &= \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} \\ &= \sqrt{x+1}-\sqrt{x} \\ f(1)+f(2)+f(3)+ \dots + f(98)+f(99) &= \sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+ \cdot + \sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99} \\ &= \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+ \cdot + \sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ &= \sqrt{100}-\sqrt{1} \\ &= 10-1 \\ &= 9 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots + \frac{2024}{2025})</math> jika <math>h(x)=\frac{3}{3+9^x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} h(x) &= \frac{3}{3+9^x} \\ h(1-x) &= \frac{3}{3+9^{1-x}} \\ &= \frac{3}{3+\frac{9}{9^x}} \\ &= \frac{9^x}{3+9^x} \\ h(x)+h(1-x) &= \frac{3}{3+9^x}+\frac{9^x}{3+9^x} \\ &= \frac{3+9^x}{3+9^x} \\ &= 1 \\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots +(1-\frac{2}{2025})+(1-\frac{1}{2025})) \\ & 5(1+1+1+ \dots +1+1) \text{ sebanyak 1012 kali } \\ & 5(1012) \\ & 5060 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>\frac{7^{2025} - 7^{2023} + 432}{7^{2024} + 7^{2023} + 72}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72} &= \frac{7^{2023}7^{2}-7^{2023} + 48 \times 9}{7^{2023}7^1+7^{2023}+8 \times 9} \\ &= \frac{7^{2023}(7^{2}-1)+48 \times 9}{7^{2023}(7^1+1)+8 \times 9} \\ &= \frac{7^{2023}(49-1)+48 \times 9}{7^{2023}(7+1) + 8 \times 9} \\ &= \frac{7^{2023} \times 48+48 \times 9}{7^{2023} \times 8+8 \times 9} \\ &= \frac{48(7^{2023}+9)}{8(7^{2023}+9)} \\ &= \frac{48}{8} \\ &= 6 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>tan (x+\frac{\pi}{4})</math> jika <math>\frac{1}{cos x}-tan x = \frac{4}{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{cos x}-tan x &= \frac{4}{5} \\ sec x-tan x &= \frac{4}{5} \\ sec^2 x-tan^2 x &= 1 \\ (sec x+tan x)(sec x-tan x) &= 1 \\ (sec x+tan x)\frac{4}{5} &= 1 \\ sec x+tan x &= \frac{5}{4} \\ \text{kedua persamaan dengan cara metode eliminasi } \\ 2 tan x &= \frac{5}{4}-\frac{4}{5} \\ 2 tan x &= \frac{9}{20} \\ tan x &= \frac{9}{40} \\ tan (x+\frac{\pi}{4}) &= \frac{tan x+tan \frac{\pi}{4}}{1-tan x \cdot tan \frac{\pi}{4}} \\ &= \frac{\frac{9}{40}+1}{1-\frac{9}{40} \cdot 1} \\ &= \frac{\frac{49}{40}}{\frac{31}{40}} \\ &= \frac{49}{31} \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>sin^3 x+csc^3 x</math> jika <math>sin x-csc x = 8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ Dengan menggunakan rumus: } (a-b)^3 &= a^3-b^3-3ab(a-b) \\ (sin x-csc x)^3 &= sin^3 x-csc^3 x-3sin x csc x(sin x-csc x) \\ 8^3 &= sin^3 x-csc^3 x-3sin x (\frac{1}{sin x})(8) \\ 512 &= sin^3 x-csc^3 x-24 \\ sin^3 x-csc^3 x &= 512+24 \\ sin^3 x-csc^3 x &= 536 \\ \end{align} </math> </div></div> <ol start=52> <li>Berapakah nilai dari <math>(sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2</math> jika <math>\frac{1}{sin x}+\frac{1}{cos x} = 10</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{sin x}+\frac{1}{cos x} &= 10 \\ \frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} &= 100 \\ (sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2 &= sin^2 x+\frac{2sin x}{cos x}+\frac{1}{cos^2 x}+cos^2 x+\frac{2cos x}{sin x}+\frac{1}{sin^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2(sin^2 x+cos^2 x)}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+100 \\ &= 101 \\ \end{align} </math> </div></div> <ol start=53> <li>Berapakah nilai dari (x-1)⁶ jika <math>x=\frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin 80^\circ &= cos 10^\circ \\ sin 80^\circ-cos 10^\circ &= 0 \\ x &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+2 sin 20^\circ cos 20^\circ}{cos 10^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+4 sin 10^\circ cos 10^\circ cos 20^\circ}{cos 10^\circ} \\ &= 4 cos 55^\circ cos 25^\circ+4 sin 10^\circ cos 20^\circ \\ &= 2(2 cos 55^\circ cos 25^\circ+2 sin 10^\circ cos 20^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ+sin (-10)^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ-sin 10^\circ) \\ &= 2(cos 80^\circ-sin 10^\circ+cos 30^\circ+sin 30^\circ) \\ &= 2(cos 80^\circ-sin (90^\circ-80^\circ)+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(cos 80^\circ-cos 80^\circ+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= \sqrt{3}+1 \\ x-1 &= \sqrt{3} \\ (x-1)^6 &= (\sqrt{3})^6 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=54> <li>Berapakah nilai dari x jika <math>x=\frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ} \\ 2x sin 20^\circ-x sin 40 ^\circ &= x sin 20^\circ-x^2 sin 10^\circ \\ x^2 sin 10^\circ+x sin 20^\circ-x sin 40 ^\circ &= 0 \\ x(x sin 10^\circ+sin 20^\circ-sin 40 ^\circ) &= 0 \\ x = 0 &\text{ atau } x sin 10^\circ+sin 20^\circ-sin 40 ^\circ = 0 \\ x sin 10^\circ+sin 20^\circ-sin 40 ^\circ &= 0 \\ x sin 10^\circ &= sin 40 ^\circ-sin 20^\circ \\ x &= \frac{sin 40 ^\circ-sin 20^\circ}{sin 10^\circ} \\ &= \frac{2 cos 30 ^\circ sin 10^\circ}{sin 10^\circ} \\ &= 2 cos 30 ^\circ \\ &= \frac{2 \sqrt{3}}{2} \\ &= \sqrt{3} \\ \end{align} </math> </div></div> <ol start=55> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x^2}{x^2-16y^2} = \frac{625}{49}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2}{x^2-16y^2} &= \frac{625}{49} \\ \frac{x^2-16y^2}{x^2} &= \frac{49}{625} \text{ (terbalik posisinya)} \\ 1-\frac{16y^2}{x^2} &= \frac{49}{625} \\ \frac{16y^2}{x^2} &= 1 - \frac{49}{625} \\ (\frac{4y}{x})^2 &= \frac{576}{625} \\ (\frac{4y}{x})^2 &= (\frac{24}{25})^2 \\ \frac{4y}{x} &= \frac{24}{25} \\ \frac{y}{x} &= \frac{6}{25} \\ \frac{x}{y} &= \frac{25}{6} \\ \end{align} </math> </div></div> <ol start=56> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x}{y}+\frac{x+10y}{y+10x} = 2</math> serta bilangan real untuk x dan y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x}{y}+\frac{x+10y}{y+10x} &= 2 \\ \frac{x}{y}+\frac{\frac{x}{y}+10}{1+10\frac{x}{y}} &= 2 \\ \text{misalkan } \frac{x}{y} = a \\ a+\frac{a+10}{1+10a} &= 2 \\ a(1+10a)+a+10 &= 2(1+10a) \\ 10a^2+a+a+10 &= 2+20a \\ 10a^2-18a+8 &= 0 \\ 5a^2-9a+4 &= 0 \\ (5a-4)(a-1) &= 0 \\ a = \frac{4}{5} &\text{ atau } a = 1 \\ \text{jadi } \frac{x}{y} = {\frac{4}{5}, 1} \\ \end{align} </math> </div></div> <ol start=57> <li>Berapakah nilai dari xy jika <math>x^4+y^4+x^2y^2=15 \text{ dan } x^2+y^2+xy=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+xy &= 5 \\ x^2+y^2 &= 5-xy \\ x^4+y^4+x^2y^2 &= 15 \\ (x^2)^2+(y^2)^2+2x^2y^2-x^2y^2 &= 15 \\ (x^2+y^2)^2-x^2y^2 &= 15 \\ (5-xy)^2-x^2y^2 &= 15 \\ 25-10xy+x^2y^2-x^2y^2 &= 15 \\ 25-10xy &= 15 \\ 10xy &= 10 \\ xy &= 1 \\ \end{align} </math> </div></div> <ol start=58> <li>Berapakah nilai dari x jika <math>4^x = 63(4^3+1)(4^6+1)(4^{12}+1)+1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 4^x &= 63(4^3+1)(4^6+1)(4^{12}+1)+1 \\ 4^x-1 &= 63(4^3+1)(4^6+1)(4^{12}+1) \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{4^3-1} \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{63} \\ &= (4^3+1)(4^6+1)(4^{12}+1)(4^3-1) \\ &= (4^3-1)(4^3+1)(4^6+1)(4^{12}+1) \\ &= (4^6-1)(4^6+1)(4^{12}+1) \\ &= (4^{12}-1)(4^{12}+1) \\ &= 4^{24}-1 \\ 4^x &= 4^{24} \\ x &= 24 \\ \end{align} </math> </div></div> <ol start=59> <li>Berapakah nilai dari <math>\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}</math> jika <math>x=\sqrt{9+4\sqrt{5}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \sqrt{9+4\sqrt{5}} \\ x &= 2+\sqrt{5} \\ x^2 &= 9+4\sqrt{5} \\ x^2-4x &= 9+4\sqrt{5}-4(2+\sqrt{5}) \\ x^2-4x &= 1 \\ x^2 &= 4x+1 \\ x^3 &= x \cdot x^2 \\ &= x(4x+1) \\ &= 4x^2+x \\ &= 4(4x+1)+x \\ &= 16x+4+x \\ &= 17x+4 \\ x^4 &= x \cdot x^3 \\ &= x(17x+4) \\ &= 17x^2+4x \\ &= 17(4x+1)+4x \\ &= 68x+17+4x \\ &= 72x+17 \\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1} &= \frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1} \\ &= \frac{72x+17-85x-20+8x+2+5x+3}{2} \\ &= \frac{2}{2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=60> <li>Berapakah nilai dari <math>\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}</math> jika 2x-1=<math>\sqrt{61}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \frac{x^3+1}{x^5-x^4-x^3+x^2} = p \\ p &= \frac{x^3+1}{x^5-x^4-x^3+x^2} \\ &= \frac{x^3+1}{x^5-x^4-(x^3-x^2)} \\ &= \frac{x^3+1}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)} \\ &= \frac{x^2-x+1}{x^2(x-1)^2} \\ &= \frac{x^2-x+1}{(x(x-1))^2} \\ &= \frac{x(x-1)+1}{(x(x-1))^2} \\ 2x-1 &= \sqrt{61} \\ x &= \frac{\sqrt{61}+1}{2} \\ x-1 &= \frac{\sqrt{61}-1}{2} \\ x(x-1) &= (\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2}) \\ &= \frac{61-1}{4} \\ &= \frac{60}{4} \\ &= 15 \\ p &= \frac{x(x-1)+1}{(x(x-1))^2} \\ &= \frac{15+1}{15^2} \\ &= \frac{16}{15^2} \\ \sqrt{p} &= \sqrt{\frac{16}{15^2}} \\ &= \frac{4}{15} \\ \end{align} </math> </div></div> <ol start=61> <li>Berapakah nilai dari <math>(\frac{x-3}{x})^{25}</math> jika <math>x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\sqrt[5]{8}+\sqrt[5]{2} &= 1+\sqrt[5]{16}+\sqrt[5]{4} \\ x+(\sqrt[5]{2})^3+\sqrt[5]{2} &= 1+(\sqrt[5]{2})^4+(\sqrt[5]{2})^2 \\ x &= (\sqrt[5]{2})^4-(\sqrt[5]{2})^3+(\sqrt[5]{2})^2-\sqrt[5]{2}+1 \\ \text{misalkan } \sqrt[5]{2} = p \\ x &= p^4-p^3+p^2-p+1 \\ x &= \frac{p^5+1}{p+1} \\ (\frac{x-3}{x})^{25} &= (1-\frac{3}{x})^{25} \\ &= (1-\frac{3}{\frac{p^5+1}{p+1}})^{25} \\ &= (1-\frac{3(p+1)}{p^5+1})^{25} \\ &= (1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2})^5+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{2+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-3\sqrt[5]{2}-3)}{3})^{25} \\ &= (-\sqrt[5]{2})^{25} \\ &= (-2)^5 \\ &= -32 \\ \end{align} </math> </div></div> <ol start=62> <li>Berapakah nilai dari <math>x^{50}+x^{49}+x^{48}+x^{47}+x^{46}</math> jika <math>x^2+x+1=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+x+1 &= 0 \\ x^2+x &= -1 \\ \frac{x^3-1}{x-1} &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46} &= x^{48}(x^2+x+1)+x^{45}(x^2+x) \\ &= x^{48}(0)+(x^3)^{15}(-1) \\ &= 0+(1)^{15}(-1) \\ &= -1 \\ \end{align} </math> </div></div> <ol start=63> <li>Berapakah 2²⁴ dari <math>8^7+8^6+8^5+8^4+8^3+8^2+8+1=A</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= A \\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1) &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8 &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= 8A+1 \\ 8^8+A &= 8A+1 \\ 8^8 &= 7A+1 \\ (2^3)^8 &= 7A+1 \\ 2^{24} &= 7A+1 \\ \end{align} </math> </div></div> <ol start=64> <li>Berapakah nilai dari <math>x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1</math> jika <math>x+\frac{1}{x}=\sqrt{3}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{x} &= \sqrt{3} \\ x^2+2+\frac{1}{x^2} &= 3 \\ x^2-1+\frac{1}{x^2} &= 0 \\ x^2(x^2-1+\frac{1}{x^2}) &= x^2(0) \\ x^4-x^2+1 &= 0 \\ (x^2+1)(x^4-x^2+1) &= (x^2+1)0 \\ x^6-x^4+x^2+x^4-x^2+1 &= 0 \\ x^6+1 &= 0 \\ x^6 &= -1 \\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1 &= {x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1 \\ &= (-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2-1+1 \\ &= -1+1-1+1-1+1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=65> <li>Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari <math>\frac{c-b}{a-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= ax^2+bx+c \\ f(2) &= a(2)^2+2b+c = 4 \\ &= 4a+2b+c = 4 \\ f(7) &= a(7)^2+7b+c = 49 \\ &= 49a+7b+c = 49 \\ 49a+7b+c &= 49 \\ 4a+2b+c &= 4 \\ 45a+5b &= 45 \text{ (f(7) dikurangi f(2)) } \\ 9a+b &= 9 \\ b &= -9a+9 \\ 4a+2b+c &= 4 \\ 4a+2(-9a+9)+c &= 4 \\ 4a-18a+18+c &= 4 \\ -14a+18+c &= 4 \\ c &= 14a-14 \\ \frac{c-b}{a-1} &= \frac{14a-14-(-9a+9)}{a-1} \\ &= \frac{14(a-1)+9(a-1)}{a-1} \\ &= \frac{(14+9)(a-1)}{a-1} \\ &= 23 \\ \end{align} </math> </div></div> <ol start=66> <li>Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+y)^3 &= x^3+y^3+3xy(x+y) \\ 11^3 &= 242+3xy(11) \text{ (dibagi 11)} \\ 11^2 &= 22+3xy \\ 121 &= 22+3xy \\ 99 &= 3xy \\ xy &= 33 \\ (x-y)^2 &= x^2+y^2-2xy \\ &= ((x+y)^2-2xy)-2xy \\ &= (x+y)^2-4xy \\ &= 11^2-4(33) \\ &= 121-132 \\ &= -11 \\ \end{align} </math> </div></div> <ol start=67> <li>Berapa f(1)+f(-1) jika <math>f(\frac{ax-b}{bx-a})</math>=x²-5x+6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ jika} f(1) = f(\frac{ax-b}{bx-a}) \\ 1 &= \frac{ax-b}{bx-a} \\ bx-a &= ax-b \\ (b-a)x &= -b+a \\ &= -(b-a) \\ &= -1 \\ f(1) &= x^2-5x+6 \\ &= (-1)^2-5(-1)+6 \\ &= 12 \\ \text{ jika} f(-1) = f(\frac{ax-b}{bx-a}) \\ -1 &= \frac{ax-b}{bx-a} \\ -(bx-a) &= ax-b \\ -bx+a &= ax-b \\ (-b-a)x &= -b-a \\ &= 1 \\ f(-1) &= x^2-5x+6 \\ &= (1)^2-5(1)+6 \\ &= 2 \\ f(1)+f(-1) &= 12+2 \\ &= 14 \\ \end{align} </math> </div></div> <ol start=68> <li>berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)-x &= f(x-1) \\ f(x)-f(x-1) &= x \\ x=1 ; f(1)-f(0) &= 1 \\ x=2 ; f(2)-f(1) &= 2 \\ x=3 ; f(3)-f(2) &= 3 \\ x=4 ; f(4)-f(3) &= 4 \\ \dots \\ x=200 ; f(200)-f(199) &= 200 \\ \text{ jumlahkan tersebut menjadi } \\ f(200)-f(0) &= 1+2+3+4+\dots+200 \\ &= \frac{200 \cdot 201}{2} \\ &= 20.100 \\ f(200)-1 &= 20.100 \\ &= 20.101 \\ \end{align} </math> </div></div> <ol start=69> <li>Misalkan f(x) adalah fungsi rekursif yang berlaku ∀x ∈ R sebagai berikut: : f(x)+f(15-x) = 2024 : f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)+f(15-x) &= 2024 \\ f(15+x) &= f(x)+2020 \\ *cara 1 \\ \text{ganti x dengan 15+x } \\ f(15+x)+f(-x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ *cara 2 \\ \text{ganti x dengan -x } \\ f(-x)+f(15+x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ \end{align} </math> </div></div> <ol start=70> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>2f(\frac{2002}{x}) + f(x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2f(\frac{2002}{x}) + f(x) &= 3x \\ \text{ganti x dengan 2 } \\ 2f(\frac{2002}{2}) + f(2) &= 3(2) \\ 2f(1001) + f(2) &= 6 \\ \text{ganti x dengan 1001 } \\ 2f(\frac{2002}{1001}) + f(1001) &= 3(1001) \\ 2f(2) + f(1001) &= 3003 \\ 2f(2) + f(1001) &= 3003 \\ f(1001) &= 3003 - 2f(2) \\ 2f(1001) + f(2) &= 6 \\ 2(3003 - 2f(2)) + f(2) &= 6 \\ 6006 - 4f(2) + f(2) &= 6 \\ 3f(2) &= 6000 \\ f(2) &= 2000 \\ \end{align} </math> </div></div> <ol start=71> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>f(\frac{1}{x}) + \frac{1}{x}f(-x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{1}{x})+\frac{1}{x}f(-x) &= 3x \\ \text{ganti x dengan 1/3 } \\ f(3)+3f(-\frac{1}{3}) &= 1 \\ \text{ganti x dengan -3 } \\ f(-\frac{1}{3}) - \frac{1}{3}f(3) &= -9 \\ \text{dikalikan 3 } \\ 3f(-\frac{1}{3})-f(3) &= -27 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ 2f(3) &= 28 \\ f(3) &= 14 \\ \end{align} </math> </div></div> <ol start=72> <li>Diketahui polinom <math>f(7^b-1)=7^{3b}-10</math>. tentukan nilai f(5)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(5) &= f(7^b-1) \\ 5 &= 7^b-1 \\ 7^b &= 6 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(6-1) &= 6^3-10 \\ f(5) &= 216-10 \\ &= 206 \\ *cara 2 \\ \text{misalkan } 7^b-1=a \text{ maka } 7^b=a+1 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(a) &= (a+1)^3-10 \\ f(5) &= (5+1)^3-10 \\ &= 6^3-10 \\ &= 216-10 \\ &= 206 \\ \end{align} </math> </div></div> <ol start=73> <li>Diketahui polinom <math>f(6^b-7)=6^{3b}-2 \cdot 6^{2b}-4</math>. tentukan nilai f(-2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(-2) &= f(6^b-7) \\ -2 &= 6^b-7 \\ 6^b &= 5 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(5-7) &= 5^3-2 \cdot 5^2-4 \\ f(-2) &= 125-50-4 \\ &= 71 \\ *cara 2 \\ \text{misalkan } 6^b-7=a \text{ maka } 6^b=a+7 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(a) &= (a+7)^3-2(a+7)^2-4 \\ f(-2) &= (-2+7)^3-2(-2+7)^2-4 \\ &= 5^3-2(5)^2-4 \\ &= 125-50-4 \\ &= 71 \\ \end{align} </math> </div></div> <ol start=74> <li>Jika <math>f(xy)=\frac{f(x)}{y}</math> dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(10) &= 7 \\ f(2 \cdot 5) &= 7 \\ f(xy) &= \frac{f(x)}{y} \\ f(2 \cdot 5) &= \frac{f(2)}{5} \\ 7 &= \frac{f(2)}{5} \\ f(2) &= 35 \\ \end{align} </math> </div></div> <ol start=75> <li>Jika <math>f(xy)=\frac{f(x+y)}{xy}</math> dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(15) &= 16 \\ f(3 \cdot 5) &= 16 \\ f(xy) &= \frac{f(x+y)}{xy} \\ f(3 \cdot 5) &= \frac{f(3+5)}{3 \cdot 5} \\ f(15) &= \frac{f(8)}{15} \\ 16 &= \frac{f(8)}{15} \\ f(8) &= 240 \\ \end{align} </math> </div></div> <ol start=76> <li>Jika <math>f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15</math> maka tentukan nilai f(16)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x+\frac{1}{x}+6) &= x^2+\frac{1}{x^2}+15 \\ &= (x+\frac{1}{x})^2-2+15 \\ &= (x+\frac{1}{x})^2+13 \\ \text{misalkan } x+\frac{1}{x} &= p \\ f(x+\frac{1}{x}+6) &= (x+\frac{1}{x})^2+13 \\ f(p+6) &= p^2+13 \\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 } \\ f(p+6) &= p^2+13 \\ f(10+6) &= 10^2+13 \\ f(16) &= 100+13 \\ &= 113 \\ \end{align} </math> </div></div> <ol start=77> <li>Tentukan nilai x jika <math>f(x)=\frac{4}{4-x}</math> dan <math>f(x \cdot f(x))^{\frac{f(4x)}{f(x)}}=256</math>!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{4}{4-x} \\ f(4x) &= \frac{4}{4-4x} \\ \frac{f(4x)}{f(x)} &= \frac{\frac{4}{4-4x}}{\frac{4}{4-x}} \\ &= \frac{4-x}{4-4x} \\ f(x \cdot f(x)) &= f(x(\frac{4}{4-x})) \\ &= f(\frac{4x}{4-x}) \\ &= \frac{4}{4-(\frac{4x}{4-x})} \\ &= \frac{4}{\frac{16-4x-4x}{4-x}} \\ &= \frac{4}{\frac{16-8x}{4-x}} \\ &= \frac{4(4-x)}{4(4-4x)} \\ &= \frac{4-x}{4-4x} \\ \text{misalkan } \frac{4-x}{4-4x} &= a \\ f(x \cdot f(x))^{\frac{f(4x)}{f(x)}} &= 256 \\ a^a &= 256 \\ a^a &= 4^4 \\ a &= 4 \\ \frac{4-x}{4-4x} &= 4 \\ 4-x &= 16-16x \\ 15x &= 12 \\ x &= \frac{4}{5} \\ \end{align} </math> </div></div> <ol start=78> <li>Fungsi <math>f(x) = \frac{kx}{2x+1} \text{dengan } x \neq -\frac{1}{2}</math>. Dengan f(f(x)) = x maka tentukan nilai k!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{kx}{2x+1} \\ f(f(x)) &= x \\ f(\frac{kx}{2x+1}) &= x \\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1} &= x \\ \frac{\frac{k^2x}{2x+1}}{\frac{2kx+2x+1}{2x+1}} &= x \\ \frac{k^2x}{2kx+2x+1} &= x \\ \frac{k^2}{2kx+2x+1} &= 1 \\ k^2 &= 2kx+2x+1 \\ k^2-2kx &= 2x+1 \\ k^2-2kx+x^2 &= x^2+2x+1 \\ (k-x)^2 &= (x+1)^2 \\ (k-x)^2-(x+1)^2 &= 0 \\ (k-x+x+1)(k-x-(x+1)) &= 0 \\ k=-1 &\text{ atau } k=2x+1 &\text{ (TM) } \\ \end{align} </math> </div></div> <ol start=79> <li>Jika n = 2023²+2024² maka berapa hasil dari <math>\sqrt{2n-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n &= 2023^2+2024^2 \\ &= 2023^2+(2023+1)^2 \\ \text{misalkan 2023 = p} \\ n &= p^2+(p+1)^2 \\ &= p^2+p^2+2p+1 \\ &= 2p^2+2p+1 \\ \sqrt{2n-1} &= \sqrt{2(2p^2+2p+1)-1} \\ &= \sqrt{4p^2+4p+2-1} \\ &= \sqrt{4p^2+4p+1} \\ &= \sqrt{(2p+1)^2} \\ &= 2p+1 \\ &= 2(2023)+1 \\ &= 4046+1 \\ &= 4047 \\ \end{align} </math> </div></div> <ol start=80> <li>Tentukan nilai dari a+b+c merupakan bilangan bulat positif jika ab = 2, bc = 3 dan ac = 6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab \cdot bc \cdot ac &= 2 \cdot 3 \cdot 6 \\ (abc)^2 &= 36 \\ abc &= \pm 6 \\ abc &= 6 \\ \frac{abc}{ab} &= c = \frac{6}{2} = 3 \\ \frac{abc}{bc} &= a = \frac{6}{3} = 2 \\ \frac{abc}{ac} &= b = \frac{6}{6} = 1 \\ a+b+c &= 6 \\ \end{align} </math> </div></div> <ol start=81> <li>Tentukan nilai dari (a-c)^b jika <math>\frac{ab}{a+b} = \frac{1}{3}</math>, <math>\frac{bc}{b+c} = \frac{1}{4}</math> dan <math>\frac{ac}{a+c} = \frac{1}{9}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{ab}{a+b} &= \frac{1}{3} \\ \frac{a+b}{ab} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{b} + \frac{1}{a} &= 3 \\ \frac{bc}{b+c} &= \frac{1}{4} \\ \frac{b+c}{bc} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{b} &= 4 \\ \frac{ac}{a+c} &= \frac{1}{9} \\ \frac{a+c}{ac} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{a} &= 9 \\ \text{Misalkan 1/a = x, 1/b = y dan 1/c = z} \\ x+y &= 3 \\ y+z &= 4 \\ x+z &= 9 \\ x+y &= 3 \\ y+z &= 4 \\ x-z &= -1 \\ x-z &= -1 \\ x+z &= 9 \\ 2x &= 8 \\ x &= 4 \\ x-z &= -1 \\ 4-z &= -1 \\ z &= 5 \\ x+y &= 3 \\ 4+y &= 3 \\ y &= -1 \\ \frac{1}{a} &= 4 \\ a &= \frac{1}{4} \\ \frac{1}{b} &= -1 \\ b &= -1 \\ \frac{1}{c} &= 5 \\ c &= \frac{1}{5} \\ (a-c)^b &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=82> <li>Tentukan nilai dari a, b dan c jika <math>\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}</math> dan a+2b+3c=28?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan k untuk semua ketiga persamaan tersebut } \\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5} &= k \\ a+b &= 2k \\ a+c &= 4k \\ b+c &= 5k \\ 2a+b+c &= 6k \\ 2a+5k &= 6k \\ k &= 2a \\ a &= \frac{k}{2} \\ b &= \frac{3k}{2} \\ c &= \frac{7k}{2} \\ a+2b+3c &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ a &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ b &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ c &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=83> <li>Tentukan nilai dari (b+c)^a jika <math>\frac{a+b+c}{2} = \sqrt{a-2}+\sqrt{b-1}+\sqrt{c}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{a+b+c}{2} &= \sqrt{a-2}+\sqrt{b-1}+\sqrt{c} \\ a+b+c &= 2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}) \\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c} &= 0 \\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1 &= 0 \\ (\sqrt{a-2}-1)^2+(\sqrt{b-1}-1)^2+(\sqrt{c}-1)^2 &= 0 \\ (\sqrt{a-2}-1)^2 &= 0 \\ \sqrt{a-2}-1 &= 0 \\ \sqrt{a-2} &= 1 \\ a-2 &= 1 \\ a &= 3 \\ (\sqrt{b-1}-1)^2 &= 0 \\ \sqrt{b-1}-1 &= 0 \\ \sqrt{b-1} &= 1 \\ b-1 &= 1 \\ b &= 1 \\ (\sqrt{c}-1)^2 &= 0 \\ \sqrt{c}-1 &= 0 \\ \sqrt{c} &= 1 \\ c &= 1 \\ (b+c)^a &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=84> <li>x dan y merupakan bilangan tak nol. Jika xy = <math>\frac{x}{y}</math> = x-y maka berapa nilai x+y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy &= \frac{x}{y} \\ y^2 &= 1 \\ y^2 - 1 &= 0 \\ (y-1)(y+1) &= 0 \\ y = 1 &\text{ atau } y = -1 \\ \frac{x}{y} &= x-y \\ x &= xy-y^2 \\ x-xy &= -y^2 \\ x(1-y) &= -y^2 \\ x &= \frac{-y^2}{1-y} \\ \text{cek y=1 } \\ x &= \frac{-1^2}{1-1} \\ \text{tidak memenuhi syarat } \\ \text{cek y=-1 } \\ x &= \frac{-(-1)^2}{1-(-1)} \\ &= \frac{-1}{2} \\ x+y &= -1-\frac{1}{2} \\ &= -\frac{3}{2} \\ \end{align} </math> </div></div> <ol start=86> <li>Berapa nilai x dari <math>(\frac{a}{b})^3+(\frac{b}{a})^3 = 2\sqrt{x}</math> jika <math>\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b} \\ \frac{a+b}{ab} &= \frac{1}{a+b} \\ (a+b)^2 &= ab \\ a^2+2ab+b^2 &= ab \\ a^2+b^2 &= -ab \\ \text{misalkan } \frac{a}{b}+\frac{b}{a} = n \\ \frac{a}{b}+\frac{b}{a} &= n \\ \frac{a^2+b^2}{ab} &= n \\ a^2+b^2 &= nab \\ n &= -1 \\ \frac{a}{b}+\frac{b}{a} &= n \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3(\frac{a}{b}+\frac{b}{a}) &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3n &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3 &= n^3-3n \\ &= (-1)^3-3(-1) \\ &= 2 \\ 2\sqrt{x} &= 2 \\ \sqrt{x} &= 1 \\ x &= 1 \\ \end{align} </math> </div></div> <ol start=87> <li>Berapa nilai m dari <math>x^2-mx-1=0</math> jika <math>\sqrt[3]{x_1}+\sqrt[3]{x_2}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x_1} &= a \\ x_1 &= a^3 \\ \sqrt[3]{x_2} &= b \\ x_2 &= b^3 \\ \sqrt[3]{x_1}+\sqrt[3]{x_2} &= 1 \\ a+b &= 1 \\ x^2-mx-1 &= 0 \\ x_1+x_2 &= m \\ x_1 \cdot x_2 &= -1 \\ x_1+x_2 &= m \\ a^3+b^3 &= m \\ x_1 \cdot x_2 &= -1 \\ a^3 \cdot b^3 &= -1 \\ (ab)^2 &= (-1)^3 \\ ab &= -1 \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ (1)^3 &= m+3(-1)(1) \\ 1 &= m-3 \\ m &= 4 \\ \end{align} </math> </div></div> <ol start=88> <li>Berapa nilai <math>\frac{x_1}{x_2}</math> dari <math>ax^2-18x-b=0</math> jika <math>ab=45</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab &= 45 \\ b &= \frac{45}{a} \\ ax^2-18x-b &= 0 \\ ax^2-18x-\frac{45}{a} &= 0 \\ a^2x^2-18ax-45 &= 0 \\ (ax-3)(ax-15) &= 0 \\ ax-3 &= 0 \\ x &= \frac{3}{a} \\ ax-15 &= 0 \\ x &= \frac{15}{a} \\ \frac{x_1}{x_2} &= \frac{\frac{3}{a}}{\frac{15}{a}} \\ &= \frac{3}{15} \\ &= \frac{1}{5} \\ \frac{x_1}{x_2} &= \frac{\frac{15}{a}}{\frac{3}{a}} \\ &= \frac{15}{3} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=89> <li>Jika <math>\frac{u_3}{u_1+u_2} = \frac{7}{8}</math> merupakan barisan aritmetika maka berapa dari <math>\frac{u_2+u_3}{u_1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{u_3}{u_1+u_2} &= \frac{7}{8} \\ \frac{a+2b}{a+a+b} &= \frac{7}{8} \\ \frac{a+2b}{2a+b} &= \frac{7}{8} \\ 8(a+2b) &= 7(2a+b) \\ 8a+16b &= 14a+7b \\ 9b &= 6a \\ b &= \frac{2a}{3} \\ \frac{u_2+u_3}{u_1} &= \frac{a+b+a+2b}{a} \\ &= \frac{2a+3b}{a} \\ &= \frac{2a+3(\frac{2a}{3})}{a} \\ &= \frac{2a+2a}{a} \\ &= \frac{4a}{a} \\ &= 4 \\ \end{align} </math> </div></div> <ol start=90> <li>Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7p+q}{2p+q} &= \frac{17p+q}{7p+q} \\ (7p+q)^2 &= (17p+q)(2p+q) \\ 49p^2+14pq+q^2 &= 34p^2+19pq+q^2 \\ 15p^2 &= 5pq \\ 3p &= q \\ \frac{7p+q}{2p+q} &= \frac{7p+3p}{2p+3p} \\ &= \frac{10p}{5p} \\ &= 2 \\ \end{align} </math> </div></div> <ol start=91> <li>Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{rataan geometris } \\ \sqrt{a \cdot b} &= b-24 \\ a \cdot b &= (b-24)^2 \\ \text{rataan aritmatik } \\ \frac{a+b}{2} &= a+15 \\ a+b &= 2(a+15) \\ a+b &= 2a+30 \\ a &= b-30 \\ a \cdot b &= (b-24)^2 \\ (b-30)b &= (b-24)^2 \\ b^2-30b &= b^2-48b+576 \\ 18b &= 576 \\ b &= 32 \\ a &= b-30 \\ &= 32-30 \\ &= 2 \\ a+b &= 32+2 \\ &= 34 \\ \end{align} </math> </div></div> <ol start=91> <li>Segitiga lancip ABC dengan <math>\frac{a^4+b^4+c^4+a^2b^2}{c^2(a^2+b^2)}=2</math>. tentukan nilai sudut C?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{syarat segitiga lancip semua sudut masing-masing kurang dari } 90^\circ \\ c^2 &= a^2+b^2-2ab cos C \\ cos C &= \frac{a^2+b^2-c^2}{2ab} \\ a^4+b^4+c^4+a^2b^2 &= 2c^2(a^2+b^2) \\ a^4+b^4+a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-2c^2(a^2+b^2)+(c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= (ab)^2 \\ a^2+b^2-c^2 &= \pm ab \\ cos C &= \pm \frac{ab}{2ab} \\ &= \pm \frac{1}{2} \\ &= \frac{1}{2} \text{ (karena sudut harus kurang dari } 90^\circ) \\ C &= 60^\circ \\ \end{align} </math> </div></div> <ol start=92> <li>Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan panjang DA dan EA } = x \text{ dan panjang AB } = y \\ \text{luas segitiga CAB } &= \frac{CA \cdot AB}{2} \\ 45 &= \frac{(x+9)(x+5)}{2} \\ 90 &= x^2+14x+45 \\ x^2+14x &= 45 \\ y^2 &= (x+9)^2+(x+5)^2 \\ &= x^2+18x+81+x^2+10x+25 \\ &= 2x^2+28x+106 \\ &= 2(x^2+14x)+106 \\ &= 2(45)+106 \\ &= 196 \\ y &= 14 \\ \end{align} </math> jadi panjang BC adalah 14 cm </div></div> <ol start=93> <li>Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kita cari ukuran GC yaitu } \\ \frac{GC}{AD} &= \frac{CE}{DE} \\ \frac{GC}{15} &= \frac{6}{18} \\ GC &= 5 \\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC } \\ &= \frac{1}{2} \cdot BC \cdot CF - \frac{1}{2} \cdot GC \cdot CE \\ &= \frac{1}{2} \cdot 15 \cdot 18 - \frac{1}{2} \cdot 5 \cdot 6 \\ &= 135 - 15 \\ &= 120 \\ \end{align} </math> jadi luas daerah BFEG adalah 120 cm² </div></div> <ol start=94> <li>Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ AI &= \frac{1}{5} AF \\ &= \frac{1}{5} 10 \\ &= 2 \\ IF &= AF-AI \\ &= 10-2 \\ &= 8 \\ \text{luas trapesium AFGD } &= \frac{(AD+EF) \cdot AF}{2} \\ &= \frac{(4+6)10}{2} \\ &= 50 \\ \text{luas segitiga AID } &= \frac{AI \cdot AF}{2} \\ &= \frac{(2)4}{2} \\ &= 4 \\ \text{luas segitiga IFG } &= \frac{IF \cdot FG}{2} \\ &= \frac{(8)6}{2} \\ &= 24 \\ \text{luas daerah segitiga IGD } &= \text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG } \\ &= 50-4-24 \\ &= 22 \\ \end{align} </math> jadi luas daerah segitiga IGD adalah 22 cm² </div></div> <ol start=96> <li>Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-<math>\pi</math>) cm³ maka berapa luas permukaan kerucut tersebut?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ \text{volume balok} \\ V_b &= x^2(12) \\ \text{volume kerucut} \\ V_b &= \frac{1}{3}\pi x^2(12) \\ &= 4\pi x^2 \\ V_{b-k} &= Vb-Vk \\ 100(3-\pi) &= 12x^2-4\pi x^2 \\ 100(3-\pi) &= 4x^2(3-\pi) \\ x^2 &= 25 \\ x &= 5 \\ s &= \sqrt{12^2+5^2} \\ &= \sqrt{144+25} \\ &= \sqrt{169} \\ &= 13 \\ \text{luas permukaan kerucut } &= \pi r(r+s) \\ &= \pi(5)(5+13) \\ &= 90\pi \\ \end{align} </math> jadi luas daerah permukaan kerucut adalah 90<math>\pi</math> cm² </div></div> <ol start=97> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 3a+1 \\ B &= 3b+2 \\ A(A+1)+3B \\ (3a+1)(3a+1+1)+3(3b+2) \\ (3a+1)(3a+2)+9b+6 \\ 9a^2+9a+2+9b+6 \\ 9a^2+9a+9b+8 \\ 9(a^2+a+b)+8 \\ \text{sisa pembagiannya adalah } 8 \\ \end{align} </math> </div></div> <ol start=98> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 9a+7 \\ B &= 9b+8 \\ A(A-5)+9B \\ (9a+7)(9a+7-5)+9(9b+8) \\ (9a+7)(9a+2)+81b+72 \\ 81a^2+81a+14+81b+72 \\ 81a^2+81a+81b+86 \\ 81a^2+81a+81b+81+5 \\ 81(a^2+a+b+1)+5 \\ \text{sisa pembagiannya adalah } 5 \\ \end{align} </math> </div></div> <ol start=99> <li>Jika <math>\begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix}</math> maka berapa hasil dari A²¹+A²⁵+A⁴⁶?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A^2 &= A \cdot A \\ &= \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \\ A^3 &= A^2 \cdot A \\ &= \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ &= - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= -I \\ A^{21}+A^{25}+A^{46} &= A^{21} \cdot (I+A^4+A^{25}) \\ &= A^{21} \cdot (I+A^3 \cdot A +A^{24} \cdot A) \\ &= (A^3)^7 \cdot (I+A^3 \cdot A +(A^3)^8 \cdot A) \\ &= (-I)^7 \cdot (I-I \cdot A +(-I)^8 \cdot A) \\ &= -I \cdot (I-A+A) \\ &= -I \cdot I \\ &= -I \\ &= -\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ \end{align} </math> </div></div> <ol start=100> <li>Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida?</li></ol> : bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15 : ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9). : ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8) : maka Ida menulis bilangan 8. <ol start=101> <li>Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9?</li></ol> : Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu. : untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja. : untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja. : Jadi banyaknya bilangan mungkin 2. <ol start=102> <li>Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli!</li></ol> ;cara 1 # 8¹ = 1 # 8² = 1 (8²=8¹x8¹ sama dengan 1x1) # 8³ = 1 (8³=8¹x8² sama dengan 1x1) # 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2) # 8⁵ = 1 # 8ⁿ = 1 (semua n untuk bilangan asli) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli ;cara 2 # 8ⁿ = b mod 7 # 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil) # (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya) # 8ⁿ = 1ⁿ mod 7 # 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli <ol start=103> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 5?</li></ol> ;cara 1 # 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 # 7¹ = 7 (sisa 1) # 7² = 49 (sisa 2) # 7³ = 343 (sisa 3) # 7⁴ = 2,401 (sisa 0) # 7⁵ = 16,807 # 7⁶ = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3 ;cara 2 :17¹ = 2 :17² = 4 :17³ = 3 :17⁴ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3 :17⁹⁹ = (17⁴)²⁴ x 17³ Untuk 17⁴ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 3. Jadi 17⁹⁹ dibagi 7 bersisa 3 ;cara 3 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 17⁴ ::17⁴ = 1 mod 5 ::(17⁴)²⁴ = 1²⁴ mod 5 ::17⁹⁶ = 1²⁴ mod 5 ::17⁹⁶ = 1 mod 5 ::17⁹⁶ x 17³ = 1 x 17³ mod 5 ::17⁹⁹ = 17³ mod 5 ::17⁹⁹ = 17 x 17 x 17 mod 5 ::17⁹⁹ = 2 x 2 x 2 mod 5 ::17⁹⁹ = 8 mod 5 ::17⁹⁹ = 3 mod 5 Jadi 17⁹⁹ dibagi 5 bersisa 3 <ol start=104> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 7?</li></ol> ;cara 1 :17¹ = 3 :17² = 2 :17³ = 6 :17⁴ = 4 :17⁵ = 5 :17⁶ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3 :17⁹⁹ = (17⁶)¹⁶ x 17³ Untuk 17⁶ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 6. Jadi 17⁹⁹ dibagi 7 bersisa 6 ;cara 2 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 17⁶ ::17⁶ = 1 mod 7 ::(17⁶)¹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1 mod 7 ::17⁹⁶ x 17³ = 1 x 17³ mod 7 ::17⁹⁹ = 17³ mod 7 ::17⁹⁹ = 17 x 17 x 17 mod 7 ::17⁹⁹ = 3 x 3 x 3 mod 7 ::17⁹⁹ = 27 mod 7 ::17⁹⁹ = 6 mod 7 Jadi 17⁹⁹ dibagi 7 bersisa 6 <ol start=105> <li>Berapa hasil sisa dari 41²⁰²⁴ dibagi 33?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 41^{2024} &= 41^{2024} \text{ mod } 33 \\ &= (33 \times 3 + 2)^{2024} \text{ mod } 33 \\ &= 2^{2024} \text{ mod } 33 \\ &= 2^{2020} 2^4 \text{ mod } 33 \\ &= (2^5)^{404} 2^4 \text{ mod } 33 \\ &= (33 - 1)^{404} 2^4 \text{ mod } 33 \\ &= (-1)^{404} 2^4 \text{ mod } 33 \\ &= 2^4 \text{ mod } 33 \\ &= 16 \text{ mod } 33 \\ \text{Jadi hasil sisa adalah } 16 \\ \end{align} </math> </div></div> <ol start=106> <li>Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 99^{99} &= (3^2 \times 11)^{99} \\ &= 3^{198} \times 11^{99} \\ 243^n &= (3^5)^n \\ &= 3^{5n} \\ \text{agar bisa membagi, maka} \\ 5n &= 198 \\ n &= 39.6 \\ \text{jadi bilangan n terbesar adalah } 39 \\ \end{align} </math> </div></div> <ol start=107> <li>Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸?</lu></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 88^{88} &= (8 \times 11)^{88} \\ &= 8^{88} \times 11^{88} \\ &= 8^{87} \times 8 \times 11^{88} \\ &= (8^3)^{29} \times 8 \times 11^{88} \\ &= 512^{29} \times 8 \times 11^{88} \\ 512^n &= 512^{29} \\ \text{jadi bilangan n terbesar adalah } 29 \\ \end{align} </math> </div></div> <ol start=108> <li>Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6!</li></ol> ; cara 1 : KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. : N dibagi 3 sisa 1 : N dibagi 5 sisa 2 : N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a : KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70 : KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21 : KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97 ; cara 2 : Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35 : kemudian ditambahkan sisa masing-masing sesuai dengan KPK. : KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, <b>97</b> : KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, <b>97</b> : KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, <b>97</b> Jadi bilangan bulat positif adalah 97 :: NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 <ol start=109> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |- | 4 || 0 || Semua isi ember B dibuang |- | 1 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. <ol start=110> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |} nah ada ember A berisi 4 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |- | 1 || 0 || Semua isi ember B dibuang |- | 0 || 1 || Tuangkan 1 l dari ember A ke B sehingga ember A kosong |- | 3 || 1 || Isikan 3 l ke ember A |- | 0 || 4 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |} nah ada ember B berisi 4 liter. # Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kemungkinan 1} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 +3 \cdot ((\frac{4}{3})^x)^2 &= 7 \cdot (\frac{4}{3})^x \\ \text{misalkan } (\frac{4}{3})^x = y \\ 4 +3y^2 &= 7y \\ 3y^2-7y+4 &= 0 \\ (3y-4)(y-1) &= 0 \\ y_1 = \frac{4}{3} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{4}{3} \\ (\frac{4}{3})^x &= \frac{4}{3} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{4}{3})^x &= 1 \\ x &= 0 \\ \text{kemungkinan 2} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot ((\frac{3}{4})^x)^2 +3 &= 7 \cdot (\frac{3}{4})^x \\ \text{misalkan } (\frac{3}{4})^x = y \\ 4y^2 +3 &= 7y \\ 4y^2-7y+3 &= 0 \\ (4y-3)(y-1) &= 0 \\ y_1 = \frac{3}{4} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{3}{4} \\ (\frac{3}{4})^x &= \frac{3}{4} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{3}{4})^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-6}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{2x-4}{3}+2x-6 &= \frac{3}{2x-4}+\frac{1}{2x-6} \\ \frac{2x-4}{3}+2x-6 &= \frac{1}{\frac{2x-4}{3}}+\frac{1}{2x-6} \\ \text{misalkan } \frac{2x-4}{3} = y \text{ dan } 2x-6 = z \\ y+z &= \frac{1}{y}+\frac{1}{z} \\ y+z &= \frac{y+z}{yz} \\ y+z-\frac{y+z}{yz} &= 0 \\ (y+z)(1-\frac{1}{yz}) &= 0 \\ *y+z=0 \\ y+z &= 0 \\ y &= -z \\ \frac{2x-4}{3} &= -(2x-6) \\ \frac{2(x-2)}{3} &= -2(x-3) \\ \frac{x-2}{3} &= -x+3 \\ x-2 &= -3x+9 \\ 4x &= 11 \\ x &= \frac{11}{4} \\ *1-\frac{1}{yz}=0 \\ 1-\frac{1}{yz} &= 0 \\ yz-1 &= 0 \\ yz &= 1 \\ (\frac{2x-4}{3})(2x-6) &= 1 \\ \frac{4x^2-20x+24}{3} &= 3 \\ 4x^2-20x+24 &= 3 \\ 4x^2-20x+21 &= 0 \\ (2x-7)(2x-3) &= 0 \\ x = \frac{7}{2} &\text{ atau } x = \frac{3}{2} \\ \end{align} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 4hjv15xmd3y8uv14exytn23s7aobqdi 117498 117497 2026-07-13T05:17:03Z Akuindo 8654 117498 wikitext text/x-wiki contoh soal <ol start=1> <li>Berapa hasil dari <math>\sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Misalkan 2020 = p} \\ \sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64} &= \sqrt{(2020-5) \cdot (2020-3) \cdot (2020+3) \cdot (2020+5) + 64} \\ &= \sqrt{(p-5) \cdot (p-3) \cdot (p+3) \cdot (p+5) + 64} \\ &= \sqrt{(p-5) \cdot (p+5) \cdot (p-3) \cdot (p+3) + 64} \\ &= \sqrt{(p^2-25) \cdot (p^2-9) + 64} \\ &= \sqrt{p^4-34p^2+ 225 + 64} \\ &= \sqrt{p^4-34p^2+ 289} \\ &= \sqrt{(p^2-17)^2} \\ &= p^2-17 \\ &= 2020^2-17 \\ &= (2000+20)^2-17 \\ &= 4.000.000+80.000+400-17 \\ &= 4.080.383 \\ \end{align} </math> </div></div> <ol start=2> <li>Berapa nilai x dari <math>\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} = \frac{9}{10}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} &= \frac{9}{10} \\ \text{misalkan untuk } \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} = p \\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} &= p \\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}} &= p^2 \\ x^2-x-p &= p^2 \\ x^2-2x+1+x-1 &= p^2+p \\ (x-1)^2+(x-1) &= p^2+p \\ x-1 &= p \\ \text{misalkan untuk } \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}} &= q^3 \\ x^2 q &= q^3 \\ x^2 &= q^2 \\ x &= q \\ \frac{x-1}{x} &= \frac{9}{10} \\ x &= 10 \\ \end{align} </math> </div></div> <ol start=3> <li>Berapa nilai x dari <math>(\frac{x}{x+10})^{x+10}=\frac{1}{1024}</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{x+10}{x})^{-(x+10)} &= (1024)^{-1} \\ (\frac{x+10}{x})^{x+10} &= 1024 \\ (\frac{x+10}{x})^{x+10} &= 2^{10} \\ (\frac{x+10}{x})^{\frac{x+10}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (\frac{1}{2})^{-1} \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (1+(-\frac{1}{2}))^{(1+(-\frac{2}{1}))} \\ \frac{10}{x} &= -\frac{1}{2} \\ x &= -20 \\ \end{align} </math> </div></div> <ol start=4> <li>Berapa nilai x dari <math>x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{2}+\sqrt{x+\frac{1}{4}} &= (\sqrt{x+\frac{1}{4}})^2+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+(\frac{1}{2})^2 \\ &= (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 \\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}} &= 4 \\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2} &= 4 \\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 4 \\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 &= 4 \\ \sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 2 \\ \sqrt{x+\frac{1}{4}} &= \frac{3}{2} \\ x+\frac{1}{4} &= \frac{9}{4} \\ x &= 2 \\ \end{align} </math> </div></div> <ol start=5> <li>Berapa nilai x dari <math>\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0</math>?</li> </ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^3}{\sqrt{8-x^2}}+x^2-8 &= 0 \\ \frac{x^3}{\sqrt{8-x^2}} &= 8-x^2 \\ x^3 &= (8-x^2)^{\frac{3}{2}} \\ x &= (8-x^2)^{\frac{1}{2}} \\ x^2 &= 8-x^2 \\ 2x^2-8 &= 0 \\ x^2-4 &= 0 \\ (x-2)(x+2) &= 0 \\ \text{membuktikan } \\ x=2 \text{ maka hasilnya 0 } \\ x=-2 \text{ maka hasilnya -8 } \\ \text{jadi } x=2 \\ \end{align} </math> </div></div> <ol start=6> <li>Berapa nilai x dari <math>\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} = 5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}} &= 5 \\ \frac{x^{50}+x^{60}+x^{70}}{31}} &= 5^5 \\ x^{50}+x^{60}+x^{70} &= 5^5 \cdot 31 \\ x^{50}(1+x^{10}+x^{20}) &= 5^5 \cdot 31 \\ (x^{10}^5)(1+x^{10}+(x^{10}^2) &= 5^5 \cdot 31 \\ \text{ misalkan } x^{10} = a \\ a^5(1+a+a^2) &= 5^5 \cdot 31 \\ a &= 5 \\ x^{10} &= 5 \\ x &= ^5 log 10 \\ \end{align} </math> </div></div> <ol start=7> <li>Berapa nilai x dari <math>\sqrt{3x+5+\sqrt{4x+5}} = x</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{3x+5+\sqrt{4x+5}} &= x \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \text{misalkan } \sqrt{4x+5}=y \text{ dan } 4x+5=y^2 \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \sqrt{y^2+y-x} &= x \\ y^2+y &= x^2+x \\ y=x \\ 4x+5 &= y^2 \\ 4x+5 &= x^2 \\ x^2-4x-5 &= 0 \\ (x-5)(x+1) &= 0 \\ x=5 &\text{ atau } x=-1 \text{ (TM) } \\ \end{align} </math> </div></div> <ol start=8> <li>Berapa nilai x dari <math>\sqrt{1+\sqrt{1+x}} = \sqrt[3]{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{1+\sqrt{1+x}} &= \sqrt[3]{x} \\ \sqrt[3]{x} &= n \\ x &= n^3 \\ \sqrt{1+\sqrt{1+n^3}} &= n \\ 1+\sqrt{1+n^3} &= n^2 \\ \sqrt{1+n^3} &= n^2-1 \\ 1+n^3 &= n^4-2n^2+1 \\ n^4-n^3-2n^2 &= 0 \\ n^2(n^2-n-2) &= 0 \\ n^2(n-2)(n+1) &= 0 \\ n=0, n=2 \text{ atau } n=-1 \\ n &= 0 \\ x &= 0^3 \\ &= 0 \\ n &= 2 \\ x &= 2^3 \\ &= 8 \\ n &= -1 \\ x &= (-1)^3 \\ &= -1 \\ \text{yang paling mungkin untuk nilai x adalah } 8 \\ \end{align} </math> </div></div> <ol start=9> <li>Berapa nilai x dari <math>\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} &= \frac{\sqrt{1+x}}{\sqrt{x}} \\ \sqrt{x}(\sqrt{1+x}+\sqrt{x}) &= (\sqrt{1+x}-\sqrt{x})\sqrt{1+x} \\ \sqrt{x(1+x)}+x &= 1+x-\sqrt{x(1+x)} \\ 2\sqrt{x(1+x)} &= 1 \\ \sqrt{x(1+x)} &= \frac{1}{2} \\ x(1+x) &= \frac{1}{4} \\ x^2+x &= \frac{1}{4} \\ 4x^2+4x &= 1 \\ 4x^2+4x-1 &= 0 \\ x &= \frac{-4 \pm \sqrt{4^2-4(4)(-1)}}{2(4)} \\ &= \frac{-4 \pm \sqrt{32}}{8} \\ &= \frac{-4 \pm 4\sqrt{2}}{8} \\ &= \frac{-1 \pm \sqrt{2}}{2} \\ \text{karena akar x harus minimal nol jadi } x = \frac{-1+\sqrt{2}}{2} \\ \end{align} </math> </div></div> <ol start=10> <li>Berapa nilai x dari <math>\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x-\sqrt{x+1}}{x+\sqrt{x+1}} &= \frac{11}{19} \\ \text{misalkan } \sqrt{x+1}=y \text{ dan } x=y^2-1 \\ \frac{y^2-1-y}{y^2-1+y} &= \frac{11}{19} \\ 19(y^2-y-1) &= 11(y^2+y-1) \\ 19y^2-19y-19 &= 11y^2+11y-11 \\ 8y^2-30y-8 &= 0 \\ 4y^2-15y-4 &= 0 \\ (4y+1)(y-4) &= 0 \\ y=-\frac{1}{4} \text{ (TM) atau } & y=4 \\ x &= 4^2-1 \\ &= 15 \\ \end{align} </math> </div></div> <ol start=11> <li>Berapa nilai x dari <math>\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} &= 98 \\ \text{misalkan } \sqrt{x^2-1}=y \\ \frac{x+y}{x-y}+\frac{x-y}{x+y} &= 98 \\ \frac{(x+y)^2+(x-y)^2}{(x-y)(x+y)} &= 98 \\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} &= 98 \\ \frac{2(x^2+y^2)}{x^2-y^2} &= 98 \\ \frac{x^2+y^2}{x^2-y^2} &= 49 \\ x^2+y^2 &= 49(x^2-y^2) \\ x^2+y^2 &= 49x^2-49y^2 \\ 48x^2 &= 50y^2 \\ 24x^2 &= 25y^2 \\ 24x^2 &= 25(\sqrt{x^2-1})^2 \\ 24x^2 &= 25(x^2-1) \\ 24x^2 &= 25x^2-25 \\ x^2 &= 25 \\ x &= \pm 5 \\ \end{align} </math> </div></div> <ol start=12> <li>Berapa nilai x dari <math>\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \sqrt{x}=y \text{ dan } x=y^2 \\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} &= \frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\sqrt{\frac{y^2}{y^2+y}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\frac{y}{\sqrt{y^2+y}} \\ y^2+y-\sqrt{(y^2+y)(y^2-y)} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^4-y^2} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^2(y^2-1)} &= \frac{5}{4}y \\ y(y+1)-y\sqrt{y^2-1} &= \frac{5}{4}y \\ y+1-\sqrt{y^2-1} &= \frac{5}{4} \\ -\sqrt{y^2-1} &= \frac{1}{4}-y \\ y^2-1 &= (\frac{1}{4}-y)^2 \\ y^2-1 &= \frac{1}{16}-\frac{1}{2}y+y^2 \\ -1 &= \frac{1}{16}-\frac{1}{2}y \\ \frac{1}{2}y &= \frac{1}{16}+1 \\ \frac{1}{2}y &= \frac{17}{16} \\ y &= \frac{17}{8} \\ x &= (\frac{17}{8})^2 \\ &= \frac{289}{64} \\ \end{align} </math> </div></div> <ol start=13> <li>Berapa nilai x dari <math>\sqrt[4]{62+x}+\sqrt[4]{275-x}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ misalkan } \sqrt[4]{62+x}=a, 62+x=a^4, \sqrt[4]{275-x}=b \text{ dan } 275-x=b^4 \\ a+b &= 7 \\ (a+b)^2 &= 49 \\ a^2+b^2+2ab &= 49 \\ a^2+b^2 &= 49-2ab \\ a^4+b^4 &= 62+x+275-x \\ (a^2+b^2)^2-2(ab)^2 &= 337 \\ (49-2ab)^2-2(ab)^2 &= 337 \\ 2401-196ab+4(ab)^2-2(ab)^2 &= 337 \\ 2(ab)^2-196ab+2064 &= 0 \\ (ab)^2-98ab+1032 &= 0 \\ (ab-12)(ab-86) &= 0 \\ ab = 12 \text{ atau } & ab = 86 \text{ (TM) karena hasil kali maksimum yaitu 12 } \\ ab =12 \text{ dan } a+b=7 \\ a+b &= 7 \\ b &= 7-a \\ ab &= 12 \\ a(7-a) &= 12 \\ -a^2+7a &= 12 \\ a^2-7a+12 &= 0 \\ (a-3)(a-4) &= 0 \\ a=3 \text{ atau } & a=4 \\ a=3, b=4 \\ 62+x &= a^4 \\ 62+x &= (3)^4 \\ 62+x &= 81 \\ x &= 19 \\ a=4, b=3 \\ 62+x &= a^4 \\ 62+x &= (4)^4 \\ 62+x &= 256 \\ x &= 194 \\ \end{align} </math> </div></div> <ol start=14> <li>Berapa nilai x dari <math>\sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2}=7</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2} &= 7 \\ (\sqrt[3]{8+x})^2-\sqrt[3]{8+x} \sqrt[3]{27-x}+(\sqrt[3]{27-x})^2 &= 7 \\ \text{misalkan } \sqrt[3]{8+x}=a, 8+x=a^3, \sqrt[3]{27-x}=b \text{ dan } 27-x=b^3 \\ a^2-ab+b^2 &= 7 \\ a^3+b^3 &= 8+x+27-x \\ &= 35 \\ a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ 35 &= (a+b)(7) \\ a+b &= 5 \\ b &= 5-a \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ 5^3 &= 35+3ab(5) \\ 125 &= 35+15ab \\ 80 &= 15ab \\ ab &= 6 \\ a(5-a) &= 6 \\ 5a-a^2 &= 6 \\ a^2-5a+6 &= 6 \\ (a-2)(a-3) &= 6 \\ a=2 &\text{ atau } a=3 \\ a=2, b=3 \text{ dan } a=3,b=2 \\ 8+x &= a^3 \\ &= 2^3 \\ &= 8 \\ x &= 0 \\ 8+x &= a^3 \\ &= 3^3 \\ &= 27 \\ x &= 19 \\ \end{align} </math> </div></div> <ol start=15> <li>Berapa nilai x dari <math>3^x+5^x-9^x+15^x-25^x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3^x+5^x-9^x+15^x-25^x &= 1 \\ 3^x+5^x-(3^2)^x+(3 \cdot 5)^x-(5^2)^x &= 1 \\ 3^x+5^x-(3^x)^2+(3^x \cdot 5^x)-(5^x)^2 &= 1 \\ \text{misalkan } 3^x=a \text{ dan } 5^x=b \\ a+b-a^2+ab-b^2 &= 1 \\ a^2-ab+b^2-a-b+1 &= 0 \\ 2a^2-2ab+2b^2-2a-2b+2 &= 0 \\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1 &= 0 \\ (a-b)^2+(a-1)^2+(b-1)^2 &= 0 \\ a-b=0; a-1=0; b-1 &= 0 \\ a=b &= 1 \\ 3^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> <ol start=16> <li>Berapa nilai x dari <math>^6log x^2+^{6x}log \frac{6}{x}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ \text{misalkan } 6x=a \text{ maka } x=\frac{a}{6} \\ ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ ^6log (\frac{a}{6})^2+^{6 \frac{a}{6}}log \frac{6}{\frac{a}{6}} &= 1 \\ ^6log \frac{a^2}{6^2}+^alog \frac{6^2}{a} &= 1 \\ ^6log a^2-^6log 6^2+^alog 6^2-^alog a &= 1 \\ 2 ^6log a-2 ^6log 6+2 ^alog 6-^alog a &= 1 \\ 2 ^6log a-2+2 \frac{1}{^6log a}-1 &= 1 \\ 2 ^6log a+2 \frac{1}{^6log a}-4 &= 0 \\ 2 ^6log^2 a-4 ^6log a+2 &= 0 \\ ^6log^2 a-2 ^6log a+1 &= 0 \\ (^6log a-1)^2 &= 0 \\ ^6log a &= 1 \\ a &= 6 \\ x &= \frac{a}{6} \\ &= \frac{6}{6} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=17> <li>Berapa nilai x dari (x+500)³+x=20?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+500)^3+x &= 20 \\ \text{misalkan } a=x+500 \text{ maka } x=a-500 \\ a^3+a-500 &= 20 \\ a^3+a &= 520 \\ a(a^2+1) &= 8 \cdot 65 \\ a(a^2+1) &= 8(64+1) \\ a(a^2+1) &= 8(8^2+1) \\ a &= 8 \\ x &= 8-500 \\ &= -492 \\ \end{align} </math> </div></div> <ol start=18> <li>Berapa nilai x dari <math>\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2} &= 0 \\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}} &= \frac{1}{2} \\ \frac{x^n+4^n}{x^n+16^n} &= (\frac{1}{2})^n \\ \frac{x^n+4^n}{x^n+16^n} &= \frac{1}{2^n} \\ 2^n(x^n+4^n) &= x^n+16^n \\ 2^n(x^n+2^{2n}) &= x^n+2^{4n} \\ 2^n \cdot x^n+2^{3n} &= x^n+2^{4n} \\ 2^n \cdot x^n-x^n &= 2^{4n}-2^{3n} \\ x^n(2^n-1) &= 2^{3n}(2^n-1) \\ x^n &= 2^{3n} \\ x^n &= (2^3)^n \\ x^n &= 8^n \\ x &= 8 \\ \end{align} </math> </div></div> <ol start=19> <li>Berapa hasil dari <math>\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ x &= \frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5 \cdot 6}+\sqrt{5 \cdot 5}+\sqrt{6 \cdot 4}+\sqrt{5 \cdot 4}}{\sqrt{5 \cdot 4}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5} \cdot \sqrt{6}+\sqrt{5} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{4}}{2 \cdot \sqrt{5}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{6} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{5}+\sqrt{5} \cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}} \\ &= \frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ &= \frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ \frac{1}{x} &= \frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}} \\ &= \frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ &= \frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1} \\ &= \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5} \\ &= \sqrt{6}-\sqrt{4} \\ &= \sqrt{6}-2 \\ x &= \frac{1}{\sqrt{6}-2} \\ &= \frac{\sqrt{6}+2}{6-4} \\ &= \frac{\sqrt{6}+2}{2} \\ &= 1+\frac{\sqrt{6}}{2} \\ \end{align} </math> </div></div> <ol start=20> <li>Berapa hasil dari <math>(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5 \\ \text{misalkan } x=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ x &= \frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ &= \frac{\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}} \\ &= \frac{\sqrt{3}+1}{4} \\ 4x &= \sqrt{3}+1 \\ 4x-1 &= \sqrt{3} \\ (4x-1)^2 &= 3 \\ 16x^2-8x+1 &= 3 \\ 16x^2 &= 8x+2 \\ 8x^2 &= 4x+1 \\ x^2 &= \frac{4x+1}{8} \\ *cara 1 \\ x^3 &= x \cdot x^2 \\ &= x(\frac{4x+1}{8}) \\ &= \frac{4x^2+x}{8} \\ &= \frac{4x^2}{8}+\frac{x}{8} \\ &= \frac{4(\frac{4x+1}{8})}{8}+\frac{x}{8} \\ &= \frac{16x+4}{64}+\frac{x}{8} \\ &= \frac{4x+1}{16}+\frac{x}{8} \\ &= \frac{4x+1+2x}{16} \\ &= \frac{6x+1}{16} \\ x^5 &= x^2 \cdot x^3 \\ &= (\frac{4x+1}{8})(\frac{6x+1}{16}) \\ &= \frac{24x^2+10x+1}{128} \\ &= \frac{24x^2}{128}+\frac{10x+1}{128} \\ &= \frac{24(\frac{4x+1}{8})}{128}+\frac{10x+1}{128} \\ &= \frac{96x+24}{1024}+\frac{10x+1}{128} \\ &= \frac{96x+24+80x+8}{1024} \\ &= \frac{176x+32}{1024} \\ &= \frac{176x}{1024}+\frac{32}{1024} \\ &= \frac{176}{1024}(\frac{\sqrt{3}+1}{4})+\frac{32}{1024} \\ &= \frac{44(\sqrt{3}+1)}{1024}+\frac{32}{1024} \\ &= \frac{44\sqrt{3}+44}{1024}+\frac{32}{1024} \\ &= \frac{76+44\sqrt{3}}{1024} \\ &= \frac{19+11\sqrt{3}}{256} \\ *cara 2 \\ x^4 &= (x^2)^2 \\ &= (\frac{4x+1}{8})^2 \\ &= \frac{16x^2+8x+1}{64} \\ &= \frac{16x^2}{64}+\frac{8x}{64}+\frac{1}{64} \\ &= \frac{x^2}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{\frac{4x+1}{8}}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{4x}{32}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{8}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{4}+\frac{3}{64} \\ x^5 &= x \cdot x^4 \\ &= (\frac{\sqrt{3}+1}{4})(\frac{x}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\frac{\sqrt{3}+1}{4}}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\sqrt{3}+1}{16}+\frac{3}{64}) \\ &= \frac{(\sqrt{3}+1)^2}{64}+(\frac{\sqrt{3}+1}{4})\frac{3}{64} \\ &= \frac{3+2\sqrt{3}+1}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{4+2\sqrt{3}}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{16+8\sqrt{3}}{256}+\frac{3\sqrt{3}+3}{256} \\ &= \frac{19+11\sqrt{3}}{256} \\ \end{align} </math> </div></div> <ol start=21> <li>Berapa hasil dari <math>\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4} &= \frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots) \\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots &= \frac{1}{1-\frac{1}{4}} \\ &= \frac{4}{3} \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{4}{3}) \\ &= \frac{1}{4}+\frac{16}{3} \\ &= \frac{67}{12} \\ x &= \frac{67}{9} \\ \end{align} </math> </div></div> <ol start=22> <li>Berapa nilai y-x jika <math>\frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} = \frac{x}{y}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} &= \frac{x}{y} \\ \frac{\frac{106 \times 107}{2}}{\frac{106}{2}(4+109)} &= \frac{x}{y} \\ \frac{53 \times 107}{53 \times 113} &= \frac{x}{y} \\ y-x &= 113-107 = 6 \\ \end{align} </math> </div></div> <ol start=23> <li>Berapa angka satuan dari hasil 17²⁰²⁴?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan angka satuannya} \\ 17^1 &= 7 \\ 17^2 &= 9 \\ 17^3 &= 3 \\ 17^4 &= 1 \\ 17^5 &= 7 \\ 17^6 &= 9 \\ 17^7 &= 3 \\ 17^8 &= 1 \\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1} \end{align} </math> </div></div> <ol start=24> <li>Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ 1! + 2! + 3! + 4! + \dots + 2024! &= 1 + (1x2) + (1x2x3) + (1x2x3x4) + \dots + 2024! \\ &= 1 + 2 + 6 + 24 + 120 + 720 + \dots + 2024! \\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka } 1+2+6+24 = 33 \text{ jadi angka satuannya adalah } 3 \end{align} </math> </div></div> <ol start=25> <li>Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ \frac{1! + 2! + 3! + 4! + \dots + 2024!}{12} &= \frac{1 + 1x2 + 1x2x3 + 1x2x3x4 + \dots + 2024!}{12} \\ &= \frac{1 + 2 + 6 + 24 + \dots + 2024!}{12} \\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi } 1+2+6 = 9 \end{align} </math> </div></div> <ol start=26> <li>Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{penjumlahan} \\ 1+1+1+1+ \dots \text{ (sebanyak 88 buah) }+x+y &= A \\ 88+x+y &= A \\ \text{perkalian} \\ 1 \times 1 \times 1 \times \dots \text{ (sebanyak 88 buah) }\times x \times y &= A \\ x \times y &= A \\ 88+x+y &= xy \\ xy-y &= 88+x \\ y(x-1) &= 88+x \\ y &= \frac{88+x}{x-1} \\ \text{uji selidiki untuk x=2} \\ y &= \frac{88+2}{2-1} \\ &= 90 \\ \text{buktikan} \\ 88+x+y &= xy \\ 88+2+90 &= 2(90) \\ 180 &= 180 \\ \text{terbukti} \\ \text{nilai A adalah } 180 \\ \end{align} </math> </div></div> <ol start=27> <li>Berapakah nilai x, y dan z dari <math>x+y-z=1, x^2+y^2-z^2=-5 \text{ dan } x^3+y^3-z^3=-53</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+y-z &= 1 \\ x+y &= z+1 \\ x^2+2xy+y^2 &= z^2+2z+1 \\ x^2+y^2-z^2 &= 2z+1-2xy \\ -5 &= 2z+1-2xy \\ 2xy &= 2z+6 \\ xy &= z+3 \\ x^2+y^2-z^2 &= -5 \\ x^2+y^2 &= z^2-5 \\ x^3+y^3-z^3 &= -53 \\ (x+y)(x^2-xy+y^2)-z^3+53 &= 0 \\ (x+y)(x^2+y^2-xy)-z^3+53 &= 0 \\ (z+1)(z^2-5-(z+3))-z^3+53 &= 0 \\ (z+1)(z^2-z-8)-z^3+53 &= 0 \\ z^3-z^2-8z+z^2-z-8-z^3+53 &= 0 \\ -9z+45 &= 0 \\ -9z &= -45 \\ z &= 5 \\ x+y &= 5+1 \\ x+y &= 6 \\ x &= 6-y \\ xy &= 5+3 \\ xy &= 8 \\ (6-y)y &= 8 \\ 6y-y^2 &= 8 \\ y^2-6y+8 &= 0 \\ (y-4)(y-2) &= 0 \\ y=4 \text{ atau } y=2 \\ \text{jika } y=4 \\ x+y &= z+1 \\ x+4 &= 5+1 \\ x &= 2 \\ \text{jika } y=2 \\ x+y &= z+1 \\ x+2 &= 5+1 \\ x &= 4 \\ \end{align} </math> </div></div> <ol start=28> <li>Berapakah nilai titik koordinat (x,y) dari <math>\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}</math> dan <math>\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x+y}+\sqrt{x-y} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+y}-\sqrt{x-y} &= \sqrt{\frac{52y}{3x}} \\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y}) &= \sqrt{\frac{432x}{13y}} \cdot \sqrt{\frac{52y}{3x}} \\ x+y-x+y &= \sqrt{\frac{432x \cdot 52y}{13y \cdot 3x}} \\ 2y &= \sqrt{144 \cdot 4} \\ 2y &= \sqrt{576} \\ 2y &= 24 \\ y &= 12 \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13(12)}} \\ x+12+x-12+2 \cdot \sqrt{x+12} \cdot \sqrt{x-12} &= \frac{36x}{13} \\ 2x+2 \sqrt{x^2-144} &= \frac{36x}{13} \\ 2(x+\sqrt{x^2-144}) &= \frac{36x}{13} \\ x+\sqrt{x^2-144} &= \frac{18x}{13} \\ \sqrt{x^2-144} &= \frac{5x}{13} \\ x^2-144 &= \frac{25x^2}{169} \\ \frac{144x^2}{169}-144 &= 0 \\ \frac{x^2}{169}-1 &= 0 \\ x^2-169 &= 0 \\ (x-13)(x+13) &= 0 \\ x_1=13 &\text{ atau } x_2=-13 \text{ (TM) karena } x>y \\ \end{align} </math> jadi titik koordinat (13,12) </div></div> <ol start=29> <li>Berapakah nilai dari <math>x^2-7x</math> jika <math>(x-2)^2+\frac{1}{(x-2)^2} = 11</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x-2)^2+\frac{1}{(x-2)^2} &= 11 \\ (x-2)^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2)^2} &= 11-2 \\ (x-2-\frac{1}{x-2})^2 &= 9 \\ x-2-\frac{1}{x-2} &= 3 \\ (x-2)^2-1 &= 3(x-2) \\ x^2-4x+4-1 &= 3x-6 \\ x^2-7x &= -9 \\ \end{align} </math> </div></div> <ol start=30> <li>Berapakah nilai dari <math>\frac{(x+y)^2(x+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)}</math> jika xy+yz+xz=1?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy+yz+xz &= 1 \\ x^2+xy+yz+xz &= x^2+1 \\ x(x+y)+z(x+y) &= x^2+1 \\ (x+y)(x+z) &= x^2+1 \\ \text{dengan pola yang sama } \\ (y+x)(y+z) &= y^2+1 \\ (x+z)(y+z) &= z^2+1 \\ \frac{(x+y)^2(y+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)} &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)} \\ &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)^2(y+z)^2(x+z)^2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=31> <li>Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} w+5 &= w+x+y+z+5 \\ x+4 &= w+x+y+z+5 \\ y+3 &= w+x+y+z+5 \\ z+2 &= w+x+y+z+5 \\ \text{jumlahkan keempat persamaan } \\ w+x+y+z+14 &= 4(w+x+y+z+5) \\ w+x+y+z+14 &= 4(w+x+y+z)+20 \\ 3(w+x+y+z) &= -6 \\ w+x+y+z &= -2 \\ \end{align} </math> </div></div> <ol start=32> <li>Berapakah nilai dari <math>\frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2}</math> jika <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3</math> dan x+y+z=xyz?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2} &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2} \\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) \\ 3^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz}) \\ 9 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz}) \\ &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2 \\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} &= 7 \\ \end{align} </math> </div></div> <ol start=33> <li>Berapakah nilai dari <math>\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}</math> jika <math>x^2+y^2+z^2 = -2(ab+bc+ac)</math>?>/li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+z^2 &= -2(xy+yz+xz) \\ x^2+y^2+z^2+2(xy+yz+xz) &= 0 \\ (x+y+z)^2 &= 0 \\ x+y+z &= 0 \\ x+y &= -z \\ x+z &= -y \\ y+z &= -x \\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z} &= \frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x} \\ &= -2-(-5)-(-7) \\ &= 10 \\ \end{align} </math> </div></div> <ol start=34> <li>Berapakah nilai dari <math>\frac{20xyz}{xy+yz+xz}</math> jika <math>16^x = 256^y = 625^z = 40</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 16^x = 256^y = 625^z &= 40 \\ 2^{4x} = 4^{4y} = 5^{4z} &= 40 \\ 2^{4x} &= 40 \\ 2 &= 40^{\frac{1}{4x}} \\ 4^{4y} &= 40 \\ 4 &= 40^{\frac{1}{4y}} \\ 5^{4z} &= 40 \\ 5 &= 40^{\frac{1}{4z}} \\ 2 \cdot 4 \cdot 5 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z}} \\ 1 &= \frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z} \\ 4 &= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{20xyz}{xy+yz+xz} &= 20 \cdot \frac{xyz}{xy+yz+xz} \\ &= 20 \cdot (\frac{xy+yz+xz}{xyz})^{-1} \\ &= 20 \cdot (\frac{1}{z} + \frac{1}{x} + \frac{1}{y})^{-1} \\ &= 20 \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{-1} \\ &= 20 \cdot (4)^{-1} \\ &= 20 \cdot \frac{1}{4} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=35> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+3x^2+1}</math> jika <math>6x^2+25x+6=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 6x^2+25x+6 &= 0 \\ 6x+25+\frac{6}{x} &= 0 \\ 6(x+\frac{1}{x}) &= -25 \\ x+\frac{1}{x} &= \frac{-25}{6} \\ (c+\frac{1}{x})^2 &= (\frac{-25}{6})^2 \\ x^2+2+\frac{1}{x^2} &= \frac{625}{36} \\ x^2+\frac{1}{x^2} &= \frac{625}{36}-2 \\ x^2+\frac{1}{x^2} &= \frac{553}{36} \\ \frac{x^2}{x^4+3x^2+1} &= \frac{1}{x^2+3+\frac{1}{x^2}} \\ &= \frac{1}{a^2+\frac{1}{x^2}+3} \\ &= \frac{1}{\frac{553}{36}+3} \\ &= \frac{1}{\frac{661}{36}} \\ &= \frac{36}{661} \\ \end{align} </math> </div></div> <ol start=36> <li>Berapakah nilai dari <math>\frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5} &= \frac{(9+2\sqrt{20})^{1013}}{((2)^3+3(2)^2(\sqrt{5})+3(2)(\sqrt{5})^2+(\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{((2+\sqrt{5})^2)^{1013}}{((2+\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{(2+\sqrt{5})^{2026}}{(2+\sqrt{5})^{2025}}+6-\sqrt{5} \\ &= 2+\sqrt{5}+6-\sqrt{5} \\ &= 8 \\ \end{align} </math> </div></div> <ol start=37> <li>Berapakah nilai dari <math>27x^3+\frac{8}{x^3}</math> jika <math>3x+\frac{2}{x}=6</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3x+\frac{2}{x} &= 6 \\ (3x+\frac{2}{x})^3 &= 6^3 \\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3} &= 216 \\ 27x^3+18(6)+\frac{8}{x^3} &= 216 \\ 27x^3+108+\frac{8}{x^3} &= 216 \\ 27x^3+\frac{8}{x^3} &= 108 \\ \end{align} </math> </div></div> <ol start=38> <li>Berapakah nilai dari <math>x^6+\frac{8}{x^3}</math> jika <math>x^3+\frac{1}{x^3}=8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3+\frac{1}{x^3} &= 8 \\ x^3 &= 8-\frac{1}{x^3} \\ x^6 &= 8x^3-1 \\ x^6+\frac{8}{x^3} &= 8x^3-1+\frac{8}{x^3} \\ &= 8x^3+\frac{8}{x^3}-1 \\ &= 8(x^3+\frac{1}{x^3})-1 \\ &= 8(8)-1 \\ &= 63 \\ \end{align} </math> </div></div> <ol start=39> <li>Berapakah nilai dari <math>4x+\frac{25}{x}</math> jika <math>2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2\sqrt{x}+\frac{5}{\sqrt{x}} &= 4x-\frac{25}{x} \\ 2\sqrt{x}+\frac{5}{\sqrt{x}} &= (2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}}) \\ 1 &= 2\sqrt{x}-\frac{5}{\sqrt{x}} \\ 1^2 &= (2\sqrt{x}-\frac{5}{\sqrt{x}})^2 \\ 1 &= 4x-20+\frac{25}{x} \\ 4x+\frac{25}{x} &= 21 \\ \end{align} </math> </div></div> <ol start=40> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2-x+1}{x^2+x+1} &= \frac{5}{6} \\ \frac{x^2+1-x}{x^2+1+x} &= \frac{5}{6} \\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} &= \frac{5}{6} \\ \text{ misalkan } x+\frac{1}{x} &= y \\ \frac{y-1}{y+1} &= \frac{5}{6} \\ 6(y-1) &= 5(y+1) \\ 6y-6 &= 5y+5 \\ y &= 11 \\ x+\frac{1}{x} &= 11 \\ \end{align} </math> </div></div> <ol start=41> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt{x}+x=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x}+x &= 1 \\ x-1 &= -\sqrt{x} \\ (x-1)^2 &= (-\sqrt{x})^2 \\ x^2-2x+1 &= x \\ x^2-3x+1 &= 0 \\ x-3+\frac{1}{x} &= 0 \\ x+\frac{1}{x} &= 3 \\ \end{align} </math> </div></div> <ol start=42> <li>Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt[3]{x}-\sqrt[3]{x-36}=3</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x}-\sqrt[3]{x-36} &= 3 \\ (\sqrt[3]{x}-\sqrt[3]{x-36})^3 &= 3^3 \\ x-(x-36)-3 \sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36}) &= 27 \\ 36-3 \sqrt[3]{x(x-36)}3 &= 27 \\ -9 \sqrt[3]{x(x-36)} &= -9 \\ \sqrt[3]{x(x-36)} &= 1 \\ x(x-36) &= 1 \\ x^2-36x-1 &= 0 \\ x-36-\frac{1}{x} &= 0 \\ x-\frac{1}{x} &= 36 \\ \end{align} </math> </div></div> <ol start=43> <li>Berapakah nilai dari <math>x+\frac{16}{x}</math> jika <math>x-3\sqrt{x}=4</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x-3\sqrt{x} &= 4 \\ x-4 &= 3\sqrt{x} \\ x^2-8x+16 &= 9x \\ x^2-17x+16 &= 0 \\ x-17+\frac{16}{x} &= 0 \\ x+\frac{16}{x} &= 17 \\ \end{align} </math> </div></div> <ol start=44> <li>Berapakah nilai dari <math>\frac{x^2}{x^4+4}</math> jika <math>x^2-7x+2=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2-7x+2 &= 0 \\ x^2+2 &= 7x \\ x+\frac{2}{x} &= 7 \\ x^2+4+\frac{4}{x^2} &= 49 \\ x^2+\frac{4}{x^2} &= 45 \\ \frac{x^4+4}{x^2} &= 45 \\ \frac{x^2}{x^4+4} &= \frac{1}{45} \\ \end{align} </math> </div></div> <ol start=45> <li>Berapakah nilai dari <math>x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}</math> jika <math>x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}} &= 25 \\ x^{\frac{1}{2}}+x^{-\frac{1}{2}} &= 23 \\ x+2+x^{-1} &= 529 \\ x+x^{-1} &= 527 \\ x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+x^{-\frac{3}{4}} &= 110 \\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1} &= x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}} \\ &= 527+110 \\ &= 637 \\ \end{align} </math> </div></div> <ol start=46> <li>Berapakah nilai dari <math>\sqrt{8x^6+x^5+x^4+5x^3+1}</math> jika <math>\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5} &= 0 \\ \frac{x^2+x+1}{x^5} &= 0 \\ x^2+x+1 &= 0 \\ x^2+x+1 &= 0 \\ (x-1)(x^2+x+1) &= 0(x-1) \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ \sqrt{8x^6+x^5+x^4+5x^3+1} &= \sqrt{(2x^3)^2+x^3x^2+x^3x+5x^3+1} \\ &= \sqrt{(2(1))^2+(1)x^2+(1)x+5(1)+1} \\ &= \sqrt{(2)^2+x^2+x+5+1} \\ &= \sqrt{4+x^2+x+1+5} \\ &= \sqrt{4+0+5} \\ &= \sqrt{9} \\ &= 3 \\ \end{align} </math> </div></div> <ol start=47> <li>Berapakah nilai dari <math>f(1)+f(2)+f(3)+ \dots + f(99)</math> jika <math>f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{1}{\sqrt{x+1}+\sqrt{x}} \\ &= \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} \\ &= \sqrt{x+1}-\sqrt{x} \\ f(1)+f(2)+f(3)+ \dots + f(98)+f(99) &= \sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+ \cdot + \sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99} \\ &= \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+ \cdot + \sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ &= \sqrt{100}-\sqrt{1} \\ &= 10-1 \\ &= 9 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots + \frac{2024}{2025})</math> jika <math>h(x)=\frac{3}{3+9^x}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} h(x) &= \frac{3}{3+9^x} \\ h(1-x) &= \frac{3}{3+9^{1-x}} \\ &= \frac{3}{3+\frac{9}{9^x}} \\ &= \frac{9^x}{3+9^x} \\ h(x)+h(1-x) &= \frac{3}{3+9^x}+\frac{9^x}{3+9^x} \\ &= \frac{3+9^x}{3+9^x} \\ &= 1 \\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots +(1-\frac{2}{2025})+(1-\frac{1}{2025})) \\ & 5(1+1+1+ \dots +1+1) \text{ sebanyak 1012 kali } \\ & 5(1012) \\ & 5060 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>\frac{7^{2025} - 7^{2023} + 432}{7^{2024} + 7^{2023} + 72}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72} &= \frac{7^{2023}7^{2}-7^{2023} + 48 \times 9}{7^{2023}7^1+7^{2023}+8 \times 9} \\ &= \frac{7^{2023}(7^{2}-1)+48 \times 9}{7^{2023}(7^1+1)+8 \times 9} \\ &= \frac{7^{2023}(49-1)+48 \times 9}{7^{2023}(7+1) + 8 \times 9} \\ &= \frac{7^{2023} \times 48+48 \times 9}{7^{2023} \times 8+8 \times 9} \\ &= \frac{48(7^{2023}+9)}{8(7^{2023}+9)} \\ &= \frac{48}{8} \\ &= 6 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>tan (x+\frac{\pi}{4})</math> jika <math>\frac{1}{cos x}-tan x = \frac{4}{5}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{cos x}-tan x &= \frac{4}{5} \\ sec x-tan x &= \frac{4}{5} \\ sec^2 x-tan^2 x &= 1 \\ (sec x+tan x)(sec x-tan x) &= 1 \\ (sec x+tan x)\frac{4}{5} &= 1 \\ sec x+tan x &= \frac{5}{4} \\ \text{kedua persamaan dengan cara metode eliminasi } \\ 2 tan x &= \frac{5}{4}-\frac{4}{5} \\ 2 tan x &= \frac{9}{20} \\ tan x &= \frac{9}{40} \\ tan (x+\frac{\pi}{4}) &= \frac{tan x+tan \frac{\pi}{4}}{1-tan x \cdot tan \frac{\pi}{4}} \\ &= \frac{\frac{9}{40}+1}{1-\frac{9}{40} \cdot 1} \\ &= \frac{\frac{49}{40}}{\frac{31}{40}} \\ &= \frac{49}{31} \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>sin^3 x+csc^3 x</math> jika <math>sin x-csc x = 8</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ Dengan menggunakan rumus: } (a-b)^3 &= a^3-b^3-3ab(a-b) \\ (sin x-csc x)^3 &= sin^3 x-csc^3 x-3sin x csc x(sin x-csc x) \\ 8^3 &= sin^3 x-csc^3 x-3sin x (\frac{1}{sin x})(8) \\ 512 &= sin^3 x-csc^3 x-24 \\ sin^3 x-csc^3 x &= 512+24 \\ sin^3 x-csc^3 x &= 536 \\ \end{align} </math> </div></div> <ol start=52> <li>Berapakah nilai dari <math>(sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2</math> jika <math>\frac{1}{sin x}+\frac{1}{cos x} = 10</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{sin x}+\frac{1}{cos x} &= 10 \\ \frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} &= 100 \\ (sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2 &= sin^2 x+\frac{2sin x}{cos x}+\frac{1}{cos^2 x}+cos^2 x+\frac{2cos x}{sin x}+\frac{1}{sin^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2(sin^2 x+cos^2 x)}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+100 \\ &= 101 \\ \end{align} </math> </div></div> <ol start=53> <li>Berapakah nilai dari (x-1)⁶ jika <math>x=\frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin 80^\circ &= cos 10^\circ \\ sin 80^\circ-cos 10^\circ &= 0 \\ x &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+2 sin 20^\circ cos 20^\circ}{cos 10^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+4 sin 10^\circ cos 10^\circ cos 20^\circ}{cos 10^\circ} \\ &= 4 cos 55^\circ cos 25^\circ+4 sin 10^\circ cos 20^\circ \\ &= 2(2 cos 55^\circ cos 25^\circ+2 sin 10^\circ cos 20^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ+sin (-10)^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ-sin 10^\circ) \\ &= 2(cos 80^\circ-sin 10^\circ+cos 30^\circ+sin 30^\circ) \\ &= 2(cos 80^\circ-sin (90^\circ-80^\circ)+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(cos 80^\circ-cos 80^\circ+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= \sqrt{3}+1 \\ x-1 &= \sqrt{3} \\ (x-1)^6 &= (\sqrt{3})^6 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=54> <li>Berapakah nilai dari x jika <math>x=\frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ} \\ 2x sin 20^\circ-x sin 40 ^\circ &= x sin 20^\circ-x^2 sin 10^\circ \\ x^2 sin 10^\circ+x sin 20^\circ-x sin 40 ^\circ &= 0 \\ x(x sin 10^\circ+sin 20^\circ-sin 40 ^\circ) &= 0 \\ x = 0 &\text{ atau } x sin 10^\circ+sin 20^\circ-sin 40 ^\circ = 0 \\ x sin 10^\circ+sin 20^\circ-sin 40 ^\circ &= 0 \\ x sin 10^\circ &= sin 40 ^\circ-sin 20^\circ \\ x &= \frac{sin 40 ^\circ-sin 20^\circ}{sin 10^\circ} \\ &= \frac{2 cos 30 ^\circ sin 10^\circ}{sin 10^\circ} \\ &= 2 cos 30 ^\circ \\ &= \frac{2 \sqrt{3}}{2} \\ &= \sqrt{3} \\ \end{align} </math> </div></div> <ol start=55> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x^2}{x^2-16y^2} = \frac{625}{49}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2}{x^2-16y^2} &= \frac{625}{49} \\ \frac{x^2-16y^2}{x^2} &= \frac{49}{625} \text{ (terbalik posisinya)} \\ 1-\frac{16y^2}{x^2} &= \frac{49}{625} \\ \frac{16y^2}{x^2} &= 1 - \frac{49}{625} \\ (\frac{4y}{x})^2 &= \frac{576}{625} \\ (\frac{4y}{x})^2 &= (\frac{24}{25})^2 \\ \frac{4y}{x} &= \frac{24}{25} \\ \frac{y}{x} &= \frac{6}{25} \\ \frac{x}{y} &= \frac{25}{6} \\ \end{align} </math> </div></div> <ol start=56> <li>Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x}{y}+\frac{x+10y}{y+10x} = 2</math> serta bilangan real untuk x dan y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x}{y}+\frac{x+10y}{y+10x} &= 2 \\ \frac{x}{y}+\frac{\frac{x}{y}+10}{1+10\frac{x}{y}} &= 2 \\ \text{misalkan } \frac{x}{y} = a \\ a+\frac{a+10}{1+10a} &= 2 \\ a(1+10a)+a+10 &= 2(1+10a) \\ 10a^2+a+a+10 &= 2+20a \\ 10a^2-18a+8 &= 0 \\ 5a^2-9a+4 &= 0 \\ (5a-4)(a-1) &= 0 \\ a = \frac{4}{5} &\text{ atau } a = 1 \\ \text{jadi } \frac{x}{y} = {\frac{4}{5}, 1} \\ \end{align} </math> </div></div> <ol start=57> <li>Berapakah nilai dari xy jika <math>x^4+y^4+x^2y^2=15 \text{ dan } x^2+y^2+xy=5</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+xy &= 5 \\ x^2+y^2 &= 5-xy \\ x^4+y^4+x^2y^2 &= 15 \\ (x^2)^2+(y^2)^2+2x^2y^2-x^2y^2 &= 15 \\ (x^2+y^2)^2-x^2y^2 &= 15 \\ (5-xy)^2-x^2y^2 &= 15 \\ 25-10xy+x^2y^2-x^2y^2 &= 15 \\ 25-10xy &= 15 \\ 10xy &= 10 \\ xy &= 1 \\ \end{align} </math> </div></div> <ol start=58> <li>Berapakah nilai dari x jika <math>4^x = 63(4^3+1)(4^6+1)(4^{12}+1)+1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 4^x &= 63(4^3+1)(4^6+1)(4^{12}+1)+1 \\ 4^x-1 &= 63(4^3+1)(4^6+1)(4^{12}+1) \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{4^3-1} \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{63} \\ &= (4^3+1)(4^6+1)(4^{12}+1)(4^3-1) \\ &= (4^3-1)(4^3+1)(4^6+1)(4^{12}+1) \\ &= (4^6-1)(4^6+1)(4^{12}+1) \\ &= (4^{12}-1)(4^{12}+1) \\ &= 4^{24}-1 \\ 4^x &= 4^{24} \\ x &= 24 \\ \end{align} </math> </div></div> <ol start=59> <li>Berapakah nilai dari <math>\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}</math> jika <math>x=\sqrt{9+4\sqrt{5}}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \sqrt{9+4\sqrt{5}} \\ x &= 2+\sqrt{5} \\ x^2 &= 9+4\sqrt{5} \\ x^2-4x &= 9+4\sqrt{5}-4(2+\sqrt{5}) \\ x^2-4x &= 1 \\ x^2 &= 4x+1 \\ x^3 &= x \cdot x^2 \\ &= x(4x+1) \\ &= 4x^2+x \\ &= 4(4x+1)+x \\ &= 16x+4+x \\ &= 17x+4 \\ x^4 &= x \cdot x^3 \\ &= x(17x+4) \\ &= 17x^2+4x \\ &= 17(4x+1)+4x \\ &= 68x+17+4x \\ &= 72x+17 \\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1} &= \frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1} \\ &= \frac{72x+17-85x-20+8x+2+5x+3}{2} \\ &= \frac{2}{2} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=60> <li>Berapakah nilai dari <math>\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}</math> jika 2x-1=<math>\sqrt{61}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \frac{x^3+1}{x^5-x^4-x^3+x^2} = p \\ p &= \frac{x^3+1}{x^5-x^4-x^3+x^2} \\ &= \frac{x^3+1}{x^5-x^4-(x^3-x^2)} \\ &= \frac{x^3+1}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)} \\ &= \frac{x^2-x+1}{x^2(x-1)^2} \\ &= \frac{x^2-x+1}{(x(x-1))^2} \\ &= \frac{x(x-1)+1}{(x(x-1))^2} \\ 2x-1 &= \sqrt{61} \\ x &= \frac{\sqrt{61}+1}{2} \\ x-1 &= \frac{\sqrt{61}-1}{2} \\ x(x-1) &= (\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2}) \\ &= \frac{61-1}{4} \\ &= \frac{60}{4} \\ &= 15 \\ p &= \frac{x(x-1)+1}{(x(x-1))^2} \\ &= \frac{15+1}{15^2} \\ &= \frac{16}{15^2} \\ \sqrt{p} &= \sqrt{\frac{16}{15^2}} \\ &= \frac{4}{15} \\ \end{align} </math> </div></div> <ol start=61> <li>Berapakah nilai dari <math>(\frac{x-3}{x})^{25}</math> jika <math>x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\sqrt[5]{8}+\sqrt[5]{2} &= 1+\sqrt[5]{16}+\sqrt[5]{4} \\ x+(\sqrt[5]{2})^3+\sqrt[5]{2} &= 1+(\sqrt[5]{2})^4+(\sqrt[5]{2})^2 \\ x &= (\sqrt[5]{2})^4-(\sqrt[5]{2})^3+(\sqrt[5]{2})^2-\sqrt[5]{2}+1 \\ \text{misalkan } \sqrt[5]{2} = p \\ x &= p^4-p^3+p^2-p+1 \\ x &= \frac{p^5+1}{p+1} \\ (\frac{x-3}{x})^{25} &= (1-\frac{3}{x})^{25} \\ &= (1-\frac{3}{\frac{p^5+1}{p+1}})^{25} \\ &= (1-\frac{3(p+1)}{p^5+1})^{25} \\ &= (1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2})^5+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{2+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-3\sqrt[5]{2}-3)}{3})^{25} \\ &= (-\sqrt[5]{2})^{25} \\ &= (-2)^5 \\ &= -32 \\ \end{align} </math> </div></div> <ol start=62> <li>Berapakah nilai dari <math>x^{50}+x^{49}+x^{48}+x^{47}+x^{46}</math> jika <math>x^2+x+1=0</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+x+1 &= 0 \\ x^2+x &= -1 \\ \frac{x^3-1}{x-1} &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46} &= x^{48}(x^2+x+1)+x^{45}(x^2+x) \\ &= x^{48}(0)+(x^3)^{15}(-1) \\ &= 0+(1)^{15}(-1) \\ &= -1 \\ \end{align} </math> </div></div> <ol start=63> <li>Berapakah 2²⁴ dari <math>8^7+8^6+8^5+8^4+8^3+8^2+8+1=A</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= A \\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1) &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8 &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= 8A+1 \\ 8^8+A &= 8A+1 \\ 8^8 &= 7A+1 \\ (2^3)^8 &= 7A+1 \\ 2^{24} &= 7A+1 \\ \end{align} </math> </div></div> <ol start=64> <li>Berapakah nilai dari <math>x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1</math> jika <math>x+\frac{1}{x}=\sqrt{3}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{x} &= \sqrt{3} \\ x^2+2+\frac{1}{x^2} &= 3 \\ x^2-1+\frac{1}{x^2} &= 0 \\ x^2(x^2-1+\frac{1}{x^2}) &= x^2(0) \\ x^4-x^2+1 &= 0 \\ (x^2+1)(x^4-x^2+1) &= (x^2+1)0 \\ x^6-x^4+x^2+x^4-x^2+1 &= 0 \\ x^6+1 &= 0 \\ x^6 &= -1 \\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1 &= {x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1 \\ &= (-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2-1+1 \\ &= -1+1-1+1-1+1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=65> <li>Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari <math>\frac{c-b}{a-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= ax^2+bx+c \\ f(2) &= a(2)^2+2b+c = 4 \\ &= 4a+2b+c = 4 \\ f(7) &= a(7)^2+7b+c = 49 \\ &= 49a+7b+c = 49 \\ 49a+7b+c &= 49 \\ 4a+2b+c &= 4 \\ 45a+5b &= 45 \text{ (f(7) dikurangi f(2)) } \\ 9a+b &= 9 \\ b &= -9a+9 \\ 4a+2b+c &= 4 \\ 4a+2(-9a+9)+c &= 4 \\ 4a-18a+18+c &= 4 \\ -14a+18+c &= 4 \\ c &= 14a-14 \\ \frac{c-b}{a-1} &= \frac{14a-14-(-9a+9)}{a-1} \\ &= \frac{14(a-1)+9(a-1)}{a-1} \\ &= \frac{(14+9)(a-1)}{a-1} \\ &= 23 \\ \end{align} </math> </div></div> <ol start=66> <li>Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+y)^3 &= x^3+y^3+3xy(x+y) \\ 11^3 &= 242+3xy(11) \text{ (dibagi 11)} \\ 11^2 &= 22+3xy \\ 121 &= 22+3xy \\ 99 &= 3xy \\ xy &= 33 \\ (x-y)^2 &= x^2+y^2-2xy \\ &= ((x+y)^2-2xy)-2xy \\ &= (x+y)^2-4xy \\ &= 11^2-4(33) \\ &= 121-132 \\ &= -11 \\ \end{align} </math> </div></div> <ol start=67> <li>Berapa f(1)+f(-1) jika <math>f(\frac{ax-b}{bx-a})</math>=x²-5x+6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ jika} f(1) = f(\frac{ax-b}{bx-a}) \\ 1 &= \frac{ax-b}{bx-a} \\ bx-a &= ax-b \\ (b-a)x &= -b+a \\ &= -(b-a) \\ &= -1 \\ f(1) &= x^2-5x+6 \\ &= (-1)^2-5(-1)+6 \\ &= 12 \\ \text{ jika} f(-1) = f(\frac{ax-b}{bx-a}) \\ -1 &= \frac{ax-b}{bx-a} \\ -(bx-a) &= ax-b \\ -bx+a &= ax-b \\ (-b-a)x &= -b-a \\ &= 1 \\ f(-1) &= x^2-5x+6 \\ &= (1)^2-5(1)+6 \\ &= 2 \\ f(1)+f(-1) &= 12+2 \\ &= 14 \\ \end{align} </math> </div></div> <ol start=68> <li>berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)-x &= f(x-1) \\ f(x)-f(x-1) &= x \\ x=1 ; f(1)-f(0) &= 1 \\ x=2 ; f(2)-f(1) &= 2 \\ x=3 ; f(3)-f(2) &= 3 \\ x=4 ; f(4)-f(3) &= 4 \\ \dots \\ x=200 ; f(200)-f(199) &= 200 \\ \text{ jumlahkan tersebut menjadi } \\ f(200)-f(0) &= 1+2+3+4+\dots+200 \\ &= \frac{200 \cdot 201}{2} \\ &= 20.100 \\ f(200)-1 &= 20.100 \\ &= 20.101 \\ \end{align} </math> </div></div> <ol start=69> <li>Misalkan f(x) adalah fungsi rekursif yang berlaku ∀x ∈ R sebagai berikut: : f(x)+f(15-x) = 2024 : f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)+f(15-x) &= 2024 \\ f(15+x) &= f(x)+2020 \\ *cara 1 \\ \text{ganti x dengan 15+x } \\ f(15+x)+f(-x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ *cara 2 \\ \text{ganti x dengan -x } \\ f(-x)+f(15+x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) &= 8 \\ \text{maka } 2f(2025)+2f(-2025) &= 8 \\ \end{align} </math> </div></div> <ol start=70> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>2f(\frac{2002}{x}) + f(x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2f(\frac{2002}{x}) + f(x) &= 3x \\ \text{ganti x dengan 2 } \\ 2f(\frac{2002}{2}) + f(2) &= 3(2) \\ 2f(1001) + f(2) &= 6 \\ \text{ganti x dengan 1001 } \\ 2f(\frac{2002}{1001}) + f(1001) &= 3(1001) \\ 2f(2) + f(1001) &= 3003 \\ 2f(2) + f(1001) &= 3003 \\ f(1001) &= 3003 - 2f(2) \\ 2f(1001) + f(2) &= 6 \\ 2(3003 - 2f(2)) + f(2) &= 6 \\ 6006 - 4f(2) + f(2) &= 6 \\ 3f(2) &= 6000 \\ f(2) &= 2000 \\ \end{align} </math> </div></div> <ol start=71> <li>Misalkan f suatu fungsi rekursif yang memenuhi <math>f(\frac{1}{x}) + \frac{1}{x}f(-x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{1}{x})+\frac{1}{x}f(-x) &= 3x \\ \text{ganti x dengan 1/3 } \\ f(3)+3f(-\frac{1}{3}) &= 1 \\ \text{ganti x dengan -3 } \\ f(-\frac{1}{3}) - \frac{1}{3}f(3) &= -9 \\ \text{dikalikan 3 } \\ 3f(-\frac{1}{3})-f(3) &= -27 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ 2f(3) &= 28 \\ f(3) &= 14 \\ \end{align} </math> </div></div> <ol start=72> <li>Diketahui polinom <math>f(7^b-1)=7^{3b}-10</math>. tentukan nilai f(5)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(5) &= f(7^b-1) \\ 5 &= 7^b-1 \\ 7^b &= 6 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(6-1) &= 6^3-10 \\ f(5) &= 216-10 \\ &= 206 \\ *cara 2 \\ \text{misalkan } 7^b-1=a \text{ maka } 7^b=a+1 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(a) &= (a+1)^3-10 \\ f(5) &= (5+1)^3-10 \\ &= 6^3-10 \\ &= 216-10 \\ &= 206 \\ \end{align} </math> </div></div> <ol start=73> <li>Diketahui polinom <math>f(6^b-7)=6^{3b}-2 \cdot 6^{2b}-4</math>. tentukan nilai f(-2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} *cara 1 \\ f(-2) &= f(6^b-7) \\ -2 &= 6^b-7 \\ 6^b &= 5 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(5-7) &= 5^3-2 \cdot 5^2-4 \\ f(-2) &= 125-50-4 \\ &= 71 \\ *cara 2 \\ \text{misalkan } 6^b-7=a \text{ maka } 6^b=a+7 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(a) &= (a+7)^3-2(a+7)^2-4 \\ f(-2) &= (-2+7)^3-2(-2+7)^2-4 \\ &= 5^3-2(5)^2-4 \\ &= 125-50-4 \\ &= 71 \\ \end{align} </math> </div></div> <ol start=74> <li>Jika <math>f(xy)=\frac{f(x)}{y}</math> dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(10) &= 7 \\ f(2 \cdot 5) &= 7 \\ f(xy) &= \frac{f(x)}{y} \\ f(2 \cdot 5) &= \frac{f(2)}{5} \\ 7 &= \frac{f(2)}{5} \\ f(2) &= 35 \\ \end{align} </math> </div></div> <ol start=75> <li>Jika <math>f(xy)=\frac{f(x+y)}{xy}</math> dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(15) &= 16 \\ f(3 \cdot 5) &= 16 \\ f(xy) &= \frac{f(x+y)}{xy} \\ f(3 \cdot 5) &= \frac{f(3+5)}{3 \cdot 5} \\ f(15) &= \frac{f(8)}{15} \\ 16 &= \frac{f(8)}{15} \\ f(8) &= 240 \\ \end{align} </math> </div></div> <ol start=76> <li>Jika <math>f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15</math> maka tentukan nilai f(16)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x+\frac{1}{x}+6) &= x^2+\frac{1}{x^2}+15 \\ &= (x+\frac{1}{x})^2-2+15 \\ &= (x+\frac{1}{x})^2+13 \\ \text{misalkan } x+\frac{1}{x} &= p \\ f(x+\frac{1}{x}+6) &= (x+\frac{1}{x})^2+13 \\ f(p+6) &= p^2+13 \\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 } \\ f(p+6) &= p^2+13 \\ f(10+6) &= 10^2+13 \\ f(16) &= 100+13 \\ &= 113 \\ \end{align} </math> </div></div> <ol start=77> <li>Tentukan nilai x jika <math>f(x)=\frac{4}{4-x}</math> dan <math>f(x \cdot f(x))^{\frac{f(4x)}{f(x)}}=256</math>!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{4}{4-x} \\ f(4x) &= \frac{4}{4-4x} \\ \frac{f(4x)}{f(x)} &= \frac{\frac{4}{4-4x}}{\frac{4}{4-x}} \\ &= \frac{4-x}{4-4x} \\ f(x \cdot f(x)) &= f(x(\frac{4}{4-x})) \\ &= f(\frac{4x}{4-x}) \\ &= \frac{4}{4-(\frac{4x}{4-x})} \\ &= \frac{4}{\frac{16-4x-4x}{4-x}} \\ &= \frac{4}{\frac{16-8x}{4-x}} \\ &= \frac{4(4-x)}{4(4-4x)} \\ &= \frac{4-x}{4-4x} \\ \text{misalkan } \frac{4-x}{4-4x} &= a \\ f(x \cdot f(x))^{\frac{f(4x)}{f(x)}} &= 256 \\ a^a &= 256 \\ a^a &= 4^4 \\ a &= 4 \\ \frac{4-x}{4-4x} &= 4 \\ 4-x &= 16-16x \\ 15x &= 12 \\ x &= \frac{4}{5} \\ \end{align} </math> </div></div> <ol start=78> <li>Fungsi <math>f(x) = \frac{kx}{2x+1} \text{dengan } x \neq -\frac{1}{2}</math>. Dengan f(f(x)) = x maka tentukan nilai k!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{kx}{2x+1} \\ f(f(x)) &= x \\ f(\frac{kx}{2x+1}) &= x \\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1} &= x \\ \frac{\frac{k^2x}{2x+1}}{\frac{2kx+2x+1}{2x+1}} &= x \\ \frac{k^2x}{2kx+2x+1} &= x \\ \frac{k^2}{2kx+2x+1} &= 1 \\ k^2 &= 2kx+2x+1 \\ k^2-2kx &= 2x+1 \\ k^2-2kx+x^2 &= x^2+2x+1 \\ (k-x)^2 &= (x+1)^2 \\ (k-x)^2-(x+1)^2 &= 0 \\ (k-x+x+1)(k-x-(x+1)) &= 0 \\ k=-1 &\text{ atau } k=2x+1 &\text{ (TM) } \\ \end{align} </math> </div></div> <ol start=79> <li>Jika n = 2023²+2024² maka berapa hasil dari <math>\sqrt{2n-1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n &= 2023^2+2024^2 \\ &= 2023^2+(2023+1)^2 \\ \text{misalkan 2023 = p} \\ n &= p^2+(p+1)^2 \\ &= p^2+p^2+2p+1 \\ &= 2p^2+2p+1 \\ \sqrt{2n-1} &= \sqrt{2(2p^2+2p+1)-1} \\ &= \sqrt{4p^2+4p+2-1} \\ &= \sqrt{4p^2+4p+1} \\ &= \sqrt{(2p+1)^2} \\ &= 2p+1 \\ &= 2(2023)+1 \\ &= 4046+1 \\ &= 4047 \\ \end{align} </math> </div></div> <ol start=80> <li>Tentukan nilai dari a+b+c merupakan bilangan bulat positif jika ab = 2, bc = 3 dan ac = 6?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab \cdot bc \cdot ac &= 2 \cdot 3 \cdot 6 \\ (abc)^2 &= 36 \\ abc &= \pm 6 \\ abc &= 6 \\ \frac{abc}{ab} &= c = \frac{6}{2} = 3 \\ \frac{abc}{bc} &= a = \frac{6}{3} = 2 \\ \frac{abc}{ac} &= b = \frac{6}{6} = 1 \\ a+b+c &= 6 \\ \end{align} </math> </div></div> <ol start=81> <li>Tentukan nilai dari (a-c)^b jika <math>\frac{ab}{a+b} = \frac{1}{3}</math>, <math>\frac{bc}{b+c} = \frac{1}{4}</math> dan <math>\frac{ac}{a+c} = \frac{1}{9}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{ab}{a+b} &= \frac{1}{3} \\ \frac{a+b}{ab} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{b} + \frac{1}{a} &= 3 \\ \frac{bc}{b+c} &= \frac{1}{4} \\ \frac{b+c}{bc} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{b} &= 4 \\ \frac{ac}{a+c} &= \frac{1}{9} \\ \frac{a+c}{ac} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{a} &= 9 \\ \text{Misalkan 1/a = x, 1/b = y dan 1/c = z} \\ x+y &= 3 \\ y+z &= 4 \\ x+z &= 9 \\ x+y &= 3 \\ y+z &= 4 \\ x-z &= -1 \\ x-z &= -1 \\ x+z &= 9 \\ 2x &= 8 \\ x &= 4 \\ x-z &= -1 \\ 4-z &= -1 \\ z &= 5 \\ x+y &= 3 \\ 4+y &= 3 \\ y &= -1 \\ \frac{1}{a} &= 4 \\ a &= \frac{1}{4} \\ \frac{1}{b} &= -1 \\ b &= -1 \\ \frac{1}{c} &= 5 \\ c &= \frac{1}{5} \\ (a-c)^b &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=82> <li>Tentukan nilai dari a, b dan c jika <math>\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}</math> dan a+2b+3c=28?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan k untuk semua ketiga persamaan tersebut } \\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5} &= k \\ a+b &= 2k \\ a+c &= 4k \\ b+c &= 5k \\ 2a+b+c &= 6k \\ 2a+5k &= 6k \\ k &= 2a \\ a &= \frac{k}{2} \\ b &= \frac{3k}{2} \\ c &= \frac{7k}{2} \\ a+2b+3c &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ a &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ b &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ c &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=83> <li>Tentukan nilai dari (b+c)^a jika <math>\frac{a+b+c}{2} = \sqrt{a-2}+\sqrt{b-1}+\sqrt{c}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{a+b+c}{2} &= \sqrt{a-2}+\sqrt{b-1}+\sqrt{c} \\ a+b+c &= 2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}) \\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c} &= 0 \\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1 &= 0 \\ (\sqrt{a-2}-1)^2+(\sqrt{b-1}-1)^2+(\sqrt{c}-1)^2 &= 0 \\ (\sqrt{a-2}-1)^2 &= 0 \\ \sqrt{a-2}-1 &= 0 \\ \sqrt{a-2} &= 1 \\ a-2 &= 1 \\ a &= 3 \\ (\sqrt{b-1}-1)^2 &= 0 \\ \sqrt{b-1}-1 &= 0 \\ \sqrt{b-1} &= 1 \\ b-1 &= 1 \\ b &= 1 \\ (\sqrt{c}-1)^2 &= 0 \\ \sqrt{c}-1 &= 0 \\ \sqrt{c} &= 1 \\ c &= 1 \\ (b+c)^a &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=84> <li>x dan y merupakan bilangan tak nol. Jika xy = <math>\frac{x}{y}</math> = x-y maka berapa nilai x+y?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy &= \frac{x}{y} \\ y^2 &= 1 \\ y^2 - 1 &= 0 \\ (y-1)(y+1) &= 0 \\ y = 1 &\text{ atau } y = -1 \\ \frac{x}{y} &= x-y \\ x &= xy-y^2 \\ x-xy &= -y^2 \\ x(1-y) &= -y^2 \\ x &= \frac{-y^2}{1-y} \\ \text{cek y=1 } \\ x &= \frac{-1^2}{1-1} \\ \text{tidak memenuhi syarat } \\ \text{cek y=-1 } \\ x &= \frac{-(-1)^2}{1-(-1)} \\ &= \frac{-1}{2} \\ x+y &= -1-\frac{1}{2} \\ &= -\frac{3}{2} \\ \end{align} </math> </div></div> <ol start=86> <li>Berapa nilai x dari <math>(\frac{a}{b})^3+(\frac{b}{a})^3 = 2\sqrt{x}</math> jika <math>\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b} \\ \frac{a+b}{ab} &= \frac{1}{a+b} \\ (a+b)^2 &= ab \\ a^2+2ab+b^2 &= ab \\ a^2+b^2 &= -ab \\ \text{misalkan } \frac{a}{b}+\frac{b}{a} = n \\ \frac{a}{b}+\frac{b}{a} &= n \\ \frac{a^2+b^2}{ab} &= n \\ a^2+b^2 &= nab \\ n &= -1 \\ \frac{a}{b}+\frac{b}{a} &= n \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3(\frac{a}{b}+\frac{b}{a}) &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3n &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3 &= n^3-3n \\ &= (-1)^3-3(-1) \\ &= 2 \\ 2\sqrt{x} &= 2 \\ \sqrt{x} &= 1 \\ x &= 1 \\ \end{align} </math> </div></div> <ol start=87> <li>Berapa nilai m dari <math>x^2-mx-1=0</math> jika <math>\sqrt[3]{x_1}+\sqrt[3]{x_2}=1</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x_1} &= a \\ x_1 &= a^3 \\ \sqrt[3]{x_2} &= b \\ x_2 &= b^3 \\ \sqrt[3]{x_1}+\sqrt[3]{x_2} &= 1 \\ a+b &= 1 \\ x^2-mx-1 &= 0 \\ x_1+x_2 &= m \\ x_1 \cdot x_2 &= -1 \\ x_1+x_2 &= m \\ a^3+b^3 &= m \\ x_1 \cdot x_2 &= -1 \\ a^3 \cdot b^3 &= -1 \\ (ab)^2 &= (-1)^3 \\ ab &= -1 \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ (1)^3 &= m+3(-1)(1) \\ 1 &= m-3 \\ m &= 4 \\ \end{align} </math> </div></div> <ol start=88> <li>Berapa nilai <math>\frac{x_1}{x_2}</math> dari <math>ax^2-18x-b=0</math> jika <math>ab=45</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab &= 45 \\ b &= \frac{45}{a} \\ ax^2-18x-b &= 0 \\ ax^2-18x-\frac{45}{a} &= 0 \\ a^2x^2-18ax-45 &= 0 \\ (ax-3)(ax-15) &= 0 \\ ax-3 &= 0 \\ x &= \frac{3}{a} \\ ax-15 &= 0 \\ x &= \frac{15}{a} \\ \frac{x_1}{x_2} &= \frac{\frac{3}{a}}{\frac{15}{a}} \\ &= \frac{3}{15} \\ &= \frac{1}{5} \\ \frac{x_1}{x_2} &= \frac{\frac{15}{a}}{\frac{3}{a}} \\ &= \frac{15}{3} \\ &= 5 \\ \end{align} </math> </div></div> <ol start=89> <li>Jika <math>\frac{u_3}{u_1+u_2} = \frac{7}{8}</math> merupakan barisan aritmetika maka berapa dari <math>\frac{u_2+u_3}{u_1}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{u_3}{u_1+u_2} &= \frac{7}{8} \\ \frac{a+2b}{a+a+b} &= \frac{7}{8} \\ \frac{a+2b}{2a+b} &= \frac{7}{8} \\ 8(a+2b) &= 7(2a+b) \\ 8a+16b &= 14a+7b \\ 9b &= 6a \\ b &= \frac{2a}{3} \\ \frac{u_2+u_3}{u_1} &= \frac{a+b+a+2b}{a} \\ &= \frac{2a+3b}{a} \\ &= \frac{2a+3(\frac{2a}{3})}{a} \\ &= \frac{2a+2a}{a} \\ &= \frac{4a}{a} \\ &= 4 \\ \end{align} </math> </div></div> <ol start=90> <li>Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7p+q}{2p+q} &= \frac{17p+q}{7p+q} \\ (7p+q)^2 &= (17p+q)(2p+q) \\ 49p^2+14pq+q^2 &= 34p^2+19pq+q^2 \\ 15p^2 &= 5pq \\ 3p &= q \\ \frac{7p+q}{2p+q} &= \frac{7p+3p}{2p+3p} \\ &= \frac{10p}{5p} \\ &= 2 \\ \end{align} </math> </div></div> <ol start=91> <li>Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{rataan geometris } \\ \sqrt{a \cdot b} &= b-24 \\ a \cdot b &= (b-24)^2 \\ \text{rataan aritmatik } \\ \frac{a+b}{2} &= a+15 \\ a+b &= 2(a+15) \\ a+b &= 2a+30 \\ a &= b-30 \\ a \cdot b &= (b-24)^2 \\ (b-30)b &= (b-24)^2 \\ b^2-30b &= b^2-48b+576 \\ 18b &= 576 \\ b &= 32 \\ a &= b-30 \\ &= 32-30 \\ &= 2 \\ a+b &= 32+2 \\ &= 34 \\ \end{align} </math> </div></div> <ol start=91> <li>Segitiga lancip ABC dengan <math>\frac{a^4+b^4+c^4+a^2b^2}{c^2(a^2+b^2)}=2</math>. tentukan nilai sudut C?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{syarat segitiga lancip semua sudut masing-masing kurang dari } 90^\circ \\ c^2 &= a^2+b^2-2ab cos C \\ cos C &= \frac{a^2+b^2-c^2}{2ab} \\ a^4+b^4+c^4+a^2b^2 &= 2c^2(a^2+b^2) \\ a^4+b^4+a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-2c^2(a^2+b^2)+(c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= (ab)^2 \\ a^2+b^2-c^2 &= \pm ab \\ cos C &= \pm \frac{ab}{2ab} \\ &= \pm \frac{1}{2} \\ &= \frac{1}{2} \text{ (karena sudut harus kurang dari } 90^\circ) \\ C &= 60^\circ \\ \end{align} </math> </div></div> <ol start=92> <li>Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm²?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan panjang DA dan EA } = x \text{ dan panjang AB } = y \\ \text{luas segitiga CAB } &= \frac{CA \cdot AB}{2} \\ 45 &= \frac{(x+9)(x+5)}{2} \\ 90 &= x^2+14x+45 \\ x^2+14x &= 45 \\ y^2 &= (x+9)^2+(x+5)^2 \\ &= x^2+18x+81+x^2+10x+25 \\ &= 2x^2+28x+106 \\ &= 2(x^2+14x)+106 \\ &= 2(45)+106 \\ &= 196 \\ y &= 14 \\ \end{align} </math> jadi panjang BC adalah 14 cm </div></div> <ol start=93> <li>Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kita cari ukuran GC yaitu } \\ \frac{GC}{AD} &= \frac{CE}{DE} \\ \frac{GC}{15} &= \frac{6}{18} \\ GC &= 5 \\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC } \\ &= \frac{1}{2} \cdot BC \cdot CF - \frac{1}{2} \cdot GC \cdot CE \\ &= \frac{1}{2} \cdot 15 \cdot 18 - \frac{1}{2} \cdot 5 \cdot 6 \\ &= 135 - 15 \\ &= 120 \\ \end{align} </math> jadi luas daerah BFEG adalah 120 cm² </div></div> <ol start=94> <li>Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ AI &= \frac{1}{5} AF \\ &= \frac{1}{5} 10 \\ &= 2 \\ IF &= AF-AI \\ &= 10-2 \\ &= 8 \\ \text{luas trapesium AFGD } &= \frac{(AD+EF) \cdot AF}{2} \\ &= \frac{(4+6)10}{2} \\ &= 50 \\ \text{luas segitiga AID } &= \frac{AI \cdot AF}{2} \\ &= \frac{(2)4}{2} \\ &= 4 \\ \text{luas segitiga IFG } &= \frac{IF \cdot FG}{2} \\ &= \frac{(8)6}{2} \\ &= 24 \\ \text{luas daerah segitiga IGD } &= \text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG } \\ &= 50-4-24 \\ &= 22 \\ \end{align} </math> jadi luas daerah segitiga IGD adalah 22 cm² </div></div> <ol start=96> <li>Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-<math>\pi</math>) cm³ maka berapa luas permukaan kerucut tersebut?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ \text{volume balok} \\ V_b &= x^2(12) \\ \text{volume kerucut} \\ V_b &= \frac{1}{3}\pi x^2(12) \\ &= 4\pi x^2 \\ V_{b-k} &= Vb-Vk \\ 100(3-\pi) &= 12x^2-4\pi x^2 \\ 100(3-\pi) &= 4x^2(3-\pi) \\ x^2 &= 25 \\ x &= 5 \\ s &= \sqrt{12^2+5^2} \\ &= \sqrt{144+25} \\ &= \sqrt{169} \\ &= 13 \\ \text{luas permukaan kerucut } &= \pi r(r+s) \\ &= \pi(5)(5+13) \\ &= 90\pi \\ \end{align} </math> jadi luas daerah permukaan kerucut adalah 90<math>\pi</math> cm² </div></div> <ol start=97> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 3a+1 \\ B &= 3b+2 \\ A(A+1)+3B \\ (3a+1)(3a+1+1)+3(3b+2) \\ (3a+1)(3a+2)+9b+6 \\ 9a^2+9a+2+9b+6 \\ 9a^2+9a+9b+8 \\ 9(a^2+a+b)+8 \\ \text{sisa pembagiannya adalah } 8 \\ \end{align} </math> </div></div> <ol start=98> <li>Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 9a+7 \\ B &= 9b+8 \\ A(A-5)+9B \\ (9a+7)(9a+7-5)+9(9b+8) \\ (9a+7)(9a+2)+81b+72 \\ 81a^2+81a+14+81b+72 \\ 81a^2+81a+81b+86 \\ 81a^2+81a+81b+81+5 \\ 81(a^2+a+b+1)+5 \\ \text{sisa pembagiannya adalah } 5 \\ \end{align} </math> </div></div> <ol start=99> <li>Jika <math>\begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix}</math> maka berapa hasil dari A²¹+A²⁵+A⁴⁶?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A^2 &= A \cdot A \\ &= \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \\ A^3 &= A^2 \cdot A \\ &= \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ &= - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= -I \\ A^{21}+A^{25}+A^{46} &= A^{21} \cdot (I+A^4+A^{25}) \\ &= A^{21} \cdot (I+A^3 \cdot A +A^{24} \cdot A) \\ &= (A^3)^7 \cdot (I+A^3 \cdot A +(A^3)^8 \cdot A) \\ &= (-I)^7 \cdot (I-I \cdot A +(-I)^8 \cdot A) \\ &= -I \cdot (I-A+A) \\ &= -I \cdot I \\ &= -I \\ &= -\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ \end{align} </math> </div></div> <ol start=100> <li>Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida?</li></ol> : bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15 : ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9). : ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8) : maka Ida menulis bilangan 8. <ol start=101> <li>Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9?</li></ol> : Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu. : untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja. : untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja. : Jadi banyaknya bilangan mungkin 2. <ol start=102> <li>Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli!</li></ol> ;cara 1 # 8¹ = 1 # 8² = 1 (8²=8¹x8¹ sama dengan 1x1) # 8³ = 1 (8³=8¹x8² sama dengan 1x1) # 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2) # 8⁵ = 1 # 8ⁿ = 1 (semua n untuk bilangan asli) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli ;cara 2 # 8ⁿ = b mod 7 # 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil) # (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya) # 8ⁿ = 1ⁿ mod 7 # 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli <ol start=103> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 5?</li></ol> ;cara 1 # 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 # 7¹ = 7 (sisa 1) # 7² = 49 (sisa 2) # 7³ = 343 (sisa 3) # 7⁴ = 2,401 (sisa 0) # 7⁵ = 16,807 # 7⁶ = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3 ;cara 2 :17¹ = 2 :17² = 4 :17³ = 3 :17⁴ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3 :17⁹⁹ = (17⁴)²⁴ x 17³ Untuk 17⁴ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 3. Jadi 17⁹⁹ dibagi 7 bersisa 3 ;cara 3 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 17⁴ ::17⁴ = 1 mod 5 ::(17⁴)²⁴ = 1²⁴ mod 5 ::17⁹⁶ = 1²⁴ mod 5 ::17⁹⁶ = 1 mod 5 ::17⁹⁶ x 17³ = 1 x 17³ mod 5 ::17⁹⁹ = 17³ mod 5 ::17⁹⁹ = 17 x 17 x 17 mod 5 ::17⁹⁹ = 2 x 2 x 2 mod 5 ::17⁹⁹ = 8 mod 5 ::17⁹⁹ = 3 mod 5 Jadi 17⁹⁹ dibagi 5 bersisa 3 <ol start=104> <li>Berapa hasil sisa dari 17⁹⁹ dibagi 7?</li></ol> ;cara 1 :17¹ = 3 :17² = 2 :17³ = 6 :17⁴ = 4 :17⁵ = 5 :17⁶ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3 :17⁹⁹ = (17⁶)¹⁶ x 17³ Untuk 17⁶ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 6. Jadi 17⁹⁹ dibagi 7 bersisa 6 ;cara 2 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 17⁶ ::17⁶ = 1 mod 7 ::(17⁶)¹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1 mod 7 ::17⁹⁶ x 17³ = 1 x 17³ mod 7 ::17⁹⁹ = 17³ mod 7 ::17⁹⁹ = 17 x 17 x 17 mod 7 ::17⁹⁹ = 3 x 3 x 3 mod 7 ::17⁹⁹ = 27 mod 7 ::17⁹⁹ = 6 mod 7 Jadi 17⁹⁹ dibagi 7 bersisa 6 <ol start=105> <li>Berapa hasil sisa dari 41²⁰²⁴ dibagi 33?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 41^{2024} &= 41^{2024} \text{ mod } 33 \\ &= (33 \times 3 + 2)^{2024} \text{ mod } 33 \\ &= 2^{2024} \text{ mod } 33 \\ &= 2^{2020} 2^4 \text{ mod } 33 \\ &= (2^5)^{404} 2^4 \text{ mod } 33 \\ &= (33 - 1)^{404} 2^4 \text{ mod } 33 \\ &= (-1)^{404} 2^4 \text{ mod } 33 \\ &= 2^4 \text{ mod } 33 \\ &= 16 \text{ mod } 33 \\ \text{Jadi hasil sisa adalah } 16 \\ \end{align} </math> </div></div> <ol start=106> <li>Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 99^{99} &= (3^2 \times 11)^{99} \\ &= 3^{198} \times 11^{99} \\ 243^n &= (3^5)^n \\ &= 3^{5n} \\ \text{agar bisa membagi, maka} \\ 5n &= 198 \\ n &= 39.6 \\ \text{jadi bilangan n terbesar adalah } 39 \\ \end{align} </math> </div></div> <ol start=107> <li>Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸?</lu></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 88^{88} &= (8 \times 11)^{88} \\ &= 8^{88} \times 11^{88} \\ &= 8^{87} \times 8 \times 11^{88} \\ &= (8^3)^{29} \times 8 \times 11^{88} \\ &= 512^{29} \times 8 \times 11^{88} \\ 512^n &= 512^{29} \\ \text{jadi bilangan n terbesar adalah } 29 \\ \end{align} </math> </div></div> <ol start=108> <li>Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6!</li></ol> ; cara 1 : KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. : N dibagi 3 sisa 1 : N dibagi 5 sisa 2 : N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a : KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70 : KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21 : KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97 ; cara 2 : Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35 : kemudian ditambahkan sisa masing-masing sesuai dengan KPK. : KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, <b>97</b> : KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, <b>97</b> : KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, <b>97</b> Jadi bilangan bulat positif adalah 97 :: NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 <ol start=109> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |- | 4 || 0 || Semua isi ember B dibuang |- | 1 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. <ol start=110> <li>Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember?</li></ol> ; cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |} nah ada ember A berisi 4 liter. ; cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |- | 1 || 0 || Semua isi ember B dibuang |- | 0 || 1 || Tuangkan 1 l dari ember A ke B sehingga ember A kosong |- | 3 || 1 || Isikan 3 l ke ember A |- | 0 || 4 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |} nah ada ember B berisi 4 liter. # Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kemungkinan 1} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 +3 \cdot ((\frac{4}{3})^x)^2 &= 7 \cdot (\frac{4}{3})^x \\ \text{misalkan } (\frac{4}{3})^x = y \\ 4 +3y^2 &= 7y \\ 3y^2-7y+4 &= 0 \\ (3y-4)(y-1) &= 0 \\ y_1 = \frac{4}{3} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{4}{3} \\ (\frac{4}{3})^x &= \frac{4}{3} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{4}{3})^x &= 1 \\ x &= 0 \\ \text{kemungkinan 2} \\ 4 \cdot 9^x+3 \cdot 16^x &= 7 \cdot 12^x \\ 4 \cdot (3^2)^x+3 \cdot (4^2)^x &= 7 \cdot (4 \cdot 3)^x \\ 4 \cdot 3^{2x}+3 \cdot 4^{2x} &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot (3^x)^2+3 \cdot (4^x)^2 &= 7 \cdot 4^x \cdot 3^x \\ 4 \cdot ((\frac{3}{4})^x)^2 +3 &= 7 \cdot (\frac{3}{4})^x \\ \text{misalkan } (\frac{3}{4})^x = y \\ 4y^2 +3 &= 7y \\ 4y^2-7y+3 &= 0 \\ (4y-3)(y-1) &= 0 \\ y_1 = \frac{3}{4} &\text{ atau } y_2 = 1 \\ *y_1 = \frac{3}{4} \\ (\frac{3}{4})^x &= \frac{3}{4} \\ x &= 1 \\ *y_2 = 1 \\ (\frac{3}{4})^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-6}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{2x-4}{3}+2x-6 &= \frac{3}{2x-4}+\frac{1}{2x-6} \\ \frac{2x-4}{3}+2x-6 &= \frac{1}{\frac{2x-4}{3}}+\frac{1}{2x-6} \\ \text{misalkan } \frac{2x-4}{3} = y \text{ dan } 2x-6 = z \\ y+z &= \frac{1}{y}+\frac{1}{z} \\ y+z &= \frac{y+z}{yz} \\ y+z-\frac{y+z}{yz} &= 0 \\ (y+z)(1-\frac{1}{yz}) &= 0 \\ *y+z=0 \\ y+z &= 0 \\ y &= -z \\ \frac{2x-4}{3} &= -(2x-6) \\ \frac{2(x-2)}{3} &= -2(x-3) \\ \frac{x-2}{3} &= -x+3 \\ x-2 &= -3x+9 \\ 4x &= 11 \\ x &= \frac{11}{4} \\ *1-\frac{1}{yz}=0 \\ 1-\frac{1}{yz} &= 0 \\ yz-1 &= 0 \\ yz &= 1 \\ (\frac{2x-4}{3})(2x-6) &= 1 \\ \frac{4x^2-20x+24}{3} &= 1 \\ 4x^2-20x+24 &= 3 \\ 4x^2-20x+21 &= 0 \\ (2x-7)(2x-3) &= 0 \\ x = \frac{7}{2} &\text{ atau } x = \frac{3}{2} \\ \end{align} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 8c93lv6k5q7m8qg3h48smrlpm8xitwp 2 24242 117481 117135 2026-07-12T15:40:08Z Rohmat Sukuriyanto 40149 117481 wikitext text/x-wiki == Rohmat Sukuriyanto == [[File:Rohmat Sukuriyanto.jpg|thumb|Rohmat Sukuriyanto penulis blog, pengelola website, dan kreator konten digital asal Indonesia]] '''Rohmat Sukuriyanto''' adalah penulis blog, pengelola website, dan kreator konten digital asal Indonesia. Ia juga dikenal dengan nama panggilan '''Domath''' dalam aktivitas digitalnya. Rohmat aktif menulis seputar blogging, SEO, teknologi digital, anime, pop culture, serta panduan ringan yang mudah dipahami pembaca umum. Rohmat mengelola beberapa situs, di antaranya: * [https://www.yukinoshita.web.id Yukinoshita.web.id] — blog yang membahas anime, teknologi digital, SEO, blogging, hiburan, dan tren pop culture. * [https://www.ruangteknik.web.id Ruang Teknik] — blog edukasi ringan seputar listrik rumah, elektronik, alat teknik, dan panduan teknis untuk pemula. * [https://www.centongdigital.com Centong Digital] — situs web yang berkaitan dengan konten tulisan dan informasi digital. Dalam aktivitas menulis, Rohmat berfokus pada penyajian informasi dengan bahasa sederhana, praktis, dan mudah dipahami. Ia juga mendalami penulisan berbasis jurnalistik, optimasi mesin pencari, serta pengembangan konten digital yang bermanfaat bagi pembaca Indonesia. == Identitas digital == Dalam aktivitas digital, Rohmat Sukuriyanto juga dikenal dengan nama '''Domath'''. Nama tersebut digunakan sebagai identitas personal dalam beberapa aktivitas online, terutama yang berkaitan dengan blogging, pengelolaan website, penulisan konten, dan pengembangan proyek digital independen. Melalui identitas digital tersebut, Rohmat membangun citra sebagai penulis blog dan pengelola website independen yang berfokus pada SEO, teknologi digital, edukasi praktis, serta pengembangan konten yang mudah dipahami oleh pembaca Indonesia. Website menjadi media utama Rohmat untuk menampilkan karya, pengalaman, dan pemikiran. Melalui situs seperti Yukinoshita.web.id dan Ruang Teknik, ia mengembangkan konten yang tidak hanya berorientasi pada informasi, tetapi juga pada nilai manfaat, kepercayaan, dan keberlanjutan karya digital. == Penghargaan dan kegiatan digital == Pada tahun 2026, website yang dikelola Rohmat, yaitu Yukinoshita.web.id, menerima penghargaan Web Excellence kategori Personal dalam ajang Indonesia Website Awards 2026.<ref>[https://www.yukinoshita.web.id/2026/06/yukinoshita-raih-penghargaan-iwa-2026.html Yukinoshita Raih Penghargaan Personal Website Excellence pada Ajang Indonesia Website Awards 2026]. Yukinoshita.web.id. Diakses 13 Juni 2026.</ref> Indonesia Website Awards 2026 merupakan ajang penghargaan website yang diselenggarakan oleh Exabytes Indonesia. Pada tahun tersebut, acara mengangkat tema “Securing Digital Trust in the AI Threat Era” dan menyoroti pentingnya kepercayaan digital, keamanan website, performa, serta kredibilitas online di tengah perkembangan kecerdasan buatan.<ref>[https://jakarta.suaramerdeka.com/teknologi/13417243112/soroti-keamanan-siber-era-ai-exabytes-indonesia-sukses-gelar-indonesia-website-awards-2026 Soroti Keamanan Siber Era AI, Exabytes Indonesia Sukses Gelar Indonesia Website Awards 2026]. Suara Merdeka Jakarta. Diakses 13 Juni 2026.</ref><ref>[https://rm.id/baca-berita/life-style/314044/exabytes-apresiasi-karya-terbaik-web-developer-di-iwa-2026 Exabytes Apresiasi Karya Terbaik Web Developer di IWA 2026]. RM.id. Diakses 13 Juni 2026.</ref><ref>[https://rri.co.id/iptek/2488134/iwa-2026-soroti-pentingnya-kepercayaan-digital-di-tengah-ancaman-ai IWA 2026 Soroti Pentingnya Kepercayaan Digital di Tengah Ancaman AI]. RRI.co.id. Diakses 13 Juni 2026.</ref> Pencapaian tersebut menjadi bagian dari perjalanan Rohmat dalam membangun website personal yang lebih kredibel, profesional, dan berkelanjutan. == Minat == * Blogging dan SEO * Teknologi digital * Pengembangan website personal * Anime dan pop culture * Listrik rumah dan peralatan teknik * Penulisan artikel informatif * Keamanan dan kepercayaan digital == Pranala luar == * [https://www.yukinoshita.web.id Yukinoshita.web.id] * [https://www.ruangteknik.web.id Ruang Teknik] * [https://www.centongdigital.com Centong Digital] == Referensi == <references/> iumo9bjzhmgdhr0vrtmelo5df51h3uc 0 24898 117482 117248 2026-07-12T16:06:19Z Nafisathallah 35868 /* Hama */ 117482 wikitext text/x-wiki [[File:Osphronemus Gourami (better).png|500px||center]] Ikan gurame atau gurami (''Osphronemus gourami'') bisa dibilang menyandang gelar sebagai rajanya ikan air tawar di Indonesia. Tingginya permintaan pasar dan harga jual yang lebih menggiurkan dibandingkan rata-rata ikan air tawar lainnya, menjadi alasan mengapa budidaya ikan satu ini digemari. Di dalam buku ini, kita akan membahas semua hal yang berhubungan dengan proses budidaya dan pemeliharaan ikan gurame. == Ciri-ciri == Anatomi ikan gurame mirip seperti ikan sepat atau ikan tembakang karena mereka masih berada dalam satu famili yang sama yaitu Osphronemidae. Ditandai dengan sirip perut yang memanjang ke bawah dan berbentuk seperti sungut. Ikan gurami memiliki beberapa varian antara lain gurame soang, gurame padang, dan gurame blorok. == Bibit == [[File:Kolam Terpal - 02.jpg|500px|center]] Kepadatan penebaran jumlah bibit pada wadah pembesaran yang disarankan yaitu sebanyak 20 ekor/m². Lebih sedikit lebih baik, karena jika lebih dari rumus yang disebutkan ini, maka efek samping yang dialami seperti pertumbuhan ikan melambat dan resiko kematian yang meningkat karena overpopulasi. Jika anda memiliki kolam dengan dimensi 3 meter x 1 meter dan tinggi permukaan air sekitar 70 cm, maka jumlah tebar benih yang direkomendasikan berjumlah 60 ekor. == Pakan == === Pakan utama === Pelet PF1000 === Pakan alternatif === ==== Daun pepaya ==== ==== Daun mengkudu/pace ==== ==== Daun pisang ==== ==== Daun pecah beling ==== ==== Daun talas ==== ==== Daun sente ==== [[File:Bira besar 1.jpg|500px]] <br>Memiliki nama lain bira besar, sesuai dengan nama artikelnya pada Wikipedia Bahasa Indonesia. ==== Kangkung ==== ==== Daun singkong ==== ==== Daun ubi rambat ==== ==== Daun caya caya ==== ==== Daun pepaya jepang ==== ==== Daun kelor ==== ==== Azolla ==== ==== Wolfina ==== == Hama == === Kini-kini === == Penyakit == === Aeromonas === [[File:Anakan ikan gurame.jpg|500px|Aeromonas virus on juvenile giant gourami|center]] Ciri-cirinya antara lain sirip (ekor, badan, perut) yang pecah-pecah, badan memerah. == Obat == Methylene blue <br>PK (permanganas kalikus)</br> == Pemijahan == Induk gurame yang akan menjalani proses pemijahan sebaiknya diberi makan taoge. == Panen == [[File:Panen ikan gurami.jpg|500px|center]] Tempatkan ikan pada wadah penampungan yang berisi air bersih sebelum dijual, ini bertujuan untuk mengosongkan saluran pencernaan ikan dari kotoran dan sisa makanan. == Penjualan == === Tengkulak === === Pasar induk === === Eceran === === Harga pasar === ==== Kota Medan ==== Tabel berikut memberikan informasi seputar riwayat harga beli ikan gurame konsumsi per kilogram di pasar {| class="wikitable" |- ! Harga !! Tanggal !! Lokasi |- | Rp34.000 (mati), Rp45.000 (hidup) || 27 Desember 2025 || Pasar Setiabudi |- | Rp48.000 (hidup) || 9 Januari 2026 || Pasar Sei Sikambing |- | - || - || Pasar Melati |} ==== Kota Surabaya ==== ==== Jakarta ==== == Bacaan lanjutan == * [https://www.researchgate.net/profile/Livia-Tanjung/publication/289857878_Ikan_Gurami_Padang_dan_Teknik_Budi_Daya_Jhonly_Pilo/links/56932b5608aec14fa55db348/Ikan-Gurami-Padang-dan-Teknik-Budi-Daya-Jhonly-Pilo.pdf| Ikan Gurame Padang dan Teknik Budi Daya Jhonly Pilo] * [https://www.researchgate.net/publication/289868186_Ikan_Gurami_Osphronemus_gouramy_Strain_Padang_Terbukti_Memiliki_Ketahanan_Alami_terhadap_Infeksi_Aeromonas| Ikan Gurami (Osphronemus gouramy) Strain Padang Terbukti Memiliki Ketahanan Alami terhadap Infeksi Aeromonas] qlq41524rjrgebz6bcj7zy0uh33lqyb 0 27063 117507 117293 2026-07-13T08:57:11Z Hendri Saleh 40599 Perbikan kalimat 117507 wikitext text/x-wiki '''Aur Cina''' adalah suatu desa yang terdapat pada Kecamatan Batang Cenaku di Kabupaten Indragiri Hulu Provinsi Riau. Desa Aur Cina pada tahun 2021-2022 mengalami kejadian sengketa lahan dan tuntutan kebun plasma kepada PT. Arvena Sepakat (PT. AS) yang beroperasi di wilayah tersebut. Warga mengklaim lahan diluar Hak Guna Usaha (HGU) perusahaan dan menuntut hak plasma. Masyarakat Desa Aur Cina menuntut PT. AS untuk mengeluarkan kebun plasma seluas 20% dari HGU perusahaan yang berada dalam wilayah Desa Aur Cina. Tuntutan tersebut sesuai dengan Peraturan Menteri Pertanian Nomor : 26/Permentan/OT.140/2007 Tentang Pedoman Perizinan Usaha Perkebunan Pasal 11 ayat 1-4, Undang-Undang Nomor 39 Tahun 2014 Tentang Perkebunan Pasal 58 ayat 1-3, Hasil Rapat Senin 13 Februari Tahun 2012 bertempat di ruang rapat Kepala Dinas Perkebunan Kabupaten Indragiri Hulu (Inhu) dan Perjanjian Tanggal 8 November Tahun 1998 Butir 2 dan 4. PT. AS mengatakan bahwa pihaknya telah membangun perkebunan plasma masyarakat seluas 677 Ha sesuai dengan tuntutan masyarakat Desa Aur Cina. Namaun masyarakat menganggap kebun yang dibangun tersebut bukanlah kebun plasma karena tidak berada dalam HGU perusahaan melainkan kebun masyarakat pribadi. Untuk itu, masyarakat menuntut perusahaan agar membuktikan bahwa kebun tersebut adalah benar berada dalam HGU perusahaan . Sebelum dapat membuktikannya, perusahaan diminta untuk tidak melakukan aktivitasnya di dalam wilayah Desa Aur Cina.<ref>https://transmediariau.com/news/detail/57429/tuntut-kebun-plasma-masyarakat-desa-aur-cina-klaim-kebun-hgu-pt-arvena-sepakat</ref>, namun perusahaan tetap beroperasi, sehingga sempat menimbulkan ricuh. Pelaksanaan pemilihan kepala desa (Pilkades) serentak 4 Desember tahun 2019. sukses digelar di desa ini. Seluruh calon yang mendaftarkan diri dalam Pilkades ini mengadakan kampanye yang emndidik, mereka saling memberikan informasi tentang visi dan misi untuk desa Aur Cina. Sedangkan petugas penyelenggara,bbekerja sesuai tugas pokok dan fungsinya dengan melibatkan masyarakat. Namun, agar demokrasi berjalan lebih baik, diperlukan partisipasi masyarakat yang lebih aktif, transparansi dari pemerintah, serta kerja sama antara pemerintah daerah dan warga. Dengan demikian, pelaksanaan demokrasi di Aur Cina dapat semakin efektif dalam mewujudkan kesejahteraan dan kepentingan bersama. [[Kategori:Pendidikan Kewarganegaraan]] [[Kategori:Kebudayaan Melayu indragiri]] 5vppagk5x8lsj8xybvyjnksl9td27x0 117508 117507 2026-07-13T08:58:02Z Hendri Saleh 40599 Referensi 117508 wikitext text/x-wiki '''Aur Cina''' adalah suatu desa yang terdapat pada Kecamatan Batang Cenaku di Kabupaten Indragiri Hulu Provinsi Riau. Desa Aur Cina pada tahun 2021-2022 mengalami kejadian sengketa lahan dan tuntutan kebun plasma kepada PT. Arvena Sepakat (PT. AS) yang beroperasi di wilayah tersebut. Warga mengklaim lahan diluar Hak Guna Usaha (HGU) perusahaan dan menuntut hak plasma. Masyarakat Desa Aur Cina menuntut PT. AS untuk mengeluarkan kebun plasma seluas 20% dari HGU perusahaan yang berada dalam wilayah Desa Aur Cina. Tuntutan tersebut sesuai dengan Peraturan Menteri Pertanian Nomor : 26/Permentan/OT.140/2007 Tentang Pedoman Perizinan Usaha Perkebunan Pasal 11 ayat 1-4, Undang-Undang Nomor 39 Tahun 2014 Tentang Perkebunan Pasal 58 ayat 1-3, Hasil Rapat Senin 13 Februari Tahun 2012 bertempat di ruang rapat Kepala Dinas Perkebunan Kabupaten Indragiri Hulu (Inhu) dan Perjanjian Tanggal 8 November Tahun 1998 Butir 2 dan 4. PT. AS mengatakan bahwa pihaknya telah membangun perkebunan plasma masyarakat seluas 677 Ha sesuai dengan tuntutan masyarakat Desa Aur Cina. Namaun masyarakat menganggap kebun yang dibangun tersebut bukanlah kebun plasma karena tidak berada dalam HGU perusahaan melainkan kebun masyarakat pribadi. Untuk itu, masyarakat menuntut perusahaan agar membuktikan bahwa kebun tersebut adalah benar berada dalam HGU perusahaan . Sebelum dapat membuktikannya, perusahaan diminta untuk tidak melakukan aktivitasnya di dalam wilayah Desa Aur Cina.<ref>https://transmediariau.com/news/detail/57429/tuntut-kebun-plasma-masyarakat-desa-aur-cina-klaim-kebun-hgu-pt-arvena-sepakat</ref>, namun perusahaan tetap beroperasi, sehingga sempat menimbulkan ricuh. Pelaksanaan pemilihan kepala desa (Pilkades) serentak 4 Desember tahun 2019. sukses digelar di desa ini. Seluruh calon yang mendaftarkan diri dalam Pilkades ini mengadakan kampanye yang emndidik, mereka saling memberikan informasi tentang visi dan misi untuk desa Aur Cina. Sedangkan petugas penyelenggara,bbekerja sesuai tugas pokok dan fungsinya dengan melibatkan masyarakat. Namun, agar demokrasi berjalan lebih baik, diperlukan partisipasi masyarakat yang lebih aktif, transparansi dari pemerintah, serta kerja sama antara pemerintah daerah dan warga. Dengan demikian, pelaksanaan demokrasi di Aur Cina dapat semakin efektif dalam mewujudkan kesejahteraan dan kepentingan bersama. == Referensi == [[Kategori:Pendidikan Kewarganegaraan]] [[Kategori:Kebudayaan Melayu indragiri]] mq1d3xeosy0jqasqfbp0hxgjpfd5exp 0 27071 117499 114499 2026-07-13T05:28:59Z Hendri Saleh 40599 Perbaikan kalimat 117499 wikitext text/x-wiki '''Desa Rawa Bangun''' merupakan salah satu desa yang terletak di Kecamatan Rengat, Kabupaten Indragiri Hulu (Inhu), Provinsi Riau. Desa ini memiliki karakteristik khas wilayah perairan dan agraris yang mencerminkan kehidupan masyarakat di sepanjang aliran sungai Indragiri. Letak geografis dan karakteristik alam Rawa Bangun antara lain kondisi wilayahnya didominasi oleh lahan basah atau rawa. Desa ini berada di dataran rendah yang sangat dipengaruhi oleh pasang surut air Sungai Indragiri. Kondisi tanah yang subur akibat endapan sungai menjadikan wilayah ini potensial untuk pengembangan sektor pertanian dan perkebunan. Mata pencaharian penduduk umumnya pada sektor-sektor pertanian, berupa tanaman sayuran (hortikultura) meskipun tantangannya banjir musiman. Sektro perkebunan, berupa komoditas kelapa sawit dan karet. Sektor perikanan, terutama mencari ikan dan kolm. Masyarakat Rawa Bangun menjunjung tinggi nilai-nilai gotong royong. Kehidupan sosialnya dipengaruhi oleh perpaduan budaya Melayu lokal dengan warga pendatang yang sudah menetap lama. Hubungan antarwarga sangat erat, terutama dalam kegiatan keagamaan dan perayaan adat. Salah satu isu yang sering dihadapi oleh Desa Rawa Bangun adalah aksesibilitas dan infrastruktur, terutama saat musim hujan di mana beberapa titik lahan rawa bisa mengalami kenaikan debit air. Namun, pemerintah daerah terus berupaya melakukan peningkatan jalan dan sarana irigasi untuk mendukung mobilitas warga dan hasil tani. Desa ini merupakan bagian penting dari penyangga ekonomi Kecamatan Rengat terutama dalam memasok hasil bumi. 16 Juni 2020 sekira pukul 02.00 Wib terjadi kebakaran di Kantor Kepaa Desa Rawa Bangun. yang terjadi pada . Peristiwa tersebut menghanguskan sebagian besar bangunan kantor desa beserta berbagai fasilitas di dalamnya. Beberapa hari pada saat warga bergotorng royong membangun Kantor Kepala Desa darurat pelayanan administrasi warga terhambat. Peristiwa ini menjadi salah satu pengalaman yang mengingatkan masyarakat akan pentingnya menjaga aset desa dan saling membantu ketika terjadi musibah. [[Kategori:Pendidikan Kewarganegaraan]] [[Kategori:Kebudayaan Melayu indragiri]] 9gns8ni8b7tsdhux8qvfujh4e0xgovg 0 27074 117506 117458 2026-07-13T08:39:34Z Hendri Saleh 40599 Perbikan kalimat 117506 wikitext text/x-wiki '''Kelayang''' adalah nama sebuah desa yang terletak di Kecamatan Rakit Kulim, Kabupaten Indragiri Hulu di Provinsi Riau. Di desa ini terdapat K''ulam Loyang'' yang merupakan kawasan situs bersejarah. Kolam Loyang merupakan peninggalan sejarah yang memiliki nilai budaya dan masih dijaga keberadaannya hingga sekarang. Kawasan bersejarah ini menjadi bagian dari sejarah lokal masyarakat Melayu Indragiri serta menjadi salah satu warisan budaya yang terus diperkenalkan kepada generasi muda. Menurut cerita yang berkembang di masyarakat, nama Kelayang berasal dari kata ''Keloyang'' atau ''Kolam Loyang''. Kawasan ini dipercaya telah dikenal sejak masa Kerajaan Indragiri dan berkaitan dengan pelaksanaan prosesi adat kerajaan. Kolam Loyang digunakan sebagai tempat pemandian Raja Narasinga II sebelum dinobatkan sebagai Raja Indragiri. Selain itu, Kolam Loyang juga dikenal melalui legenda Mahligai Keloyang yang mengisahkan pertemuan Datuk Sati dengan para bidadari., di kolam pemandian ini Di kawasan Kolam Loyang pernah ditemukan beberapa benda yang diduga merupakan peninggalan masa lampau, seperti guci, keramik, tombak, meriam, dan peralatan rumah tangga tradisional. Dalam kehidupan masyarakat Desa Kelayang, nilai-nilai demokrasi dapat dilihat melalui kebiasaan bermusyawarah dalam mengambil keputusan bersama, seperti menentukan jadwal gotong royong membersihkan kawasan Kolam Loyang, memilih pengurus kegiatan masyarakat atau pemuda, serta membahas pelaksanaan kegiatan adat di desa. Setiap warga diberikan kesempatan untuk menyampaikan pendapat sebelum keputusan diambil secara mufakat. rxuxbqvezswu2s4vumtnsdry2f4y5ef 0 27078 117505 117295 2026-07-13T08:28:42Z Hendri Saleh 40599 Perbikan kalimat 117505 wikitext text/x-wiki '''Batu Sawar''' adalah salah satu desa yang terdapat pada Kecamatan Kelayang di Kabupaten Indragiri Indragiri Hulu, Provinsi Riau. Sekitar tahun 2012 di desa ini banyak terdapat Penambangan Emas Tanpa Izin (PETI) di sepanjang aliran Sungai Indragiri yang berdampak terhadap kerusakan lingkungan dan ekosistem sungai serta kesehatan masyarakat. Penambang liar beroperasi setiap hari mulai dari pagi hingga malam. Menggunakan kapal pompong kayu bermesin ''dompeng'' (masyarakat Kelayang menyebutnya ''bocai'') mereka menyedot pasir dengan pompa hisap lalu disaring dan dipisah dengan bahan kimia ''Merkur'''''i.''' Bahan kimia untuk mencuci atau mendulang emas tersebut dilepas ke aliran sungaiI Indragiri, sementara masyarakat masih menggunakan sungai mandi, cuci dan kakus (MCK) bahkan untuk memasak. Pada sisi lain ikan yang dikonsumsi masyarakat sebagian besar masih berasal dari Sungai Indragiri. Akibatnya masyarakat ada yang terdampak penyakit. [[Kategori:Pendidikan Kewarganegaraan]] [[Kategori:Kebudayaan Melayu indragiri]] t8b4bh7v1vgihwwrh2gveqcgexrw2ln 0 27085 117504 117300 2026-07-13T08:13:23Z Hendri Saleh 40599 Perbikan kalimat 117504 wikitext text/x-wiki '''Lubuk Cabau''' adalah nama sebuah desa yang terletak di Kecamatan V Koto, Kabupaten Mukomuko, Provinsi Bengkulu. Lubuk Cabau kental akan budaya dan nilai-nilai luhurnya, Salah satu budayanya adalah ''mayi yutang'' (bayar hutang) kepada leluhur didesa tersebut. Tradisi ini dilakukan setiap satu tahun sekali ketika menyambut bulan puasa. Kegiatan ini bentuk ucapan terimakasi kepada leluhur (alam) atas perlindungan dan rezeki yang berlimpah dari alam seperti panen padi, perkebunan yang berlimpah dan lain sebagainya. Kegiatan ini ditandai dengan berdoa, dan makan bersama. Pelaksanaan demokrasi di Desa Lubuk Cabau pada umumnya telah berjalan dengan baik melalui pelaksanaan Pemilu maupun pemilihan lainnya, seperti pemilihan kepala desa. Masyarakat diberikan hak yang sama untuk memilih pemimpin sesuai dengan hati nurani mereka. Hal ini menunjukkan bahwa prinsip demokrasi, yaitu kebebasan berpendapat dan berpartisipasi dalam pemerintahan, telah diterapkan di tingkat desa. Namun demikian, sosialisasi mengenai pentingnya menggunakan hak pilih, peningkatan partisipasi masyarakat, terutama pemilih muda, serta menjaga netralitas seluruh pihak agar proses demokrasi benar-benar berjalan secara adil dan transparan., perlu digesa lagi di desa ini. [[Kategori:Pendidikan Kewarganegaraan]] [[Kategori:Kebudayaan Melayu]] jwzzr7ld9hbt95dgkxz9fep5ih60szt 0 27088 117501 117460 2026-07-13T06:12:24Z Hendri Saleh 40599 Perbikan kalimat 117501 wikitext text/x-wiki Kampung Pulau adalah salah satu desa yang terdapat pada Kecamatan Rengat di Kabupaten Indragiri Hulu, Provinsi Riau. [[Kategori:Pendidikan Kewarganegaraan]] [[Kategori:Kebudayaan Melayu Indragiri]] Desa Kampung Pulau merupakan desa yang hidup dengan semangat gotong royong dan saling membantu. Sebagian besar penduduk bekerja sebagai petani, pekebun, dan pedagang kecil. Kehidupan bermasyarakat di desa ini cukup rukun karena masyarakat berasal dari berbagai suku dan tetap menjaga persatuan. Salah satu peristiwa penting yang pernah terjadi adalah ketika banjir akibat curah hujan tinggi menggenangi beberapa rumah dan lahan pertanian. Pada saat itu masyarakat bersama pemerintah desa dan warga,terutama pemuda dan remaja bekerja sama mengevakuasi warga, menyalurkan bantuan, serta membersihkan lingkungan setelah banjir surut. Pada sisi lain banjir tersebut menyebabkan desa ini menjadi destinasi wisata dadakan. Banyak warga dari desa lain datang kemari untuk main air. Semula kedatangn mereka untuk menjala ikan. Karena banjir, ikan-ikan dari Sungai Indragiri masuk ke sungai-sungai kecil selain ikan dari kolam warga yang jebol. Saking ramainya orang datang, pada titik-titik tertentu berdiri lapak-lapak dadakan penjual gorengan, mie instan dan makanan lainnya. Bentuk demokrasi yang diterapkan di Desa Kampung Pulau terlihat melalui pelaksanaan musyawarah desa. Dalam musyawarah tersebut, masyarakat diberi kesempatan untuk menyampaikan pendapat, usulan, dan kebutuhan desa. Selain itu, warga juga berpartisipasi dalam pemilihan kepala desa secara langsung sesuai dengan ketentuan yang berlaku. Dengan cara ini, keputusan yang diambil mencerminkan aspirasi masyarakat dan menjadi wujud pelaksanaan demokrasi di tingkat desa o903ar7jl641f68i54u203qierdko04 0 27733 117503 117456 2026-07-13T07:06:32Z Hendri Saleh 40599 Daftar Isi 117503 wikitext text/x-wiki __PAKSADAFTARISI__ '''Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri''' adalah sebuah Proyek (Proma) Mahasiswa berupa kajian empiris yang dilakukan oleh Dosen Pengampu Matakuliah Pendidikan Kewarganegaraan pada Institut Teknologi dan Bisnis Indragiri, Hendri. S, MA bersama dengan mahasiswa yang mengikuti matakuliah dimaksud tahun 2026 pada Program Studi S1 Agribisnis, S1 Kebidananan dan S1 Teknik Sipil. Mahasiswa merekam dan menuliskan secara singkat dan padat data dan fakta yang terdapat di desa Promanya masing-masing, sesuai silabus/Rencana Pembelajaran Semester (RPS) yang diturunkan Dosen Pengampu. Perbaikan informasi (terkadang juga EYD kalimat) dilakukan oleh Dosen Pengampu. == Pendahuluan == Sesuai Undang-Undang No, 20 Tahun 2003 tentang Sistem Pendidikan Nasional (Sisdiknas), pendidikan adalah suatu usaha yang dilakukan secara sadar dan terencana untuk mewujudkan suasana belajar dan proses pembelajaran agar peserta didik secara aktif mengembangkan potensi dirinya untuk memiliki kekuatan spiritual keagamaan, pengendalian diri, kepribadian, kecerdasan, akhlak mulia, serta keterampilan yang diperlukan dirinya, masyarakat, bangsa dan negara. Dari defenisi pendidikan tersebut dapat diketahui bahwa dalam proses pendidikan terdapat dua aspek, yaitu pendidikan dan pengajaran. Pendidikan akan menghasilkan kekuatan spiritual keagamaan, pengendalian diri, kepribadian, dan akhlak mulia. Sedangkan pengajaran akan menghasilkan kecerdasan dan keterampilan. Kata Kewarganegaraan, adalah kata benda yang diberi imbukan ke dan akhiran an, sehingga maknanya berobah menjadi kumpulan sifat-sifat. Kata dasarnya adalah warga dan negara.. Warga mengandung arti peserta, anggota atau member dari suatu organisasi atau komunitas yang terikat dengan aturan (termasuk tujuan organisasi tersebut). Istilah "negara" berasal dari bahasa Sanskerta yaitu ''nagari'' atau ''nagara'' yang berarti wilayah, kota, atau penguasa. Ilmuwan politik modern mendefinisikan negara sebagai suatu organisasi terbesar dan tertinggi dalam suatu wilayah yang memiliki kekuasaan tertinggi yang sah dan ditaati oleh rakyatnya. Jadi, warga negara arinya individu dari masyarakat suatu negara yang patuh terhadap aturan hukum dan norma-norma yang berlaku dalam negara tersebut. DAS Indragiri adalah kawasan yang menjadi daerah kekuasaan Kesultanan Indragiri. Wilayahnya meliputi tiga kabupaten di Riau saat ini, yaitu Kabupaten Indragiri Hulu, Kabupaten Indragiri Hulir dan Kabupaten Kuantan Singingi. Batas wilayahnya adalah bagian Selatan dengan Provinsi Jambi, Sebelah Barat berbatasan dengan Kab. Sawahlunto-Sijunjung (Sumatera Barat). Bagian Utara berbatasan Kabupaten Kampar (Riau) dan sebelah Timur berbatasan dengan Selat Berhala. Ciri khas DAS Indragiri adalah; di sebelah Timur berawa-rawa dan sebelah Barat berbukit-bukit.Wiayahnya dibelah oleh Sungai Indragiri, hulunya di Danau Singkarak bernama Batang Ombilin, memasuki Riau, sungai ini bernama Batang Kuantan, dan memasuki wilayah Sei. Lalak (Kab. Indragiri Hulu) bernama Sungai Indragiri, hingga bermuara di laut, di kawasan Kab. Indragiri Hilir. Penduduk yang mendiami DAS Indragiri mayoritas etnis Melayu dengan kebudayaan Melayu Indragiri. == Referensi == # <nowiki>https://pusmendik.kemdikbud.go.id/pdf/file-154</nowiki> # <nowiki>https://brainly.co.id/tugas/3405045</nowiki> # <nowiki>https://papuapegunungan.kpu.go.id/blog/read/2851_negara-adalah-pengertian-unsur-dan-fungsinya</nowiki> # Ahmad Yusuf dkk., ''Sejarah Kesultanan Inderagiri, Unri Press, Pekanbaru, 2003.'' == Daftar Isi == * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Air Putih Lubuk Batu Jaya|Aiir Putih Lubuk Batu Jaya]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Aur Cina Batang Cenaku|Aur Cina Batang Cenaku]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Batu Sawar Kelayang|Batu Sawar Kelayang]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Bayas Jaya Kempas|Bayas Jaya Kempas]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Bongkal Malang Kelayang|Bongkal Malang Kelayang]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/danau baru rengat barat riau|Danau Baru Rengat Barat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Dusun Tua Kelayang|Dusun Tua Kelayang]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Dusun Tua Pelang Kelayang|Dusun Tua Pelang Kelayang]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Kampung Besar Kota Rengat|Kampung Besar Kota Rengat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Kampung Besar Seberang Rengat|Kampung Besar Seberang Rengat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Kampung Dagang Rengat|Kampung Dagang Rengat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Kampung Pulau Rengat|Kampung Pulau Rengat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Keloyang Rakit Kulim|Keloyang Rakit Kulim]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Kota Baru|Kota Baru]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Kuala Gading Batang Cenaku|Kuala Gading Batang Cenaku]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Kuala Kilan Batang Cenaku|Kuala Kilan Batang Cenaku]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Kuantan Babu Rengat|Kuantan Babu Rengat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Kuantan Tenang Rakit Kulim|Kuantan Tenang Rakit Kulim]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Kulam Loyang Rakit Kulim|Kulam Loyang Rakit Kulim]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Lubuk Cabau V Koto|Lubuk Cabau V Koto]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Pasir Kelampaian Sei. Lala|Pasir Kelampaian Sei. Lala]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Pasir Kemilu Rengat|Pasir Kemilu Rengat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Pasir Ringgit Lirik|Pasir Ringgit Lirik]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Pelangko Kelayang|Pelangko Kelayang]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Pematang Manggis|Pematang Manggis]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Pematang Reba|Pematang Reba]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Peranap|Peranap]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Desa Perkebunan Sungai Parit|Perkebunan Sungai Parit]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Petaling Jaya Batang Cenaku|Petaling Jaya Batang Cenaku]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Rantau Mapesai|Rantau Mapesai]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Rawa Bangun Rengat|Rawa Bangun Rengat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Rawa Sekip Rengat|Rawa Sekip Rengat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/ Desa Rimpian Lubuk Batu Jaya|Rimpian Lubuk Batu Jaya]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Rumbai Jaya Tempuling|Rumbai Jaya Tempuling]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Redang Seko Lirik|Redang Seko Lirik]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Sekip Hilir Rengat|Sekip Hilir Rengat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Sekip Hulu Rengat|Sekip Hulu Rengat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Seresam Seberida|Seresam Seberida]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Siambul Batang Gansal|Siambul Batang Gansal]] * [[Tinjuan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Sibabat|Sibabat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri / Simpang Kota Medan|Simpang Kota Medan]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Sungai Banyak Ikan Kelayang|Sungai Banyak Ikan Kelayang]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Sungai Baung Rengat Barat|Sungai Baung Rengat Barat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Sungai Beringin Rengat|Sungai Beringin Rengat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Sungai Dawu Rengat Barat|Sungai Dawu Rengat Barat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Sungai Parit Sei Lala|Sungai Parit Sei Lala]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Sungai Raya Rengat|Sungai Raya Rengat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Sungai Sagu Lirik|Sungai Sagu Lirik]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Tanah Datar Rengat Barat|Tanah Datar Rengat Barat]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Desa Teluk Kabung|Teluk Kabung]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri//Teluk Pinang Jaya|Teluk Pinang Jaya]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Tembilahan|Tembilahan]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Desa Teluk Kabung|Desa Teluk Kabung]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/Dusun Tua Pelang Kelayang]] * [[Tinjauan Pendidikan Kewarganegaraan di Daerah Aliran Sungai (DAS) Indragiri/ Desa Rimpian Lubuk Batu Jaya]] 6oim9byp6zxeyh87dodgcahiu56x3tz 0 27740 117495 117462 2026-07-13T05:10:49Z Hendri Saleh 40599 Perbaikan kalimat 117495 wikitext text/x-wiki '''Danau Baru''' merupakan salah satu desa yang berada di Kecamatan Rengat Barat, Kabupaten Indragiri Hulu (Inhu), Provinsi Riau. Saat ini desa ini dipimpin oleh M Ridwan. Danau Baru memiliki letak yang strategis di kawasan aliran Sungai Indragiri sehingga kehidupan masyarakatnya sangat dipengaruhi oleh kondisi sungai dan sumber daya alam di sekitarnya. Selain memiliki potensi di sektor pertanian dan perikanan, Desa Danau Baru juga dikenal memiliki kehidupan sosial yang harmonis serta semangat gotong royong yang masih terjaga.penduduk nya menggunakan bahasa melayu dan saling menjaga adat di desa nya.Masyarakat di sini berkomunikasi dengan berbicara secara langsung atau tatap muka, dengan sopan santun, dan menggunakan bahasa melayu, dan manfaat agar masyarakat bisa lebih akrab. Salah satu peristiwa penting yang pernah terjadi di desa danau baru adalah abrasi atau pengikisan tebing Sungai Indragiri. Pada 22 Maret 2017. Abrasi semakin parah akibat tingginya debit air sungai. Jarak antara bibir sungai dan rumah warga hanya sekitar satu meter sehingga mengancam permukiman, jalan desa, serta arena Pacu Jalur. Masyarakat meminta pemerintah segera membangun turap untuk mencegah kerusakan yang lebih besar. Mayoritas penduduk Desa Danau Baru bekerja sebagai petani, nelayan, buruh, dan tukang. Sektor pertanian menjadi penopang utama perekonomian masyarakat karena didukung oleh lahan yang cukup subur. Selain itu, hasil perikanan dari Sungai Indragiri juga menjadi sumber penghasilan bagi sebagian warga. Aktivitas ekonomi desa terus berkembang melalui berbagai program pemberdayaan masyarakat dan pembangunan desa. biasa nya masyarakat bertani sawit, dalam 2 minggu sekali petani akan memanen buah sawit tersebut. BPD (Badan Permusyawaratan Desa) tujuannya itu agar setiap urusan fasilitas di desa berjalan dengan lancar, segala urusan yang ada di desa berjalan dengan tertib. Masyarakat di sini sering memasak ''tempoyak'' ikan patin, untuk tempoyak masyarakat akan menunggu musim durian tiba. edangkan ikan patin, biasanya dicari di sungai indragiri. Desa Danau Baru memiliki tugu. Pembangunan tugu Desa danau baru itu berada di pinggiran sungai indragiri berbatasan antara Desa Alang Kepayang dan Kota Lama. Bentuk tugu, berupa orang yang memegang dayung. Tugu ini melambangkan budaya ''Pacu Jalur'', karena Danau Baru dikenal sebagai esa atlit ''Pacu Jalur''. Penentuan lokasi menempatkan jaring patin di Desa Danau Baru caranya adalah r48nrzjvjgfpkrfryzu6sg7bfbus4b5 0 27769 117500 117461 2026-07-13T06:02:50Z Hendri Saleh 40599 Perbaikan kalimat dan penambahan rapek panitia, tungganai 117500 wikitext text/x-wiki '''Peranap''' merupakan salah satu desa yang berada di Kecamatan Peranap, Kabupaten Indragiri Hulu, Provinsi Riau. Wilayah ini dikenal memiliki potensi alam yang melimpah, terutama di bidang pertanian dan perkebunan. Sebagian besar masyarakatnya bekerja sebagai petani karet dan kelapa sawit. Salah satu peristiwa penting yang pernah dialami masyarakat Peranap adalah terjadinya banjir akibat meluapnya Sungai Indragiri pada musim hujan. Banjir menyebabkan beberapa rumah dan lahan pertanian terendam sehingga aktivitas masyarakat terganggu. Namun, masyarakat menunjukkan semangat kebersamaan dengan saling membantu mengevakuasi warga, mendirikan posko bantuan, serta bergotong royong membersihkan lingkungan setelah banjir surut. Peristiwa tersebut mempererat persatuan dan kepedulian antarwarga. Bentuk demokrasi yang diterapkan masyarakat Peranap terlihat dalam musyawarah untuk mencapai mufakat. Misalnya, ketika akan mengadakan kegiatan gotong royong membersihkan lingkungan, memperbaiki jalan, atau menyelenggarakan peringatan Hari Kemerdekaan, warga berkumpul untuk berdiskusi. Setiap orang diberi kesempatan menyampaikan pendapat, usul, maupun saran. Setelah itu, keputusan diambil berdasarkan kesepakatan bersama dan dilaksanakan secara gotong royong. Sikap saling menghargai pendapat, bekerja sama, dan menerima hasil musyawarah merupakan wujud penerapan nilai-nilai demokrasi dalam kehidupan bermasyarakat. Musyawarah menjelang kegiatan ini disebut ''rapek panitia.'' biasanya dipimpin oleh ''Tungganai'' (orang yang dituakan). 5k6untsb4tixpqsi6p0fp9ibp5tw0sa 0 27774 117502 2026-07-13T07:04:31Z Hendri Saleh 40599 Membuat halaman Kota Baru 117502 wikitext text/x-wiki '''Kota Baru''' adalah sebuah desa yang terletak di Kecamatan Rakit Kulim Kabupaten Indragiri Hulu, Provinsi Riau. Desa ini memiliki kode pos '''29359''' dan dihuni oleh masyarakat yang sebagian besar berlatar belakang suku Melayu serta merupakan salah satu wilayah yang didiami oleh komunitas adat suku Talang Mamak. Sekitar tahun 2022 terjadi banjir besar di Kota Baru menyebabkan banyak warga harus mengungsi ke tempat yang lebih aman. Hujan deras yang turun selama beberapa hari membuat sungai meluap hingga merendam rumah-rumah, jalan, dan fasilitas umum. Ketinggian air di beberapa wilayah mencapai lebih dari satu meter, sehingga aktivitas masyarakat menjadi terhenti. Akibat banjir yang sangat dalam, banyak warga tidak dapat berangkat bekerja karena jalan tertutup genangan air dan kendaraan tidak bisa melintas. Para pedagang terpaksa menutup usahanya, pekerja harian kehilangan penghasilan, dan anak-anak tidak dapat pergi ke sekolah. Kondisi ini membuat kehidupan masyarakat menjadi semakin sulit. Di tempat pengungsian, warga saling membantu dan berharap air segera surut agar mereka dapat kembali ke rumah dan menjalani aktivitas seperti biasa. Bantuan berupa makanan, air bersih, pakaian, dan layanan kesehatan sangat dibutuhkan untuk meringankan beban para korban banjir. l0itrx62kt61seah0a6yxqe6ayegq10