Wikibuku idwikibooks https://id.wikibooks.org/wiki/Halaman_Utama MediaWiki 1.47.0-wmf.24 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 8539 121438 120751 2026-10-23T06:11:19Z Bennylin 425 Mengembalikan suntingan oleh [[Special:Contributions/Berusa69|Berusa69]] ([[User talk:Berusa69|bicara]]) ke revisi terakhir oleh [[User:Bennylin|Bennylin]] 31124 wikitext text/x-wiki ===Gembala yang baik ({{Alkitab|Yohanes 10:1-21}})=== [[Berkas:The Good Shepherd by Gabrielptdedi.jpg|thumb|''Bon pastor'': Aku datang, supaya mereka mempunyai hidup, dan mempunyainya dalam segala kelimpahan. ]] Aku berkata kepadamu: Sesungguhnya siapa yang masuk ke dalam kandang domba dengan tidak melalui pintu, tetapi dengan memanjat tembok, ia adalah seorang pencuri dan seorang perampok; tetapi siapa yang masuk melalui pintu, ia adalah gembala domba. Untuk dia penjaga membuka pintu dan domba-domba mendengarkan suaranya dan ia memanggil domba-dombanya masing-masing menurut namanya dan menuntunnya ke luar. Jika semua dombanya telah dibawanya ke luar, ia berjalan di depan mereka dan domba-domba itu mengikuti dia, karena mereka mengenal suaranya. Tetapi seorang asing pasti tidak mereka ikuti, malah mereka lari dari padanya, karena suara orang-orang asing tidak mereka kenal. Aku berkata kepadamu, sesungguhnya Akulah pintu ke domba-domba itu. Semua orang yang datang sebelum Aku, adalah pencuri dan perampok, dan domba-domba itu tidak mendengarkan mereka. {{rquote||Akulah pintu; barangsiapa masuk melalui Aku, ia akan selamat dan ia akan masuk dan keluar dan menemukan padang rumput.}} Pencuri datang hanya untuk mencuri dan membunuh dan membinasakan; Aku datang, supaya mereka mempunyai hidup, dan mempunyainya dalam segala kelimpahan. {{rquote||Akulah gembala yang baik. Gembala yang baik memberikan nyawanya bagi domba-dombanya;}} sedangkan seorang upahan yang bukan gembala, dan yang bukan pemilik domba-domba itu sendiri, ketika melihat serigala datang, meninggalkan domba-domba itu lalu lari, sehingga serigala itu menerkam dan mencerai-beraikan domba-domba itu. Ia lari karena ia seorang upahan dan tidak memperhatikan domba-domba itu. {{pquote|Akulah gembala yang baik dan Aku mengenal domba-domba-Ku dan domba-domba-Ku mengenal Aku sama seperti Bapa mengenal Aku dan Aku mengenal Bapa, dan Aku memberikan nyawa-Ku bagi domba-domba-Ku. }} Ada lagi pada-Ku domba-domba lain, yang bukan dari kandang ini; domba-domba itu harus Kutuntun juga dan mereka akan mendengarkan suara-Ku dan mereka akan menjadi satu kawanan dengan satu gembala. Bapa mengasihi Aku, oleh karena Aku memberikan nyawa-Ku untuk menerimanya kembali. Tidak seorangpun mengambilnya dari pada-Ku, melainkan Aku memberikannya menurut kehendak-Ku sendiri. Aku berkuasa memberikannya dan berkuasa mengambilnya kembali. Inilah tugas yang Kuterima dari Bapa-Ku. <noinclude>{{navigation|Book=Perumpamaan Yesus |current= Gembala yang baik |previous= Gadis-gadis yang bijaksana dan gadis-gadis yang bodoh |next= Hakim yang tidak benar}}</noinclude> [[Kategori:Perumpamaan Yesus|{{SUBPAGENAME}}]] q9l01dxr0htub8c8msypsglkiv5t10i 0 23568 121430 121429 2026-10-23T01:09:03Z Akuindo 8654 121430 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>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-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{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> <ol start=7> <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} = y \\ y^5(1+y+y^2) &= 5^5 \cdot 31 \\ y &= 5 \\ x^{10} &= 5 \\ x &= ^5log \, 10 \\ \end{align} </math> </div></div> <ol start=8> <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=9> <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=10> <li>Berapa nilai x dari <math>\frac{\sqrt{(^4log \, x)^2+ ^2log \, x-4}}{16- ^2log \, x- (^4log \, x)^2} = \frac{1}{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} ^4 log \, x &= \frac{^2log \, x}{2} \\ \frac{\sqrt{(\frac{^2log \, x}{2})^2+ ^2log \, x-4}}{16- ^2log \, x- (\frac{^2log \, x}{2})^2} &= \frac{1}{4} \\ \text{misalkan } y &= ^2log \, x \\ \frac{\sqrt{(\frac{y}{2})^2+y-4}}{16-y-(\frac{y}{2})^2} &= \frac{1}{4} \\ \frac{\sqrt{(\frac{y}{2})^2+y-4}}{12-(-4+y+(\frac{y}{2})^2)} &= \frac{1}{4} \\ \text{misalkan } a &= (\frac{y}{2})^2+y-4 \\ \frac{\sqrt{a}}{12-a} &= \frac{1}{4}, a \ge 0 \\ \text{misalkan } \sqrt{a} &= b, b \ge 0 \\ \frac{b}{12-b^2} &= \frac{1}{4} \\ 4b &= 12-b^2 \\ b^2+4b-12 &= 0 \\ (b-2)(b+6) &= 0 \\ b = 2 &\text{ atau } b = -6 (TM) \\ \sqrt{a} &= 2 \\ a &= 4 \\ (\frac{y}{2})^2+y-4 &= 4 \\ \frac{y^2}{4}+y-8 &= 0 \\ y^2+4y-32 &= 0 \\ (y-4)(y+8) &= 0 \\ y = 4 &\text{ atau } y=-8 \\ ^2 log \, x &= 4 \\ x &= 16 \\ ^2 log \, x = -8 \\ x &= \frac{1}{256} \\ \end{align} </math> </div></div> <ol start=11> <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=12> <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=13> <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=14> <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=15> <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=16> <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=17> <li>Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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> <ol start=18> <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=19> <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=20> <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=21> <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=22> <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=23> <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=24> <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=25> <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=26> <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=27> <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=28> <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=29> <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=30> <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=31> <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=32> <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=33> <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=34> <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=35> <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=36> <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=37> <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=38> <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=39> <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=40> <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=41> <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=42> <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=43> <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=44> <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=45> <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=46> <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=47> <li>Berapakah nilai dari <math>5x+\frac{1}{4x}</math> jika <math>25x^2+\frac{1}{16x^2}=\frac{13}{2}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (5x+\frac{1}{4x})^2 &= 25x^2+\frac{5}{2}+\frac{16}{x^2} \\ &= 25x^2+\frac{16}{x^2}+\frac{5}{2} \\ &= \frac{13}{2}+\frac{5}{2} \\ &= 9 \\ &= \sqrt{9} \\ &= \pm 3 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>x^{28}+x^{25}+x^{23}+x^{20}</math> jika <math>x+\frac{1}{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} x+\frac{1}{x} &= 1 \\ x-1+\frac{1}{x} &= 0 \\ x^2-x+1 &= 0 \\ (x+1)(x^2-x+1) &= (x+1)0 \\ x^3+1 &= 0 \\ x^3 &= -1 \\ x &= -1 \\ x^{28}+x^{25}+x^{23}+x^{20} &= (-1)^{28}+(-1)^{25}+(-1)^{23}+(-1)^{20} \\ &= 1-1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>x^{28}-x^{25}+x^{23}-x^{20}</math> jika <math>x+\frac{1}{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} x+\frac{1}{x} &= -1 \\ x+1+\frac{1}{x} &= 0 \\ x^2-x+1 &= 0 \\ (x-1)(x^2-x+1) &= (x-1)0 \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{28}-x^{25}+x^{23}-x^{20} &= x^{25}(x^{3}-1)+x^{20}(x^{3}-1) &= x^{25}(0)+x^{20}(0) \\ &= 0 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>\frac{3}{2x-3}+\frac{4}{2y-4}+\frac{5}{2z-5}</math> jika <math>\frac{2x}{2x-3}+\frac{2y}{2y-4}+\frac{2z}{2z-5}=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} \frac{2x}{2x-3}+\frac{2y}{2y-4}+\frac{2z}{2z-5} &= 7 \\ \frac{2x-3+3}{2x-3}+\frac{2y-4+4}{2y-4}+\frac{2z-5+5}{2z-5} &= 7 \\ 1+\frac{3}{2x-3}+1+\frac{4}{2y-4}+1+\frac{5}{2z-5} &= 7 \\ \frac{3}{2x-3}+\frac{4}{2y-4}+\frac{5}{2z-5} &= 4 \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>\frac{4x}{4x-7}+\frac{5y}{5y-7}+\frac{6z}{6z-7}</math> jika <math>\frac{7}{4x-7}+\frac{7}{5y-7}+\frac{7}{6z-7}=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{7}{4x-7}+\frac{7}{5y-7}+\frac{7}{6z-7} &= 10 \\ \frac{7+4x-4x}{4x-7}+\frac{7+5y-5y}{5y-7}+\frac{7+6z-6z}{6z-7} &= 10 \\ \frac{4x-(4x-7)}{4x-7}+\frac{5y-(5y-7)}{5y-7}+\frac{6z-(6z-7)}{6z-7} &= 10 \\ \frac{4x}{4x-7}-1+\frac{5y}{5y-7}-1+\frac{6z}{6z-7}-1 &= 10 \\ \frac{4x}{4x-7}+\frac{5y}{5y-7}+\frac{6z}{6z-7} &= 13 \\ \end{align} </math> </div></div> <ol start=52> <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=53> <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=54> <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=55> <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=56> <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=57> <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=58> <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=59> <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-3 sin \, x csc \, x(sin \, x-csc \, x) \\ 8^3 &= sin^3 \, x-csc^3 \, x-3 sin \, 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=60> <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{2 sin \, 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=61> <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=62> <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=63> <li>Buktikan bahwa <math>\frac{sin \, x}{1-cos \, x} = cot \, \frac{y}{4}</math> jika <math>cot \, x = \frac{1+cos \, y}{sin \, 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} cot \, x &= \frac{2 cos^2 \, \frac{y}{2}}{2 sin \, \frac{y}{2} \, cos \, \frac{y}{2}} \\ &= \frac{cos \frac{y}{2}}{sin \, \frac{y}{2}} \\ &= cot \, \frac{y}{2} \\ x &=\frac{y}{2} \\ \frac{sin \, x}{1-cos \, x} &= \frac{sin \, \frac{y}{2}}{1-cos \, \frac{y}{2}} \\ &= \frac{2 sin \, \frac{y}{4} \, cos \, \frac{y}{4}}{2 sin^2 \, \frac{y}{4}} \\ &= \frac{cos \, \frac{y}{4}}{sin \, \frac{y}{4}} \\ &= cot \, \frac{y}{4} \end{align} </math> </div></div> <ol start=64> <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=65> <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=66> <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=67> <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=68> <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=69> <li>Berapakah nilai dari <math>x^5-3x^4-x^3+9x^2-11x+3</math> jika <math>x+\frac{1}{x}=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} *cara 1 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2-3x+1 &= 0 \\ (x^3-2x+3)(x^2-3x+1 &= 0(x^3-2x+3) \\ x^5-3x^4-x^3+9x^2-11x+3 &= 0 \\ *cara 2 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2-3x+1 &= 0 \\ x^5-3x^4-x^3+9x^2-11x+3 &= x^5-3x^4+x^3-2x^3+6x^2+3x^2-2x-9x+3 \\ &= x^5-3x^4+x^3-2x^3+6x^2-2x+3x^2-9x+3 \\ &= x^3(x^2-3x+1)-2x(x^2-3x+1)+3(x^2-3x+1) \\ &= x^3(0)-2x(0)+3(0) \\ &= 0 \\ *cara 3 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2 &= 3x-1 \\ x^2-3x+1 &= 0 \\ x^3 &= x \cdot x^2 \\ &= x (3x-1) \\ &= 3x^2-x x^4 &= x \cdot x^3 \\ &= x(3x^2-x) \\ &= 3x^3-x^2 \\ x^5 &= x \cdot x^4 \\ &= x(3x^3-x^2) \\ &= 3x^4-x^3 \\ x^5-3x^4-x^3+9x^2-11x+3 &= 3x^4-x^3-3(3x^3-x^2)-(3x^2-x)+9(3x-1)-11x+3 \\ \end{align} </math> </div></div> <ol start=70> <li>Berapakah nilai dari <math>\sqrt{x-7}</math> jika x=<math>16+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 &= 16+4\sqrt{5} \\ x-7 &= 16-7+4\sqrt{5} \\ &= 9+4\sqrt{5} \\ &= (2+\sqrt{5})^2 \\ \sqrt{x-7} &= 2+\sqrt{5} \\ \end{align} </math> </div></div> <ol start=71> <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=72> <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=73> <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=74> <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=75> <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=76> <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=77> <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=78> <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=79> <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=80> <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=81> <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=82> <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=83> <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=84> <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=85> <li>Jika <math>f(x) \cdot f(y) = f(x+y)</math> dan f(18)= 5 maka tentukan nilai f(72)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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) \cdot f(y) &= f(x+y) \\ f(18) \cdot f(18) &= f(18+18) \\ 5 \cdot 5 &= f(36) \\ f(36) &= 25 \\ f(36) \cdot f(36) &= f(36+36) \\ 25 \cdot 25 &= f(72) \\ f(72) &= 625 \\ \end{align} </math> </div></div> <ol start=86> <li>Jika <math>f(x) \cdot f(y) = f(xy)</math> dan f(4)= 16 maka tentukan nilai f(17)-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) \cdot f(y) &= f(xy) \\ f(4) \cdot f(4) &= f(4 \cdot 4) \\ 16 \cdot 16 &= f(16) \\ 16^2 &= f(16) \\ f(x) &= x^2 \\ f(17) &= 17^2 = 289 \\ f(17)-f(16) &= 289-256 = 33 \\ \end{align} </math> </div></div> <ol start=87> <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=88> <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=89> <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=90> <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=91> <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=92> <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=93> <li>Berapakah nilai xyz dan <math>\frac{xy+xz}{yz}</math> dari x+y=14, x+z=13 dan y+z=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} \text{ketiga persamaan tersebut dijumlahkan menjadi } \\ 2x+2y+2z &= 30 \\ x+y+z &= 15 \\ \text{untuk mencari x, y dan z masing-masing } \\ z &= 1 \\ y &= 2 \\ x &= 12 \\ xyz &= 24 \\ \frac{xy+xz}{yz} &= \frac{1 \cdot 2 + 1 \cdot 12}{2 \cdot 12} \\ &= \frac{14}{24} \\ &= \frac{7}{12} \\ \end{align} </math> </div></div> <ol start=94> <li>Tentukan nilai dari x+y+z dan yz(xy-xz) merupakan bilangan bulat positif jika xy = 2, xz = 6 dan yz = 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} xy \cdot yz \cdot xz &= 2 \cdot 3 \cdot 6 \\ (xyz)^2 &= 36 \\ xyz &= \pm 6 \\ xyz &= 6 \\ \frac{xyz}{xy} &= z = \frac{6}{2} = 3 \\ \frac{xyz}{yz} &= x = \frac{6}{3} = 2 \\ \frac{xyz}{xz} &= y = \frac{6}{6} = 1 \\ x+y+z &= 6 \\ yz(xy-xz) &= 3(2-6) = 3(-4) = -12 \\ \end{align} </math> </div></div> <ol start=95> <li>Tentukan nilai dari (x-z)^y jika <math>\frac{xy}{x+y} = \frac{1}{3}</math>, <math>\frac{yz}{y+z} = \frac{1}{4}</math> dan <math>\frac{xz}{x+z} = \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{xy}{x+y} &= \frac{1}{3} \\ \frac{x+y}{xy} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{y} + \frac{1}{x} &= 3 \\ \frac{yz}{y+z} &= \frac{1}{4} \\ \frac{y+z}{yz} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{z} + \frac{1}{y} &= 4 \\ \frac{xz}{x+z} &= \frac{1}{9} \\ \frac{x+z}{xz} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{z} + \frac{1}{x} &= 9 \\ \text{Misalkan 1/x = a, 1/y = b dan 1/z = c} \\ a+b &= 3 \\ b+c &= 4 \\ a+c &= 9 \\ a+b &= 3 \\ b+c &= 4 \\ a-c &= -1 \\ a-c &= -1 \\ a+c &= 9 \\ 2a &= 8 \\ a &= 4 \\ a-c &= -1 \\ 4-c &= -1 \\ c &= 5 \\ a+b &= 3 \\ 4+b &= 3 \\ b &= -1 \\ \frac{1}{x} &= 4 \\ x &= \frac{1}{4} \\ \frac{1}{y} &= -1 \\ y &= -1 \\ \frac{1}{z} &= 5 \\ z &= \frac{1}{5} \\ (x-z)^y &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=96> <li>Tentukan nilai dari (y+z)^x jika <math>\frac{x+y+z}{2} = \sqrt{x-2}+\sqrt{y-1}+\sqrt{z}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="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+z}{2} &= \sqrt{x-2}+\sqrt{y-1}+\sqrt{z} \\ x+y+z &= 2(\sqrt{x-2}+\sqrt{y-1}+\sqrt{z}) \\ x-2\sqrt{x-2}+y-2\sqrt{y-1}+z-2\sqrt{z} &= 0 \\ x-2-2\sqrt{x-2}+1+y-1-2\sqrt{y-1}+1+z-2\sqrt{z}+1 &= 0 \\ (\sqrt{x-2}-1)^2+(\sqrt{y-1}-1)^2+(\sqrt{z}-1)^2 &= 0 \\ (\sqrt{x-2}-1)^2 &= 0 \\ \sqrt{x-2}-1 &= 0 \\ \sqrt{x-2} &= 1 \\ x-2 &= 1 \\ x &= 3 \\ (\sqrt{y-1}-1)^2 &= 0 \\ \sqrt{y-1}-1 &= 0 \\ \sqrt{y-1} &= 1 \\ y-1 &= 1 \\ y &= 2 \\ (\sqrt{z}-1)^2 &= 0 \\ \sqrt{z}-1 &= 0 \\ \sqrt{z} &= 1 \\ z &= 1 \\ (y+z)^x &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=97> <li>Tentukan nilai dari x, y dan z jika <math>\frac{x+y}{2}=\frac{x+z}{4}=\frac{y+z}{5}</math> dan x+2y+3z=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{x+y}{2}=\frac{x+z}{4}=\frac{y+z}{5} &= k \\ x+y &= 2k \\ x+z &= 4k \\ y+z &= 5k \\ 2x+y+z &= 6k \\ 2x+5k &= 6k \\ k &= 2x \\ x &= \frac{k}{2} \\ y &= \frac{3k}{2} \\ z &= \frac{7k}{2} \\ x+2y+3z &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ x &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ y &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ z &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=98> <li>Tentukan nilai dari 2^x jika <math>2^{(x+y)}=40, 2^{(y+z)}=20, 2^{(x+z)}=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} 2^{(x+y)}=40, 2^{(y+z)}=20, 2^{(x+z)}=10 \\ *cara 1 \\ 2^{(x+y)} \cdot 2^{(y+z)} \cdot 2^{(x+z)} &= 40 \cdot 20 \cdot 10 \\ 2^{2(x+y+z)} &= 8000 \\ 2^{(x+y+z)} &= 40\sqrt{5} \\ 2^x \cdot 2^{(y+z)} &= 40\sqrt{5} \\ 2^x \cdot 20 &= 40\sqrt{5} \\ 2^x = 2\sqrt{5} \\ *cara 2 \\ \frac{2^{(x+y)}}{2^{(y+z)}} &= \frac{40}{20} \\ 2^{(x-z)} &= 2 \\ 2^{(x+z)} \cdot 2^{(x-z)} = 10 \cdot 2 \\ 2^{2x} &= 20 \\ 2^{x} &= 2\sqrt{5} \\ *cara 3 \\ \frac{2^{(x+y)} \cdot 2^{(x+z)}}{2^{(y+z)}} &= \frac{40(10)}{20} \\ 2^{2x} &= 20 \\ 2^x &= 2\sqrt{5} \\ *cara 4 \\ 2^{(x+y)}=40 &\text{ menjadi } x+y = ^2log 40 \\ 2^{(y+z)}=20 &\text{ menjadi } y+z = ^2log 20 \\ 2^{(x+z)}=10 &\text{ menjadi } x+z = ^2log 10 \\ x+y+x+z-(y+z) &= ^2log 40+^2log 10-^2log 20 \\ 2x &= ^2log 20 \\ 2^{2x} &= 20 \\ 2^x &= 2\sqrt{5} \\ \end{align} </math> </div></div> <ol start=98> <li>Tentukan nilai dari <math>\frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z}</math> jika <math>\frac{x^2}{by+cz} = \frac{y^2}{cz+ax} = \frac{z^2}{ax+by} = 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{x^2}{by+cz} = \frac{y^2}{cz+ax} = \frac{z^2}{ax+by} &= 1 \\ \frac{x^2}{by+cz} &= 1 \\ x^2 &= by+cz \\ \frac{y^2}{cz+ax} &= 1 \\ y^2 &= cz+ax \\ \frac{z^2}{ax+by} &= 1 \\ z^2 &= ax+by \\ \frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z} &= \frac{ax}{(a+x)x} + \frac{by}{(b+y)y} + \frac{cz}{(c+z)z} \\ &= \frac{ax}{ax+x^2} + \frac{by}{by+y^2} + \frac{cz}{cz+z^2} \\ &= \frac{ax}{ax+by+cz} + \frac{by}{by+cz+ax} + \frac{cz}{cz+ax+by} \\ &= \frac{ax+by+cz}{ax+by+cz} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=99> <li>Tentukan nilai dari <math>\frac{1}{xy+z-1} + \frac{1}{yz+x-1} + \frac{1}{xz+y-1}</math> jika x+y+z=2, <math>x^2+y^2+z^2=6</math> serta xyz = 4?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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 &= 2 \\ x &= 2-y-z \\ y &= 2-x-z \\ z &= 2-x-y \\ \frac{1}{xy+z-1} + \frac{1}{yz+x-1} + \frac{1}{xz+y-1} &= \frac{1}{xy+2-x-y-1} + \frac{1}{yz+2-y-z-1} + \frac{1}{xz+2-x-z-1} \\ &= \frac{1}{xy-x-y+1} + \frac{1}{yz-y-z+1} + \frac{1}{xz-x-z+1} \\ &= \frac{1}{x(y-1)-(y-1)} + \frac{1}{y(z-1)-(z-1)} + \frac{1}{x(z-1)-(z-1)} \\ &= \frac{1}{(x-1)(y-1)} + \frac{1}{(y-1)(z-1)} + \frac{1}{(x-1)(z-1)} \\ &= \frac{z-1+x-1+y-1}{(x-1)(y-1)(z-1)} \\ &= \frac{x+y+z-3}{(x-1)(y-1)(z-1)} \\ &= \frac{x+y+z-3}{xyz-(xy+yz+xz)+(x+y+z)-1} \\ (x+y+z)^2 &= x^2+y^2+z^2+2(xy+yz+xz) \\ 2^2 &= 6+2(xy+yz+xz) \\ 4 &= 6+2(xy+yz+xz) \\ -2 &= 2(xy+yz+xz) \\ xy+yz+xz &= -1 \\ \frac{x+y+z-3}{xyz-(xy+yz+xz)+(x+y+z)-1} &= \frac{2-3}{4-(-1)+2-1} \\ &= \frac{-1}{6} \\ &= -\frac{1}{6} \\ \end{align} </math> </div></div> <ol start=100> <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=101> <li>x, y dan z merupakan bilangan tak nol. Jika <math>x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}</math> maka berapa nilai 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} x+\frac{1}{y} &= y+\frac{1}{z} \\ x-y &= \frac{1}{z}-\frac{1}{y} \\ &= \frac{y-z}{yz} \\ y+\frac{1}{z} &= z+\frac{1}{x} \\ y-z &= \frac{1}{x}-\frac{1}{z} \\ &= \frac{z-x}{xz} \\ z+\frac{1}{x} &= x+\frac{1}{y} \\ z-x &= \frac{1}{y}-\frac{1}{x} \\ &= \frac{x-y}{xy} \\ (x-y)(y-z)(z-x) &= (\frac{y-z}{yz}) \cdot (\frac{z-x}{xz}) \cdot (\frac{x-y}{xy}) \\ x^2y^2z^2 &= 1 \\ (xyz)^2 &= 1 \\ xyz &= \pm 1 \\ \end{align} </math> </div></div> <ol start=102> <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=103> <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=104> <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=105> <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=106> <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=107> <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=108> <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=109> <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=110> <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=111> <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=112> <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=113> <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=114> <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=115> <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=116> <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=117> <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=118> <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=119> <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=120> <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=121> <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=122> <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=123> <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=124> <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=125> <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=126> <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. [[Kategori:Soal-Soal Matematika]] makcmmgc1cpxo95um4e0aa2c243c7xp 121431 121430 2026-10-23T01:12:12Z Akuindo 8654 121431 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>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-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{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> <ol start=7> <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} = y \\ y^5(1+y+y^2) &= 5^5 \cdot 31 \\ y &= 5 \\ x^{10} &= 5 \\ x &= ^5log \, 10 \\ \end{align} </math> </div></div> <ol start=8> <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=9> <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=10> <li>Berapa nilai x dari <math>\frac{\sqrt{(^4log \, x)^2+ ^2log \, x-4}}{16- ^2log \, x- (^4log \, x)^2} = \frac{1}{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} ^4 log \, x &= \frac{^2log \, x}{2} \\ \frac{\sqrt{(\frac{^2log \, x}{2})^2+ ^2log \, x-4}}{16- ^2log \, x- (\frac{^2log \, x}{2})^2} &= \frac{1}{4} \\ \text{misalkan } y &= ^2log \, x \\ \frac{\sqrt{(\frac{y}{2})^2+y-4}}{16-y-(\frac{y}{2})^2} &= \frac{1}{4} \\ \frac{\sqrt{(\frac{y}{2})^2+y-4}}{12-(-4+y+(\frac{y}{2})^2)} &= \frac{1}{4} \\ \text{misalkan } a &= (\frac{y}{2})^2+y-4 \\ \frac{\sqrt{a}}{12-a} &= \frac{1}{4}, a \ge 0 \\ \text{misalkan } \sqrt{a} &= b, b \ge 0 \\ \frac{b}{12-b^2} &= \frac{1}{4} \\ 4b &= 12-b^2 \\ b^2+4b-12 &= 0 \\ (b-2)(b+6) &= 0 \\ b = 2 &\text{ atau } b = -6 (TM) \\ \sqrt{a} &= 2 \\ a &= 4 \\ (\frac{y}{2})^2+y-4 &= 4 \\ \frac{y^2}{4}+y-8 &= 0 \\ y^2+4y-32 &= 0 \\ (y-4)(y+8) &= 0 \\ y = 4 &\text{ atau } y=-8 \\ ^2 log \, x &= 4 \\ x &= 16 \\ ^2 log \, x = -8 \\ x &= \frac{1}{256} \\ \end{align} </math> </div></div> <ol start=11> <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=12> <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=13> <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=14> <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=15> <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=16> <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=17> <li>Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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> <ol start=18> <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=19> <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=20> <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=21> <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=22> <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=23> <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=24> <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=25> <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=26> <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=27> <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=28> <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=29> <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=30> <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=31> <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=32> <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=33> <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=34> <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=35> <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=36> <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=37> <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=38> <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=39> <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=40> <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=41> <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=42> <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=43> <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=44> <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=45> <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=46> <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=47> <li>Berapakah nilai dari <math>5x+\frac{1}{4x}</math> jika <math>25x^2+\frac{1}{16x^2}=\frac{13}{2}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (5x+\frac{1}{4x})^2 &= 25x^2+\frac{5}{2}+\frac{16}{x^2} \\ &= 25x^2+\frac{16}{x^2}+\frac{5}{2} \\ &= \frac{13}{2}+\frac{5}{2} \\ &= 9 \\ &= \sqrt{9} \\ &= \pm 3 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>x^{28}+x^{25}+x^{23}+x^{20}</math> jika <math>x+\frac{1}{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} x+\frac{1}{x} &= 1 \\ x-1+\frac{1}{x} &= 0 \\ x^2-x+1 &= 0 \\ (x+1)(x^2-x+1) &= (x+1)0 \\ x^3+1 &= 0 \\ x^3 &= -1 \\ x &= -1 \\ x^{28}+x^{25}+x^{23}+x^{20} &= (-1)^{28}+(-1)^{25}+(-1)^{23}+(-1)^{20} \\ &= 1-1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>x^{28}-x^{25}+x^{23}-x^{20}</math> jika <math>x+\frac{1}{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} x+\frac{1}{x} &= -1 \\ x+1+\frac{1}{x} &= 0 \\ x^2-x+1 &= 0 \\ (x-1)(x^2-x+1) &= (x-1)0 \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{28}-x^{25}+x^{23}-x^{20} &= x^{25}(x^{3}-1)+x^{20}(x^{3}-1) &= x^{25}(0)+x^{20}(0) \\ &= 0 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>\frac{3}{2x-3}+\frac{4}{2y-4}+\frac{5}{2z-5}</math> jika <math>\frac{2x}{2x-3}+\frac{2y}{2y-4}+\frac{2z}{2z-5}=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} \frac{2x}{2x-3}+\frac{2y}{2y-4}+\frac{2z}{2z-5} &= 7 \\ \frac{2x-3+3}{2x-3}+\frac{2y-4+4}{2y-4}+\frac{2z-5+5}{2z-5} &= 7 \\ 1+\frac{3}{2x-3}+1+\frac{4}{2y-4}+1+\frac{5}{2z-5} &= 7 \\ \frac{3}{2x-3}+\frac{4}{2y-4}+\frac{5}{2z-5} &= 4 \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>\frac{4x}{4x-7}+\frac{5y}{5y-7}+\frac{6z}{6z-7}</math> jika <math>\frac{7}{4x-7}+\frac{7}{5y-7}+\frac{7}{6z-7}=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{7}{4x-7}+\frac{7}{5y-7}+\frac{7}{6z-7} &= 10 \\ \frac{7+4x-4x}{4x-7}+\frac{7+5y-5y}{5y-7}+\frac{7+6z-6z}{6z-7} &= 10 \\ \frac{4x-(4x-7)}{4x-7}+\frac{5y-(5y-7)}{5y-7}+\frac{6z-(6z-7)}{6z-7} &= 10 \\ \frac{4x}{4x-7}-1+\frac{5y}{5y-7}-1+\frac{6z}{6z-7}-1 &= 10 \\ \frac{4x}{4x-7}+\frac{5y}{5y-7}+\frac{6z}{6z-7} &= 13 \\ \end{align} </math> </div></div> <ol start=52> <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=53> <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=54> <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=55> <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=56> <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=57> <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=58> <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=59> <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-3 sin \, x csc \, x(sin \, x-csc \, x) \\ 8^3 &= sin^3 \, x-csc^3 \, x-3 sin \, 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=60> <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{2 sin \, 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=61> <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=62> <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=63> <li>Buktikan bahwa <math>\frac{sin \, x}{1-cos \, x} = cot \, \frac{y}{4}</math> jika <math>cot \, x = \frac{1+cos \, y}{sin \, 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} cot \, x &= \frac{2 cos^2 \, \frac{y}{2}}{2 sin \, \frac{y}{2} \, cos \, \frac{y}{2}} \\ &= \frac{cos \frac{y}{2}}{sin \, \frac{y}{2}} \\ &= cot \, \frac{y}{2} \\ x &=\frac{y}{2} \\ \frac{sin \, x}{1-cos \, x} &= \frac{sin \, \frac{y}{2}}{1-cos \, \frac{y}{2}} \\ &= \frac{2 sin \, \frac{y}{4} \, cos \, \frac{y}{4}}{2 sin^2 \, \frac{y}{4}} \\ &= \frac{cos \, \frac{y}{4}}{sin \, \frac{y}{4}} \\ &= cot \, \frac{y}{4} \end{align} </math> </div></div> <ol start=64> <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=65> <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=66> <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=67> <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=68> <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=69> <li>Berapakah nilai dari <math>x^5-3x^4-x^3+9x^2-11x+3</math> jika <math>x+\frac{1}{x}=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} *cara 1 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2-3x+1 &= 0 \\ (x^3-2x+3)(x^2-3x+1 &= 0(x^3-2x+3) \\ x^5-3x^4-x^3+9x^2-11x+3 &= 0 \\ *cara 2 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2-3x+1 &= 0 \\ x^5-3x^4-x^3+9x^2-11x+3 &= x^5-3x^4+x^3-2x^3+6x^2+3x^2-2x-9x+3 \\ &= x^5-3x^4+x^3-2x^3+6x^2-2x+3x^2-9x+3 \\ &= x^3(x^2-3x+1)-2x(x^2-3x+1)+3(x^2-3x+1) \\ &= x^3(0)-2x(0)+3(0) \\ &= 0 \\ *cara 3 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2 &= 3x-1 \\ x^2-3x+1 &= 0 \\ x^3 &= x \cdot x^2 \\ &= x (3x-1) \\ &= 3x^2-x x^4 &= x \cdot x^3 \\ &= x(3x^2-x) \\ &= 3x^3-x^2 \\ x^5 &= x \cdot x^4 \\ &= x(3x^3-x^2) \\ &= 3x^4-x^3 \\ x^5-3x^4-x^3+9x^2-11x+3 &= 3x^4-x^3-3(3x^3-x^2)-(3x^2-x)+9(3x-1)-11x+3 \\ \end{align} </math> </div></div> <ol start=70> <li>Berapakah nilai dari <math>\sqrt{x-7}</math> jika x=<math>16+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 &= 16+4\sqrt{5} \\ x-7 &= 16-7+4\sqrt{5} \\ &= 9+4\sqrt{5} \\ &= (2+\sqrt{5})^2 \\ \sqrt{x-7} &= 2+\sqrt{5} \\ \end{align} </math> </div></div> <ol start=71> <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=72> <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=73> <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=74> <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=75> <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=76> <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=77> <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=78> <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=79> <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=80> <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=81> <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=82> <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=83> <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=84> <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=85> <li>Jika <math>f(x) \cdot f(y) = f(x+y)</math> dan f(18)= 5 maka tentukan nilai f(72)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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) \cdot f(y) &= f(x+y) \\ f(18) \cdot f(18) &= f(18+18) \\ 5 \cdot 5 &= f(36) \\ f(36) &= 25 \\ f(36) \cdot f(36) &= f(36+36) \\ 25 \cdot 25 &= f(72) \\ f(72) &= 625 \\ \end{align} </math> </div></div> <ol start=86> <li>Jika <math>f(x) \cdot f(y) = f(xy)</math> dan f(4)= 16 maka tentukan nilai f(17)-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) \cdot f(y) &= f(xy) \\ f(4) \cdot f(4) &= f(4 \cdot 4) \\ 16 \cdot 16 &= f(16) \\ 16^2 &= f(16) \\ f(x) &= x^2 \\ f(17) &= 17^2 = 289 \\ f(17)-f(16) &= 289-256 = 33 \\ \end{align} </math> </div></div> <ol start=87> <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=88> <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=89> <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=90> <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=91> <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=92> <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=93> <li>Berapakah nilai xyz dan <math>\frac{xy+xz}{yz}</math> dari x+y=14, x+z=13 dan y+z=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} \text{ketiga persamaan tersebut dijumlahkan menjadi } \\ 2x+2y+2z &= 30 \\ x+y+z &= 15 \\ \text{untuk mencari x, y dan z masing-masing } \\ z &= 1 \\ y &= 2 \\ x &= 12 \\ xyz &= 24 \\ \frac{xy+xz}{yz} &= \frac{1 \cdot 2 + 1 \cdot 12}{2 \cdot 12} \\ &= \frac{14}{24} \\ &= \frac{7}{12} \\ \end{align} </math> </div></div> <ol start=94> <li>Tentukan nilai dari x+y+z dan yz(xy-xz) merupakan bilangan bulat positif jika xy = 2, xz = 6 dan yz = 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} xy \cdot yz \cdot xz &= 2 \cdot 3 \cdot 6 \\ (xyz)^2 &= 36 \\ xyz &= \pm 6 \\ xyz &= 6 \\ \frac{xyz}{xy} &= z = \frac{6}{2} = 3 \\ \frac{xyz}{yz} &= x = \frac{6}{3} = 2 \\ \frac{xyz}{xz} &= y = \frac{6}{6} = 1 \\ x+y+z &= 6 \\ yz(xy-xz) &= 3(2-6) = 3(-4) = -12 \\ \end{align} </math> </div></div> <ol start=95> <li>Tentukan nilai dari (x-z)^y jika <math>\frac{xy}{x+y} = \frac{1}{3}</math>, <math>\frac{yz}{y+z} = \frac{1}{4}</math> dan <math>\frac{xz}{x+z} = \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{xy}{x+y} &= \frac{1}{3} \\ \frac{x+y}{xy} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{y} + \frac{1}{x} &= 3 \\ \frac{yz}{y+z} &= \frac{1}{4} \\ \frac{y+z}{yz} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{z} + \frac{1}{y} &= 4 \\ \frac{xz}{x+z} &= \frac{1}{9} \\ \frac{x+z}{xz} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{z} + \frac{1}{x} &= 9 \\ \text{Misalkan 1/x = a, 1/y = b dan 1/z = c} \\ a+b &= 3 \\ b+c &= 4 \\ a+c &= 9 \\ a+b &= 3 \\ b+c &= 4 \\ a-c &= -1 \\ a-c &= -1 \\ a+c &= 9 \\ 2a &= 8 \\ a &= 4 \\ a-c &= -1 \\ 4-c &= -1 \\ c &= 5 \\ a+b &= 3 \\ 4+b &= 3 \\ b &= -1 \\ \frac{1}{x} &= 4 \\ x &= \frac{1}{4} \\ \frac{1}{y} &= -1 \\ y &= -1 \\ \frac{1}{z} &= 5 \\ z &= \frac{1}{5} \\ (x-z)^y &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=96> <li>Tentukan nilai dari (y+z)^x jika <math>\frac{x+y+z}{2} = \sqrt{x-2}+\sqrt{y-1}+\sqrt{z}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="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+z}{2} &= \sqrt{x-2}+\sqrt{y-1}+\sqrt{z} \\ x+y+z &= 2(\sqrt{x-2}+\sqrt{y-1}+\sqrt{z}) \\ x-2\sqrt{x-2}+y-2\sqrt{y-1}+z-2\sqrt{z} &= 0 \\ x-2-2\sqrt{x-2}+1+y-1-2\sqrt{y-1}+1+z-2\sqrt{z}+1 &= 0 \\ (\sqrt{x-2}-1)^2+(\sqrt{y-1}-1)^2+(\sqrt{z}-1)^2 &= 0 \\ (\sqrt{x-2}-1)^2 &= 0 \\ \sqrt{x-2}-1 &= 0 \\ \sqrt{x-2} &= 1 \\ x-2 &= 1 \\ x &= 3 \\ (\sqrt{y-1}-1)^2 &= 0 \\ \sqrt{y-1}-1 &= 0 \\ \sqrt{y-1} &= 1 \\ y-1 &= 1 \\ y &= 2 \\ (\sqrt{z}-1)^2 &= 0 \\ \sqrt{z}-1 &= 0 \\ \sqrt{z} &= 1 \\ z &= 1 \\ (y+z)^x &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=97> <li>Tentukan nilai dari x, y dan z jika <math>\frac{x+y}{2}=\frac{x+z}{4}=\frac{y+z}{5}</math> dan x+2y+3z=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{x+y}{2}=\frac{x+z}{4}=\frac{y+z}{5} &= k \\ x+y &= 2k \\ x+z &= 4k \\ y+z &= 5k \\ 2x+y+z &= 6k \\ 2x+5k &= 6k \\ k &= 2x \\ x &= \frac{k}{2} \\ y &= \frac{3k}{2} \\ z &= \frac{7k}{2} \\ x+2y+3z &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ x &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ y &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ z &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=98> <li>Tentukan nilai dari 2^x jika <math>2^{(x+y)}=40, 2^{(y+z)}=20, 2^{(x+z)}=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} *cara 1 \\ 2^{(x+y)} \cdot 2^{(y+z)} \cdot 2^{(x+z)} &= 40 \cdot 20 \cdot 10 \\ 2^{2(x+y+z)} &= 8000 \\ 2^{(x+y+z)} &= 40\sqrt{5} \\ 2^x \cdot 2^{(y+z)} &= 40\sqrt{5} \\ 2^x \cdot 20 &= 40\sqrt{5} \\ 2^x &= 2\sqrt{5} \\ *cara 2 \\ \frac{2^{(x+y)}}{2^{(y+z)}} &= \frac{40}{20} \\ 2^{(x-z)} &= 2 \\ 2^{(x+z)} \cdot 2^{(x-z)} &= 10 \cdot 2 \\ 2^{2x} &= 20 \\ 2^{x} &= 2\sqrt{5} \\ *cara 3 \\ \frac{2^{(x+y)} \cdot 2^{(x+z)}}{2^{(y+z)}} &= \frac{40(10)}{20} \\ 2^{2x} &= 20 \\ 2^x &= 2\sqrt{5} \\ *cara 4 \\ 2^{(x+y)}=40 &\text{ menjadi } x+y = ^2log 40 \\ 2^{(y+z)}=20 &\text{ menjadi } y+z = ^2log 20 \\ 2^{(x+z)}=10 &\text{ menjadi } x+z = ^2log 10 \\ x+y+x+z-(y+z) &= ^2log 40+^2log 10-^2log 20 \\ 2x &= ^2log 20 \\ 2^{2x} &= 20 \\ 2^x &= 2\sqrt{5} \\ \end{align} </math> </div></div> <ol start=98> <li>Tentukan nilai dari <math>\frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z}</math> jika <math>\frac{x^2}{by+cz} = \frac{y^2}{cz+ax} = \frac{z^2}{ax+by} = 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{x^2}{by+cz} = \frac{y^2}{cz+ax} = \frac{z^2}{ax+by} &= 1 \\ \frac{x^2}{by+cz} &= 1 \\ x^2 &= by+cz \\ \frac{y^2}{cz+ax} &= 1 \\ y^2 &= cz+ax \\ \frac{z^2}{ax+by} &= 1 \\ z^2 &= ax+by \\ \frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z} &= \frac{ax}{(a+x)x} + \frac{by}{(b+y)y} + \frac{cz}{(c+z)z} \\ &= \frac{ax}{ax+x^2} + \frac{by}{by+y^2} + \frac{cz}{cz+z^2} \\ &= \frac{ax}{ax+by+cz} + \frac{by}{by+cz+ax} + \frac{cz}{cz+ax+by} \\ &= \frac{ax+by+cz}{ax+by+cz} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=99> <li>Tentukan nilai dari <math>\frac{1}{xy+z-1} + \frac{1}{yz+x-1} + \frac{1}{xz+y-1}</math> jika x+y+z=2, <math>x^2+y^2+z^2=6</math> serta xyz = 4?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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 &= 2 \\ x &= 2-y-z \\ y &= 2-x-z \\ z &= 2-x-y \\ \frac{1}{xy+z-1} + \frac{1}{yz+x-1} + \frac{1}{xz+y-1} &= \frac{1}{xy+2-x-y-1} + \frac{1}{yz+2-y-z-1} + \frac{1}{xz+2-x-z-1} \\ &= \frac{1}{xy-x-y+1} + \frac{1}{yz-y-z+1} + \frac{1}{xz-x-z+1} \\ &= \frac{1}{x(y-1)-(y-1)} + \frac{1}{y(z-1)-(z-1)} + \frac{1}{x(z-1)-(z-1)} \\ &= \frac{1}{(x-1)(y-1)} + \frac{1}{(y-1)(z-1)} + \frac{1}{(x-1)(z-1)} \\ &= \frac{z-1+x-1+y-1}{(x-1)(y-1)(z-1)} \\ &= \frac{x+y+z-3}{(x-1)(y-1)(z-1)} \\ &= \frac{x+y+z-3}{xyz-(xy+yz+xz)+(x+y+z)-1} \\ (x+y+z)^2 &= x^2+y^2+z^2+2(xy+yz+xz) \\ 2^2 &= 6+2(xy+yz+xz) \\ 4 &= 6+2(xy+yz+xz) \\ -2 &= 2(xy+yz+xz) \\ xy+yz+xz &= -1 \\ \frac{x+y+z-3}{xyz-(xy+yz+xz)+(x+y+z)-1} &= \frac{2-3}{4-(-1)+2-1} \\ &= \frac{-1}{6} \\ &= -\frac{1}{6} \\ \end{align} </math> </div></div> <ol start=100> <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=101> <li>x, y dan z merupakan bilangan tak nol. Jika <math>x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}</math> maka berapa nilai 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} x+\frac{1}{y} &= y+\frac{1}{z} \\ x-y &= \frac{1}{z}-\frac{1}{y} \\ &= \frac{y-z}{yz} \\ y+\frac{1}{z} &= z+\frac{1}{x} \\ y-z &= \frac{1}{x}-\frac{1}{z} \\ &= \frac{z-x}{xz} \\ z+\frac{1}{x} &= x+\frac{1}{y} \\ z-x &= \frac{1}{y}-\frac{1}{x} \\ &= \frac{x-y}{xy} \\ (x-y)(y-z)(z-x) &= (\frac{y-z}{yz}) \cdot (\frac{z-x}{xz}) \cdot (\frac{x-y}{xy}) \\ x^2y^2z^2 &= 1 \\ (xyz)^2 &= 1 \\ xyz &= \pm 1 \\ \end{align} </math> </div></div> <ol start=102> <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=103> <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=104> <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=105> <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=106> <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=107> <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=108> <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=109> <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=110> <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=111> <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=112> <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=113> <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=114> <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=115> <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=116> <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=117> <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=118> <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=119> <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=120> <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=121> <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=122> <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=123> <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=124> <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=125> <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=126> <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. [[Kategori:Soal-Soal Matematika]] hvvemdhqcwtwuif8h34iyhykzldqs94 121432 121431 2026-10-23T02:25:27Z Akuindo 8654 121432 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>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-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{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> <ol start=7> <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} = y \\ y^5(1+y+y^2) &= 5^5 \cdot 31 \\ y &= 5 \\ x^{10} &= 5 \\ x &= ^5log \, 10 \\ \end{align} </math> </div></div> <ol start=8> <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=9> <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=10> <li>Berapa nilai x dari <math>\frac{\sqrt{(^4log \, x)^2+ ^2log \, x-4}}{16- ^2log \, x- (^4log \, x)^2} = \frac{1}{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} ^4 log \, x &= \frac{^2log \, x}{2} \\ \frac{\sqrt{(\frac{^2log \, x}{2})^2+ ^2log \, x-4}}{16- ^2log \, x- (\frac{^2log \, x}{2})^2} &= \frac{1}{4} \\ \text{misalkan } y &= ^2log \, x \\ \frac{\sqrt{(\frac{y}{2})^2+y-4}}{16-y-(\frac{y}{2})^2} &= \frac{1}{4} \\ \frac{\sqrt{(\frac{y}{2})^2+y-4}}{12-(-4+y+(\frac{y}{2})^2)} &= \frac{1}{4} \\ \text{misalkan } a &= (\frac{y}{2})^2+y-4 \\ \frac{\sqrt{a}}{12-a} &= \frac{1}{4}, a \ge 0 \\ \text{misalkan } \sqrt{a} &= b, b \ge 0 \\ \frac{b}{12-b^2} &= \frac{1}{4} \\ 4b &= 12-b^2 \\ b^2+4b-12 &= 0 \\ (b-2)(b+6) &= 0 \\ b = 2 &\text{ atau } b = -6 (TM) \\ \sqrt{a} &= 2 \\ a &= 4 \\ (\frac{y}{2})^2+y-4 &= 4 \\ \frac{y^2}{4}+y-8 &= 0 \\ y^2+4y-32 &= 0 \\ (y-4)(y+8) &= 0 \\ y = 4 &\text{ atau } y=-8 \\ ^2 log \, x &= 4 \\ x &= 16 \\ ^2 log \, x = -8 \\ x &= \frac{1}{256} \\ \end{align} </math> </div></div> <ol start=11> <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=12> <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=13> <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=14> <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=15> <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=16> <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=17> <li>Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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> <ol start=18> <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=19> <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=20> <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=21> <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=22> <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=23> <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=24> <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=25> <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=26> <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=27> <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=28> <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=29> <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=30> <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=31> <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=32> <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=33> <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=34> <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=35> <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=36> <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=37> <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=38> <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=39> <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=40> <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=41> <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=42> <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=43> <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=44> <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=45> <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=46> <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=47> <li>Berapakah nilai dari <math>5x+\frac{1}{4x}</math> jika <math>25x^2+\frac{1}{16x^2}=\frac{13}{2}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (5x+\frac{1}{4x})^2 &= 25x^2+\frac{5}{2}+\frac{16}{x^2} \\ &= 25x^2+\frac{16}{x^2}+\frac{5}{2} \\ &= \frac{13}{2}+\frac{5}{2} \\ &= 9 \\ &= \sqrt{9} \\ &= \pm 3 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>x^{28}+x^{25}+x^{23}+x^{20}</math> jika <math>x+\frac{1}{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} x+\frac{1}{x} &= 1 \\ x-1+\frac{1}{x} &= 0 \\ x^2-x+1 &= 0 \\ (x+1)(x^2-x+1) &= (x+1)0 \\ x^3+1 &= 0 \\ x^3 &= -1 \\ x &= -1 \\ x^{28}+x^{25}+x^{23}+x^{20} &= (-1)^{28}+(-1)^{25}+(-1)^{23}+(-1)^{20} \\ &= 1-1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>x^{28}-x^{25}+x^{23}-x^{20}</math> jika <math>x+\frac{1}{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} x+\frac{1}{x} &= -1 \\ x+1+\frac{1}{x} &= 0 \\ x^2-x+1 &= 0 \\ (x-1)(x^2-x+1) &= (x-1)0 \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{28}-x^{25}+x^{23}-x^{20} &= x^{25}(x^{3}-1)+x^{20}(x^{3}-1) &= x^{25}(0)+x^{20}(0) \\ &= 0 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>\frac{3}{2x-3}+\frac{4}{2y-4}+\frac{5}{2z-5}</math> jika <math>\frac{2x}{2x-3}+\frac{2y}{2y-4}+\frac{2z}{2z-5}=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} \frac{2x}{2x-3}+\frac{2y}{2y-4}+\frac{2z}{2z-5} &= 7 \\ \frac{2x-3+3}{2x-3}+\frac{2y-4+4}{2y-4}+\frac{2z-5+5}{2z-5} &= 7 \\ 1+\frac{3}{2x-3}+1+\frac{4}{2y-4}+1+\frac{5}{2z-5} &= 7 \\ \frac{3}{2x-3}+\frac{4}{2y-4}+\frac{5}{2z-5} &= 4 \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>\frac{4x}{4x-7}+\frac{5y}{5y-7}+\frac{6z}{6z-7}</math> jika <math>\frac{7}{4x-7}+\frac{7}{5y-7}+\frac{7}{6z-7}=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{7}{4x-7}+\frac{7}{5y-7}+\frac{7}{6z-7} &= 10 \\ \frac{7+4x-4x}{4x-7}+\frac{7+5y-5y}{5y-7}+\frac{7+6z-6z}{6z-7} &= 10 \\ \frac{4x-(4x-7)}{4x-7}+\frac{5y-(5y-7)}{5y-7}+\frac{6z-(6z-7)}{6z-7} &= 10 \\ \frac{4x}{4x-7}-1+\frac{5y}{5y-7}-1+\frac{6z}{6z-7}-1 &= 10 \\ \frac{4x}{4x-7}+\frac{5y}{5y-7}+\frac{6z}{6z-7} &= 13 \\ \end{align} </math> </div></div> <ol start=52> <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=53> <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=54> <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=55> <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=56> <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=57> <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=58> <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=59> <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-3 sin \, x csc \, x(sin \, x-csc \, x) \\ 8^3 &= sin^3 \, x-csc^3 \, x-3 sin \, 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=60> <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{2 sin \, 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=61> <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=62> <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=63> <li>Berapa nilsi dari sin (x+y) dan cos (x+y) jika sin x + sin y = a serta cos x + cos y = 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} sin \, x+sin \, y &= 2 \, sin \, \frac{x+y}{2} \, cos \, \frac{x-y}{2} \\ cos \, x+cos \, y &= 2 \, cos \, \frac{x+y}{2} \, cos \, \frac{x-y}{2} \\ \frac{2 \, sin \, \frac{x+y}{2} \, cos \, \frac{x-y}{2}}{2 \, cos \, \frac{x+y}{2} \, cos \, \frac{x-y}{2}} &= \frac{sin \, \frac{x+y}{2}}{cos \frac{x+y}{2}} \\ &= tan \, \frac{x+y}{2} \\ \text{misalkan } p &= \frac{x+y}{2} \text{ maka } 2p &= x+y \\ tan \, \frac{x+y}{2} &= tan \, p \\ &= \frac{a}{b} \\ sin (x+y) = sin 2p &= \frac{2 \, tan \, p}{1+tan^2 \, p} \\ &= \frac{2 \, (\frac{a}{b})}{1+(\frac{a}{b})^2} \\ &= \frac{2 \, (\frac{a}{b})}{\frac{b^2+a^2}{b^2}} \\ &= \frac{2 \, (\frac{a}{b})}{\frac{b^2+a^2}{b^2}} \\ &= \frac{2ab}{a^2+b^2} \end{align} </math> </div></div> <ol start=64> <li>Buktikan bahwa <math>\frac{sin \, x}{1-cos \, x} = cot \, \frac{y}{4}</math> jika <math>cot \, x = \frac{1+cos \, y}{sin \, 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} cot \, x &= \frac{2 cos^2 \, \frac{y}{2}}{2 sin \, \frac{y}{2} \, cos \, \frac{y}{2}} \\ &= \frac{cos \frac{y}{2}}{sin \, \frac{y}{2}} \\ &= cot \, \frac{y}{2} \\ x &=\frac{y}{2} \\ \frac{sin \, x}{1-cos \, x} &= \frac{sin \, \frac{y}{2}}{1-cos \, \frac{y}{2}} \\ &= \frac{2 sin \, \frac{y}{4} \, cos \, \frac{y}{4}}{2 sin^2 \, \frac{y}{4}} \\ &= \frac{cos \, \frac{y}{4}}{sin \, \frac{y}{4}} \\ &= cot \, \frac{y}{4} \end{align} </math> </div></div> <ol start=64> <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=65> <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=66> <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=67> <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=68> <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=69> <li>Berapakah nilai dari <math>x^5-3x^4-x^3+9x^2-11x+3</math> jika <math>x+\frac{1}{x}=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} *cara 1 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2-3x+1 &= 0 \\ (x^3-2x+3)(x^2-3x+1 &= 0(x^3-2x+3) \\ x^5-3x^4-x^3+9x^2-11x+3 &= 0 \\ *cara 2 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2-3x+1 &= 0 \\ x^5-3x^4-x^3+9x^2-11x+3 &= x^5-3x^4+x^3-2x^3+6x^2+3x^2-2x-9x+3 \\ &= x^5-3x^4+x^3-2x^3+6x^2-2x+3x^2-9x+3 \\ &= x^3(x^2-3x+1)-2x(x^2-3x+1)+3(x^2-3x+1) \\ &= x^3(0)-2x(0)+3(0) \\ &= 0 \\ *cara 3 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2 &= 3x-1 \\ x^2-3x+1 &= 0 \\ x^3 &= x \cdot x^2 \\ &= x (3x-1) \\ &= 3x^2-x x^4 &= x \cdot x^3 \\ &= x(3x^2-x) \\ &= 3x^3-x^2 \\ x^5 &= x \cdot x^4 \\ &= x(3x^3-x^2) \\ &= 3x^4-x^3 \\ x^5-3x^4-x^3+9x^2-11x+3 &= 3x^4-x^3-3(3x^3-x^2)-(3x^2-x)+9(3x-1)-11x+3 \\ \end{align} </math> </div></div> <ol start=70> <li>Berapakah nilai dari <math>\sqrt{x-7}</math> jika x=<math>16+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 &= 16+4\sqrt{5} \\ x-7 &= 16-7+4\sqrt{5} \\ &= 9+4\sqrt{5} \\ &= (2+\sqrt{5})^2 \\ \sqrt{x-7} &= 2+\sqrt{5} \\ \end{align} </math> </div></div> <ol start=71> <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=72> <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=73> <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=74> <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=75> <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=76> <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=77> <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=78> <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=79> <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=80> <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=81> <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=82> <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=83> <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=84> <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=85> <li>Jika <math>f(x) \cdot f(y) = f(x+y)</math> dan f(18)= 5 maka tentukan nilai f(72)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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) \cdot f(y) &= f(x+y) \\ f(18) \cdot f(18) &= f(18+18) \\ 5 \cdot 5 &= f(36) \\ f(36) &= 25 \\ f(36) \cdot f(36) &= f(36+36) \\ 25 \cdot 25 &= f(72) \\ f(72) &= 625 \\ \end{align} </math> </div></div> <ol start=86> <li>Jika <math>f(x) \cdot f(y) = f(xy)</math> dan f(4)= 16 maka tentukan nilai f(17)-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) \cdot f(y) &= f(xy) \\ f(4) \cdot f(4) &= f(4 \cdot 4) \\ 16 \cdot 16 &= f(16) \\ 16^2 &= f(16) \\ f(x) &= x^2 \\ f(17) &= 17^2 = 289 \\ f(17)-f(16) &= 289-256 = 33 \\ \end{align} </math> </div></div> <ol start=87> <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=88> <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=89> <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=90> <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=91> <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=92> <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=93> <li>Berapakah nilai xyz dan <math>\frac{xy+xz}{yz}</math> dari x+y=14, x+z=13 dan y+z=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} \text{ketiga persamaan tersebut dijumlahkan menjadi } \\ 2x+2y+2z &= 30 \\ x+y+z &= 15 \\ \text{untuk mencari x, y dan z masing-masing } \\ z &= 1 \\ y &= 2 \\ x &= 12 \\ xyz &= 24 \\ \frac{xy+xz}{yz} &= \frac{1 \cdot 2 + 1 \cdot 12}{2 \cdot 12} \\ &= \frac{14}{24} \\ &= \frac{7}{12} \\ \end{align} </math> </div></div> <ol start=94> <li>Tentukan nilai dari x+y+z dan yz(xy-xz) merupakan bilangan bulat positif jika xy = 2, xz = 6 dan yz = 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} xy \cdot yz \cdot xz &= 2 \cdot 3 \cdot 6 \\ (xyz)^2 &= 36 \\ xyz &= \pm 6 \\ xyz &= 6 \\ \frac{xyz}{xy} &= z = \frac{6}{2} = 3 \\ \frac{xyz}{yz} &= x = \frac{6}{3} = 2 \\ \frac{xyz}{xz} &= y = \frac{6}{6} = 1 \\ x+y+z &= 6 \\ yz(xy-xz) &= 3(2-6) = 3(-4) = -12 \\ \end{align} </math> </div></div> <ol start=95> <li>Tentukan nilai dari (x-z)^y jika <math>\frac{xy}{x+y} = \frac{1}{3}</math>, <math>\frac{yz}{y+z} = \frac{1}{4}</math> dan <math>\frac{xz}{x+z} = \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{xy}{x+y} &= \frac{1}{3} \\ \frac{x+y}{xy} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{y} + \frac{1}{x} &= 3 \\ \frac{yz}{y+z} &= \frac{1}{4} \\ \frac{y+z}{yz} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{z} + \frac{1}{y} &= 4 \\ \frac{xz}{x+z} &= \frac{1}{9} \\ \frac{x+z}{xz} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{z} + \frac{1}{x} &= 9 \\ \text{Misalkan 1/x = a, 1/y = b dan 1/z = c} \\ a+b &= 3 \\ b+c &= 4 \\ a+c &= 9 \\ a+b &= 3 \\ b+c &= 4 \\ a-c &= -1 \\ a-c &= -1 \\ a+c &= 9 \\ 2a &= 8 \\ a &= 4 \\ a-c &= -1 \\ 4-c &= -1 \\ c &= 5 \\ a+b &= 3 \\ 4+b &= 3 \\ b &= -1 \\ \frac{1}{x} &= 4 \\ x &= \frac{1}{4} \\ \frac{1}{y} &= -1 \\ y &= -1 \\ \frac{1}{z} &= 5 \\ z &= \frac{1}{5} \\ (x-z)^y &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=96> <li>Tentukan nilai dari (y+z)^x jika <math>\frac{x+y+z}{2} = \sqrt{x-2}+\sqrt{y-1}+\sqrt{z}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="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+z}{2} &= \sqrt{x-2}+\sqrt{y-1}+\sqrt{z} \\ x+y+z &= 2(\sqrt{x-2}+\sqrt{y-1}+\sqrt{z}) \\ x-2\sqrt{x-2}+y-2\sqrt{y-1}+z-2\sqrt{z} &= 0 \\ x-2-2\sqrt{x-2}+1+y-1-2\sqrt{y-1}+1+z-2\sqrt{z}+1 &= 0 \\ (\sqrt{x-2}-1)^2+(\sqrt{y-1}-1)^2+(\sqrt{z}-1)^2 &= 0 \\ (\sqrt{x-2}-1)^2 &= 0 \\ \sqrt{x-2}-1 &= 0 \\ \sqrt{x-2} &= 1 \\ x-2 &= 1 \\ x &= 3 \\ (\sqrt{y-1}-1)^2 &= 0 \\ \sqrt{y-1}-1 &= 0 \\ \sqrt{y-1} &= 1 \\ y-1 &= 1 \\ y &= 2 \\ (\sqrt{z}-1)^2 &= 0 \\ \sqrt{z}-1 &= 0 \\ \sqrt{z} &= 1 \\ z &= 1 \\ (y+z)^x &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=97> <li>Tentukan nilai dari x, y dan z jika <math>\frac{x+y}{2}=\frac{x+z}{4}=\frac{y+z}{5}</math> dan x+2y+3z=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{x+y}{2}=\frac{x+z}{4}=\frac{y+z}{5} &= k \\ x+y &= 2k \\ x+z &= 4k \\ y+z &= 5k \\ 2x+y+z &= 6k \\ 2x+5k &= 6k \\ k &= 2x \\ x &= \frac{k}{2} \\ y &= \frac{3k}{2} \\ z &= \frac{7k}{2} \\ x+2y+3z &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ x &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ y &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ z &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=98> <li>Tentukan nilai dari 2^x jika <math>2^{(x+y)}=40, 2^{(y+z)}=20, 2^{(x+z)}=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} *cara 1 \\ 2^{(x+y)} \cdot 2^{(y+z)} \cdot 2^{(x+z)} &= 40 \cdot 20 \cdot 10 \\ 2^{2(x+y+z)} &= 8000 \\ 2^{(x+y+z)} &= 40\sqrt{5} \\ 2^x \cdot 2^{(y+z)} &= 40\sqrt{5} \\ 2^x \cdot 20 &= 40\sqrt{5} \\ 2^x &= 2\sqrt{5} \\ *cara 2 \\ \frac{2^{(x+y)}}{2^{(y+z)}} &= \frac{40}{20} \\ 2^{(x-z)} &= 2 \\ 2^{(x+z)} \cdot 2^{(x-z)} &= 10 \cdot 2 \\ 2^{2x} &= 20 \\ 2^{x} &= 2\sqrt{5} \\ *cara 3 \\ \frac{2^{(x+y)} \cdot 2^{(x+z)}}{2^{(y+z)}} &= \frac{40(10)}{20} \\ 2^{2x} &= 20 \\ 2^x &= 2\sqrt{5} \\ *cara 4 \\ 2^{(x+y)}=40 &\text{ menjadi } x+y = ^2log 40 \\ 2^{(y+z)}=20 &\text{ menjadi } y+z = ^2log 20 \\ 2^{(x+z)}=10 &\text{ menjadi } x+z = ^2log 10 \\ x+y+x+z-(y+z) &= ^2log 40+^2log 10-^2log 20 \\ 2x &= ^2log 20 \\ 2^{2x} &= 20 \\ 2^x &= 2\sqrt{5} \\ \end{align} </math> </div></div> <ol start=98> <li>Tentukan nilai dari <math>\frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z}</math> jika <math>\frac{x^2}{by+cz} = \frac{y^2}{cz+ax} = \frac{z^2}{ax+by} = 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{x^2}{by+cz} = \frac{y^2}{cz+ax} = \frac{z^2}{ax+by} &= 1 \\ \frac{x^2}{by+cz} &= 1 \\ x^2 &= by+cz \\ \frac{y^2}{cz+ax} &= 1 \\ y^2 &= cz+ax \\ \frac{z^2}{ax+by} &= 1 \\ z^2 &= ax+by \\ \frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z} &= \frac{ax}{(a+x)x} + \frac{by}{(b+y)y} + \frac{cz}{(c+z)z} \\ &= \frac{ax}{ax+x^2} + \frac{by}{by+y^2} + \frac{cz}{cz+z^2} \\ &= \frac{ax}{ax+by+cz} + \frac{by}{by+cz+ax} + \frac{cz}{cz+ax+by} \\ &= \frac{ax+by+cz}{ax+by+cz} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=99> <li>Tentukan nilai dari <math>\frac{1}{xy+z-1} + \frac{1}{yz+x-1} + \frac{1}{xz+y-1}</math> jika x+y+z=2, <math>x^2+y^2+z^2=6</math> serta xyz = 4?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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 &= 2 \\ x &= 2-y-z \\ y &= 2-x-z \\ z &= 2-x-y \\ \frac{1}{xy+z-1} + \frac{1}{yz+x-1} + \frac{1}{xz+y-1} &= \frac{1}{xy+2-x-y-1} + \frac{1}{yz+2-y-z-1} + \frac{1}{xz+2-x-z-1} \\ &= \frac{1}{xy-x-y+1} + \frac{1}{yz-y-z+1} + \frac{1}{xz-x-z+1} \\ &= \frac{1}{x(y-1)-(y-1)} + \frac{1}{y(z-1)-(z-1)} + \frac{1}{x(z-1)-(z-1)} \\ &= \frac{1}{(x-1)(y-1)} + \frac{1}{(y-1)(z-1)} + \frac{1}{(x-1)(z-1)} \\ &= \frac{z-1+x-1+y-1}{(x-1)(y-1)(z-1)} \\ &= \frac{x+y+z-3}{(x-1)(y-1)(z-1)} \\ &= \frac{x+y+z-3}{xyz-(xy+yz+xz)+(x+y+z)-1} \\ (x+y+z)^2 &= x^2+y^2+z^2+2(xy+yz+xz) \\ 2^2 &= 6+2(xy+yz+xz) \\ 4 &= 6+2(xy+yz+xz) \\ -2 &= 2(xy+yz+xz) \\ xy+yz+xz &= -1 \\ \frac{x+y+z-3}{xyz-(xy+yz+xz)+(x+y+z)-1} &= \frac{2-3}{4-(-1)+2-1} \\ &= \frac{-1}{6} \\ &= -\frac{1}{6} \\ \end{align} </math> </div></div> <ol start=100> <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=101> <li>x, y dan z merupakan bilangan tak nol. Jika <math>x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}</math> maka berapa nilai 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} x+\frac{1}{y} &= y+\frac{1}{z} \\ x-y &= \frac{1}{z}-\frac{1}{y} \\ &= \frac{y-z}{yz} \\ y+\frac{1}{z} &= z+\frac{1}{x} \\ y-z &= \frac{1}{x}-\frac{1}{z} \\ &= \frac{z-x}{xz} \\ z+\frac{1}{x} &= x+\frac{1}{y} \\ z-x &= \frac{1}{y}-\frac{1}{x} \\ &= \frac{x-y}{xy} \\ (x-y)(y-z)(z-x) &= (\frac{y-z}{yz}) \cdot (\frac{z-x}{xz}) \cdot (\frac{x-y}{xy}) \\ x^2y^2z^2 &= 1 \\ (xyz)^2 &= 1 \\ xyz &= \pm 1 \\ \end{align} </math> </div></div> <ol start=102> <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=103> <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=104> <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=105> <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=106> <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=107> <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=108> <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=109> <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=110> <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=111> <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=112> <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=113> <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=114> <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=115> <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=116> <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=117> <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=118> <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=119> <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=120> <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=121> <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=122> <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=123> <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=124> <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=125> <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=126> <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. [[Kategori:Soal-Soal Matematika]] dwqzami1lotvrlgep7lt98gkdn1gl2g 121433 121432 2026-10-23T02:29:06Z Akuindo 8654 121433 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>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-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{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> <ol start=7> <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} = y \\ y^5(1+y+y^2) &= 5^5 \cdot 31 \\ y &= 5 \\ x^{10} &= 5 \\ x &= ^5log \, 10 \\ \end{align} </math> </div></div> <ol start=8> <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=9> <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=10> <li>Berapa nilai x dari <math>\frac{\sqrt{(^4log \, x)^2+ ^2log \, x-4}}{16- ^2log \, x- (^4log \, x)^2} = \frac{1}{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} ^4 log \, x &= \frac{^2log \, x}{2} \\ \frac{\sqrt{(\frac{^2log \, x}{2})^2+ ^2log \, x-4}}{16- ^2log \, x- (\frac{^2log \, x}{2})^2} &= \frac{1}{4} \\ \text{misalkan } y &= ^2log \, x \\ \frac{\sqrt{(\frac{y}{2})^2+y-4}}{16-y-(\frac{y}{2})^2} &= \frac{1}{4} \\ \frac{\sqrt{(\frac{y}{2})^2+y-4}}{12-(-4+y+(\frac{y}{2})^2)} &= \frac{1}{4} \\ \text{misalkan } a &= (\frac{y}{2})^2+y-4 \\ \frac{\sqrt{a}}{12-a} &= \frac{1}{4}, a \ge 0 \\ \text{misalkan } \sqrt{a} &= b, b \ge 0 \\ \frac{b}{12-b^2} &= \frac{1}{4} \\ 4b &= 12-b^2 \\ b^2+4b-12 &= 0 \\ (b-2)(b+6) &= 0 \\ b = 2 &\text{ atau } b = -6 (TM) \\ \sqrt{a} &= 2 \\ a &= 4 \\ (\frac{y}{2})^2+y-4 &= 4 \\ \frac{y^2}{4}+y-8 &= 0 \\ y^2+4y-32 &= 0 \\ (y-4)(y+8) &= 0 \\ y = 4 &\text{ atau } y=-8 \\ ^2 log \, x &= 4 \\ x &= 16 \\ ^2 log \, x = -8 \\ x &= \frac{1}{256} \\ \end{align} </math> </div></div> <ol start=11> <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=12> <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=13> <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=14> <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=15> <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=16> <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=17> <li>Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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> <ol start=18> <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=19> <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=20> <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=21> <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=22> <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=23> <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=24> <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=25> <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=26> <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=27> <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=28> <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=29> <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=30> <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=31> <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=32> <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=33> <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=34> <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=35> <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=36> <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=37> <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=38> <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=39> <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=40> <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=41> <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=42> <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=43> <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=44> <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=45> <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=46> <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=47> <li>Berapakah nilai dari <math>5x+\frac{1}{4x}</math> jika <math>25x^2+\frac{1}{16x^2}=\frac{13}{2}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (5x+\frac{1}{4x})^2 &= 25x^2+\frac{5}{2}+\frac{16}{x^2} \\ &= 25x^2+\frac{16}{x^2}+\frac{5}{2} \\ &= \frac{13}{2}+\frac{5}{2} \\ &= 9 \\ &= \sqrt{9} \\ &= \pm 3 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>x^{28}+x^{25}+x^{23}+x^{20}</math> jika <math>x+\frac{1}{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} x+\frac{1}{x} &= 1 \\ x-1+\frac{1}{x} &= 0 \\ x^2-x+1 &= 0 \\ (x+1)(x^2-x+1) &= (x+1)0 \\ x^3+1 &= 0 \\ x^3 &= -1 \\ x &= -1 \\ x^{28}+x^{25}+x^{23}+x^{20} &= (-1)^{28}+(-1)^{25}+(-1)^{23}+(-1)^{20} \\ &= 1-1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>x^{28}-x^{25}+x^{23}-x^{20}</math> jika <math>x+\frac{1}{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} x+\frac{1}{x} &= -1 \\ x+1+\frac{1}{x} &= 0 \\ x^2-x+1 &= 0 \\ (x-1)(x^2-x+1) &= (x-1)0 \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{28}-x^{25}+x^{23}-x^{20} &= x^{25}(x^{3}-1)+x^{20}(x^{3}-1) &= x^{25}(0)+x^{20}(0) \\ &= 0 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>\frac{3}{2x-3}+\frac{4}{2y-4}+\frac{5}{2z-5}</math> jika <math>\frac{2x}{2x-3}+\frac{2y}{2y-4}+\frac{2z}{2z-5}=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} \frac{2x}{2x-3}+\frac{2y}{2y-4}+\frac{2z}{2z-5} &= 7 \\ \frac{2x-3+3}{2x-3}+\frac{2y-4+4}{2y-4}+\frac{2z-5+5}{2z-5} &= 7 \\ 1+\frac{3}{2x-3}+1+\frac{4}{2y-4}+1+\frac{5}{2z-5} &= 7 \\ \frac{3}{2x-3}+\frac{4}{2y-4}+\frac{5}{2z-5} &= 4 \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>\frac{4x}{4x-7}+\frac{5y}{5y-7}+\frac{6z}{6z-7}</math> jika <math>\frac{7}{4x-7}+\frac{7}{5y-7}+\frac{7}{6z-7}=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{7}{4x-7}+\frac{7}{5y-7}+\frac{7}{6z-7} &= 10 \\ \frac{7+4x-4x}{4x-7}+\frac{7+5y-5y}{5y-7}+\frac{7+6z-6z}{6z-7} &= 10 \\ \frac{4x-(4x-7)}{4x-7}+\frac{5y-(5y-7)}{5y-7}+\frac{6z-(6z-7)}{6z-7} &= 10 \\ \frac{4x}{4x-7}-1+\frac{5y}{5y-7}-1+\frac{6z}{6z-7}-1 &= 10 \\ \frac{4x}{4x-7}+\frac{5y}{5y-7}+\frac{6z}{6z-7} &= 13 \\ \end{align} </math> </div></div> <ol start=52> <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=53> <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=54> <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=55> <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=56> <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=57> <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=58> <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=59> <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-3 sin \, x csc \, x(sin \, x-csc \, x) \\ 8^3 &= sin^3 \, x-csc^3 \, x-3 sin \, 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=60> <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{2 sin \, 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=61> <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=62> <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=63> <li>Berapa nilsi dari sin (x+y) dan cos (x+y) jika sin x + sin y = a serta cos x + cos y = 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} sin \, x+sin \, y &= 2 \, sin \, (\frac{x+y}{2}) \, cos \, (\frac{x-y}{2}) \\ cos \, x+cos \, y &= 2 \, cos \, (\frac{x+y}{2}) \, cos \, (\frac{x-y}{2}) \\ \frac{2 \, sin \, (\frac{x+y}{2}) \, cos \, (\frac{x-y}{2})}{2 \, cos \, (\frac{x+y}{2}) \, cos \, (\frac{x-y}{2})} &= \frac{sin \, (\frac{x+y}{2})}{cos (\frac{x+y}{2})} \\ &= tan \, (\frac{x+y}{2}) \\ \text{misalkan } p &= \frac{x+y}{2} \text{ maka } 2p = x+y \\ tan \, (\frac{x+y}{2}) &= tan \, p \\ &= \frac{a}{b} \\ sin (x+y) = sin 2p &= \frac{2 \, tan \, p}{1+tan^2 \, p} \\ &= \frac{2 \, (\frac{a}{b})}{1+(\frac{a}{b})^2} \\ &= \frac{2 \, (\frac{a}{b})}{\frac{b^2+a^2}{b^2}} \\ &= \frac{2 \, (\frac{a}{b})}{\frac{b^2+a^2}{b^2}} \\ &= \frac{2ab}{a^2+b^2} \end{align} </math> </div></div> <ol start=64> <li>Buktikan bahwa <math>\frac{sin \, x}{1-cos \, x} = cot \, \frac{y}{4}</math> jika <math>cot \, x = \frac{1+cos \, y}{sin \, 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} cot \, x &= \frac{2 cos^2 \, \frac{y}{2}}{2 sin \, \frac{y}{2} \, cos \, \frac{y}{2}} \\ &= \frac{cos \frac{y}{2}}{sin \, \frac{y}{2}} \\ &= cot \, \frac{y}{2} \\ x &=\frac{y}{2} \\ \frac{sin \, x}{1-cos \, x} &= \frac{sin \, \frac{y}{2}}{1-cos \, \frac{y}{2}} \\ &= \frac{2 sin \, \frac{y}{4} \, cos \, \frac{y}{4}}{2 sin^2 \, \frac{y}{4}} \\ &= \frac{cos \, \frac{y}{4}}{sin \, \frac{y}{4}} \\ &= cot \, \frac{y}{4} \end{align} </math> </div></div> <ol start=64> <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=65> <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=66> <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=67> <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=68> <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=69> <li>Berapakah nilai dari <math>x^5-3x^4-x^3+9x^2-11x+3</math> jika <math>x+\frac{1}{x}=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} *cara 1 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2-3x+1 &= 0 \\ (x^3-2x+3)(x^2-3x+1 &= 0(x^3-2x+3) \\ x^5-3x^4-x^3+9x^2-11x+3 &= 0 \\ *cara 2 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2-3x+1 &= 0 \\ x^5-3x^4-x^3+9x^2-11x+3 &= x^5-3x^4+x^3-2x^3+6x^2+3x^2-2x-9x+3 \\ &= x^5-3x^4+x^3-2x^3+6x^2-2x+3x^2-9x+3 \\ &= x^3(x^2-3x+1)-2x(x^2-3x+1)+3(x^2-3x+1) \\ &= x^3(0)-2x(0)+3(0) \\ &= 0 \\ *cara 3 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2 &= 3x-1 \\ x^2-3x+1 &= 0 \\ x^3 &= x \cdot x^2 \\ &= x (3x-1) \\ &= 3x^2-x x^4 &= x \cdot x^3 \\ &= x(3x^2-x) \\ &= 3x^3-x^2 \\ x^5 &= x \cdot x^4 \\ &= x(3x^3-x^2) \\ &= 3x^4-x^3 \\ x^5-3x^4-x^3+9x^2-11x+3 &= 3x^4-x^3-3(3x^3-x^2)-(3x^2-x)+9(3x-1)-11x+3 \\ \end{align} </math> </div></div> <ol start=70> <li>Berapakah nilai dari <math>\sqrt{x-7}</math> jika x=<math>16+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 &= 16+4\sqrt{5} \\ x-7 &= 16-7+4\sqrt{5} \\ &= 9+4\sqrt{5} \\ &= (2+\sqrt{5})^2 \\ \sqrt{x-7} &= 2+\sqrt{5} \\ \end{align} </math> </div></div> <ol start=71> <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=72> <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=73> <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=74> <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=75> <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=76> <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=77> <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=78> <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=79> <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=80> <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=81> <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=82> <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=83> <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=84> <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=85> <li>Jika <math>f(x) \cdot f(y) = f(x+y)</math> dan f(18)= 5 maka tentukan nilai f(72)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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) \cdot f(y) &= f(x+y) \\ f(18) \cdot f(18) &= f(18+18) \\ 5 \cdot 5 &= f(36) \\ f(36) &= 25 \\ f(36) \cdot f(36) &= f(36+36) \\ 25 \cdot 25 &= f(72) \\ f(72) &= 625 \\ \end{align} </math> </div></div> <ol start=86> <li>Jika <math>f(x) \cdot f(y) = f(xy)</math> dan f(4)= 16 maka tentukan nilai f(17)-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) \cdot f(y) &= f(xy) \\ f(4) \cdot f(4) &= f(4 \cdot 4) \\ 16 \cdot 16 &= f(16) \\ 16^2 &= f(16) \\ f(x) &= x^2 \\ f(17) &= 17^2 = 289 \\ f(17)-f(16) &= 289-256 = 33 \\ \end{align} </math> </div></div> <ol start=87> <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=88> <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=89> <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=90> <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=91> <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=92> <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=93> <li>Berapakah nilai xyz dan <math>\frac{xy+xz}{yz}</math> dari x+y=14, x+z=13 dan y+z=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} \text{ketiga persamaan tersebut dijumlahkan menjadi } \\ 2x+2y+2z &= 30 \\ x+y+z &= 15 \\ \text{untuk mencari x, y dan z masing-masing } \\ z &= 1 \\ y &= 2 \\ x &= 12 \\ xyz &= 24 \\ \frac{xy+xz}{yz} &= \frac{1 \cdot 2 + 1 \cdot 12}{2 \cdot 12} \\ &= \frac{14}{24} \\ &= \frac{7}{12} \\ \end{align} </math> </div></div> <ol start=94> <li>Tentukan nilai dari x+y+z dan yz(xy-xz) merupakan bilangan bulat positif jika xy = 2, xz = 6 dan yz = 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} xy \cdot yz \cdot xz &= 2 \cdot 3 \cdot 6 \\ (xyz)^2 &= 36 \\ xyz &= \pm 6 \\ xyz &= 6 \\ \frac{xyz}{xy} &= z = \frac{6}{2} = 3 \\ \frac{xyz}{yz} &= x = \frac{6}{3} = 2 \\ \frac{xyz}{xz} &= y = \frac{6}{6} = 1 \\ x+y+z &= 6 \\ yz(xy-xz) &= 3(2-6) = 3(-4) = -12 \\ \end{align} </math> </div></div> <ol start=95> <li>Tentukan nilai dari (x-z)^y jika <math>\frac{xy}{x+y} = \frac{1}{3}</math>, <math>\frac{yz}{y+z} = \frac{1}{4}</math> dan <math>\frac{xz}{x+z} = \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{xy}{x+y} &= \frac{1}{3} \\ \frac{x+y}{xy} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{y} + \frac{1}{x} &= 3 \\ \frac{yz}{y+z} &= \frac{1}{4} \\ \frac{y+z}{yz} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{z} + \frac{1}{y} &= 4 \\ \frac{xz}{x+z} &= \frac{1}{9} \\ \frac{x+z}{xz} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{z} + \frac{1}{x} &= 9 \\ \text{Misalkan 1/x = a, 1/y = b dan 1/z = c} \\ a+b &= 3 \\ b+c &= 4 \\ a+c &= 9 \\ a+b &= 3 \\ b+c &= 4 \\ a-c &= -1 \\ a-c &= -1 \\ a+c &= 9 \\ 2a &= 8 \\ a &= 4 \\ a-c &= -1 \\ 4-c &= -1 \\ c &= 5 \\ a+b &= 3 \\ 4+b &= 3 \\ b &= -1 \\ \frac{1}{x} &= 4 \\ x &= \frac{1}{4} \\ \frac{1}{y} &= -1 \\ y &= -1 \\ \frac{1}{z} &= 5 \\ z &= \frac{1}{5} \\ (x-z)^y &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=96> <li>Tentukan nilai dari (y+z)^x jika <math>\frac{x+y+z}{2} = \sqrt{x-2}+\sqrt{y-1}+\sqrt{z}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="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+z}{2} &= \sqrt{x-2}+\sqrt{y-1}+\sqrt{z} \\ x+y+z &= 2(\sqrt{x-2}+\sqrt{y-1}+\sqrt{z}) \\ x-2\sqrt{x-2}+y-2\sqrt{y-1}+z-2\sqrt{z} &= 0 \\ x-2-2\sqrt{x-2}+1+y-1-2\sqrt{y-1}+1+z-2\sqrt{z}+1 &= 0 \\ (\sqrt{x-2}-1)^2+(\sqrt{y-1}-1)^2+(\sqrt{z}-1)^2 &= 0 \\ (\sqrt{x-2}-1)^2 &= 0 \\ \sqrt{x-2}-1 &= 0 \\ \sqrt{x-2} &= 1 \\ x-2 &= 1 \\ x &= 3 \\ (\sqrt{y-1}-1)^2 &= 0 \\ \sqrt{y-1}-1 &= 0 \\ \sqrt{y-1} &= 1 \\ y-1 &= 1 \\ y &= 2 \\ (\sqrt{z}-1)^2 &= 0 \\ \sqrt{z}-1 &= 0 \\ \sqrt{z} &= 1 \\ z &= 1 \\ (y+z)^x &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=97> <li>Tentukan nilai dari x, y dan z jika <math>\frac{x+y}{2}=\frac{x+z}{4}=\frac{y+z}{5}</math> dan x+2y+3z=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{x+y}{2}=\frac{x+z}{4}=\frac{y+z}{5} &= k \\ x+y &= 2k \\ x+z &= 4k \\ y+z &= 5k \\ 2x+y+z &= 6k \\ 2x+5k &= 6k \\ k &= 2x \\ x &= \frac{k}{2} \\ y &= \frac{3k}{2} \\ z &= \frac{7k}{2} \\ x+2y+3z &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ x &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ y &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ z &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=98> <li>Tentukan nilai dari 2^x jika <math>2^{(x+y)}=40, 2^{(y+z)}=20, 2^{(x+z)}=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} *cara 1 \\ 2^{(x+y)} \cdot 2^{(y+z)} \cdot 2^{(x+z)} &= 40 \cdot 20 \cdot 10 \\ 2^{2(x+y+z)} &= 8000 \\ 2^{(x+y+z)} &= 40\sqrt{5} \\ 2^x \cdot 2^{(y+z)} &= 40\sqrt{5} \\ 2^x \cdot 20 &= 40\sqrt{5} \\ 2^x &= 2\sqrt{5} \\ *cara 2 \\ \frac{2^{(x+y)}}{2^{(y+z)}} &= \frac{40}{20} \\ 2^{(x-z)} &= 2 \\ 2^{(x+z)} \cdot 2^{(x-z)} &= 10 \cdot 2 \\ 2^{2x} &= 20 \\ 2^{x} &= 2\sqrt{5} \\ *cara 3 \\ \frac{2^{(x+y)} \cdot 2^{(x+z)}}{2^{(y+z)}} &= \frac{40(10)}{20} \\ 2^{2x} &= 20 \\ 2^x &= 2\sqrt{5} \\ *cara 4 \\ 2^{(x+y)}=40 &\text{ menjadi } x+y = ^2log 40 \\ 2^{(y+z)}=20 &\text{ menjadi } y+z = ^2log 20 \\ 2^{(x+z)}=10 &\text{ menjadi } x+z = ^2log 10 \\ x+y+x+z-(y+z) &= ^2log 40+^2log 10-^2log 20 \\ 2x &= ^2log 20 \\ 2^{2x} &= 20 \\ 2^x &= 2\sqrt{5} \\ \end{align} </math> </div></div> <ol start=98> <li>Tentukan nilai dari <math>\frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z}</math> jika <math>\frac{x^2}{by+cz} = \frac{y^2}{cz+ax} = \frac{z^2}{ax+by} = 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{x^2}{by+cz} = \frac{y^2}{cz+ax} = \frac{z^2}{ax+by} &= 1 \\ \frac{x^2}{by+cz} &= 1 \\ x^2 &= by+cz \\ \frac{y^2}{cz+ax} &= 1 \\ y^2 &= cz+ax \\ \frac{z^2}{ax+by} &= 1 \\ z^2 &= ax+by \\ \frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z} &= \frac{ax}{(a+x)x} + \frac{by}{(b+y)y} + \frac{cz}{(c+z)z} \\ &= \frac{ax}{ax+x^2} + \frac{by}{by+y^2} + \frac{cz}{cz+z^2} \\ &= \frac{ax}{ax+by+cz} + \frac{by}{by+cz+ax} + \frac{cz}{cz+ax+by} \\ &= \frac{ax+by+cz}{ax+by+cz} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=99> <li>Tentukan nilai dari <math>\frac{1}{xy+z-1} + \frac{1}{yz+x-1} + \frac{1}{xz+y-1}</math> jika x+y+z=2, <math>x^2+y^2+z^2=6</math> serta xyz = 4?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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 &= 2 \\ x &= 2-y-z \\ y &= 2-x-z \\ z &= 2-x-y \\ \frac{1}{xy+z-1} + \frac{1}{yz+x-1} + \frac{1}{xz+y-1} &= \frac{1}{xy+2-x-y-1} + \frac{1}{yz+2-y-z-1} + \frac{1}{xz+2-x-z-1} \\ &= \frac{1}{xy-x-y+1} + \frac{1}{yz-y-z+1} + \frac{1}{xz-x-z+1} \\ &= \frac{1}{x(y-1)-(y-1)} + \frac{1}{y(z-1)-(z-1)} + \frac{1}{x(z-1)-(z-1)} \\ &= \frac{1}{(x-1)(y-1)} + \frac{1}{(y-1)(z-1)} + \frac{1}{(x-1)(z-1)} \\ &= \frac{z-1+x-1+y-1}{(x-1)(y-1)(z-1)} \\ &= \frac{x+y+z-3}{(x-1)(y-1)(z-1)} \\ &= \frac{x+y+z-3}{xyz-(xy+yz+xz)+(x+y+z)-1} \\ (x+y+z)^2 &= x^2+y^2+z^2+2(xy+yz+xz) \\ 2^2 &= 6+2(xy+yz+xz) \\ 4 &= 6+2(xy+yz+xz) \\ -2 &= 2(xy+yz+xz) \\ xy+yz+xz &= -1 \\ \frac{x+y+z-3}{xyz-(xy+yz+xz)+(x+y+z)-1} &= \frac{2-3}{4-(-1)+2-1} \\ &= \frac{-1}{6} \\ &= -\frac{1}{6} \\ \end{align} </math> </div></div> <ol start=100> <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=101> <li>x, y dan z merupakan bilangan tak nol. Jika <math>x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}</math> maka berapa nilai 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} x+\frac{1}{y} &= y+\frac{1}{z} \\ x-y &= \frac{1}{z}-\frac{1}{y} \\ &= \frac{y-z}{yz} \\ y+\frac{1}{z} &= z+\frac{1}{x} \\ y-z &= \frac{1}{x}-\frac{1}{z} \\ &= \frac{z-x}{xz} \\ z+\frac{1}{x} &= x+\frac{1}{y} \\ z-x &= \frac{1}{y}-\frac{1}{x} \\ &= \frac{x-y}{xy} \\ (x-y)(y-z)(z-x) &= (\frac{y-z}{yz}) \cdot (\frac{z-x}{xz}) \cdot (\frac{x-y}{xy}) \\ x^2y^2z^2 &= 1 \\ (xyz)^2 &= 1 \\ xyz &= \pm 1 \\ \end{align} </math> </div></div> <ol start=102> <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=103> <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=104> <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=105> <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=106> <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=107> <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=108> <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=109> <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=110> <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=111> <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=112> <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=113> <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=114> <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=115> <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=116> <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=117> <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=118> <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=119> <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=120> <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=121> <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=122> <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=123> <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=124> <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=125> <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=126> <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. [[Kategori:Soal-Soal Matematika]] gn8en8szphvgz1jzz07efcskeyqscss 121434 121433 2026-10-23T02:34:24Z Akuindo 8654 121434 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>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-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{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> <ol start=7> <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} = y \\ y^5(1+y+y^2) &= 5^5 \cdot 31 \\ y &= 5 \\ x^{10} &= 5 \\ x &= ^5log \, 10 \\ \end{align} </math> </div></div> <ol start=8> <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=9> <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=10> <li>Berapa nilai x dari <math>\frac{\sqrt{(^4log \, x)^2+ ^2log \, x-4}}{16- ^2log \, x- (^4log \, x)^2} = \frac{1}{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} ^4 log \, x &= \frac{^2log \, x}{2} \\ \frac{\sqrt{(\frac{^2log \, x}{2})^2+ ^2log \, x-4}}{16- ^2log \, x- (\frac{^2log \, x}{2})^2} &= \frac{1}{4} \\ \text{misalkan } y &= ^2log \, x \\ \frac{\sqrt{(\frac{y}{2})^2+y-4}}{16-y-(\frac{y}{2})^2} &= \frac{1}{4} \\ \frac{\sqrt{(\frac{y}{2})^2+y-4}}{12-(-4+y+(\frac{y}{2})^2)} &= \frac{1}{4} \\ \text{misalkan } a &= (\frac{y}{2})^2+y-4 \\ \frac{\sqrt{a}}{12-a} &= \frac{1}{4}, a \ge 0 \\ \text{misalkan } \sqrt{a} &= b, b \ge 0 \\ \frac{b}{12-b^2} &= \frac{1}{4} \\ 4b &= 12-b^2 \\ b^2+4b-12 &= 0 \\ (b-2)(b+6) &= 0 \\ b = 2 &\text{ atau } b = -6 (TM) \\ \sqrt{a} &= 2 \\ a &= 4 \\ (\frac{y}{2})^2+y-4 &= 4 \\ \frac{y^2}{4}+y-8 &= 0 \\ y^2+4y-32 &= 0 \\ (y-4)(y+8) &= 0 \\ y = 4 &\text{ atau } y=-8 \\ ^2 log \, x &= 4 \\ x &= 16 \\ ^2 log \, x = -8 \\ x &= \frac{1}{256} \\ \end{align} </math> </div></div> <ol start=11> <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=12> <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=13> <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=14> <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=15> <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=16> <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=17> <li>Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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> <ol start=18> <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=19> <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=20> <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=21> <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=22> <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=23> <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=24> <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=25> <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=26> <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=27> <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=28> <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=29> <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=30> <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=31> <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=32> <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=33> <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=34> <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=35> <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=36> <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=37> <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=38> <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=39> <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=40> <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=41> <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=42> <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=43> <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=44> <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=45> <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=46> <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=47> <li>Berapakah nilai dari <math>5x+\frac{1}{4x}</math> jika <math>25x^2+\frac{1}{16x^2}=\frac{13}{2}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (5x+\frac{1}{4x})^2 &= 25x^2+\frac{5}{2}+\frac{16}{x^2} \\ &= 25x^2+\frac{16}{x^2}+\frac{5}{2} \\ &= \frac{13}{2}+\frac{5}{2} \\ &= 9 \\ &= \sqrt{9} \\ &= \pm 3 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>x^{28}+x^{25}+x^{23}+x^{20}</math> jika <math>x+\frac{1}{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} x+\frac{1}{x} &= 1 \\ x-1+\frac{1}{x} &= 0 \\ x^2-x+1 &= 0 \\ (x+1)(x^2-x+1) &= (x+1)0 \\ x^3+1 &= 0 \\ x^3 &= -1 \\ x &= -1 \\ x^{28}+x^{25}+x^{23}+x^{20} &= (-1)^{28}+(-1)^{25}+(-1)^{23}+(-1)^{20} \\ &= 1-1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>x^{28}-x^{25}+x^{23}-x^{20}</math> jika <math>x+\frac{1}{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} x+\frac{1}{x} &= -1 \\ x+1+\frac{1}{x} &= 0 \\ x^2-x+1 &= 0 \\ (x-1)(x^2-x+1) &= (x-1)0 \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{28}-x^{25}+x^{23}-x^{20} &= x^{25}(x^{3}-1)+x^{20}(x^{3}-1) &= x^{25}(0)+x^{20}(0) \\ &= 0 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>\frac{3}{2x-3}+\frac{4}{2y-4}+\frac{5}{2z-5}</math> jika <math>\frac{2x}{2x-3}+\frac{2y}{2y-4}+\frac{2z}{2z-5}=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} \frac{2x}{2x-3}+\frac{2y}{2y-4}+\frac{2z}{2z-5} &= 7 \\ \frac{2x-3+3}{2x-3}+\frac{2y-4+4}{2y-4}+\frac{2z-5+5}{2z-5} &= 7 \\ 1+\frac{3}{2x-3}+1+\frac{4}{2y-4}+1+\frac{5}{2z-5} &= 7 \\ \frac{3}{2x-3}+\frac{4}{2y-4}+\frac{5}{2z-5} &= 4 \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>\frac{4x}{4x-7}+\frac{5y}{5y-7}+\frac{6z}{6z-7}</math> jika <math>\frac{7}{4x-7}+\frac{7}{5y-7}+\frac{7}{6z-7}=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{7}{4x-7}+\frac{7}{5y-7}+\frac{7}{6z-7} &= 10 \\ \frac{7+4x-4x}{4x-7}+\frac{7+5y-5y}{5y-7}+\frac{7+6z-6z}{6z-7} &= 10 \\ \frac{4x-(4x-7)}{4x-7}+\frac{5y-(5y-7)}{5y-7}+\frac{6z-(6z-7)}{6z-7} &= 10 \\ \frac{4x}{4x-7}-1+\frac{5y}{5y-7}-1+\frac{6z}{6z-7}-1 &= 10 \\ \frac{4x}{4x-7}+\frac{5y}{5y-7}+\frac{6z}{6z-7} &= 13 \\ \end{align} </math> </div></div> <ol start=52> <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=53> <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=54> <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=55> <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=56> <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=57> <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=58> <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=59> <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-3 sin \, x csc \, x(sin \, x-csc \, x) \\ 8^3 &= sin^3 \, x-csc^3 \, x-3 sin \, 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=60> <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{2 sin \, 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=61> <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=62> <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=63> <li>Berapa nilsi dari sin (x+y) dan cos (x+y) jika sin x + sin y = a serta cos x + cos y = 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} sin \, x+sin \, y &= 2 \, sin \, (\frac{x+y}{2}) \, cos \, (\frac{x-y}{2}) \\ cos \, x+cos \, y &= 2 \, cos \, (\frac{x+y}{2}) \, cos \, (\frac{x-y}{2}) \\ \frac{2 \, sin \, (\frac{x+y}{2}) \, cos \, (\frac{x-y}{2})}{2 \, cos \, (\frac{x+y}{2}) \, cos \, (\frac{x-y}{2})} &= \frac{sin \, (\frac{x+y}{2})}{cos (\frac{x+y}{2})} \\ &= tan \, (\frac{x+y}{2}) \\ \text{misalkan } p &= \frac{x+y}{2} \text{ maka } 2p = x+y \\ tan \, (\frac{x+y}{2}) &= tan \, p \\ &= \frac{a}{b} \\ sin (x+y) = sin 2p &= \frac{2 \, tan \, p}{1+tan^2 \, p} \\ &= \frac{2 \, (\frac{a}{b})}{1+(\frac{a}{b})^2} \\ &= \frac{2 \, (\frac{a}{b})}{\frac{b^2+a^2}{b^2}} \\ &= \frac{2 \, (\frac{a}{b})}{\frac{b^2+a^2}{b^2}} \\ &= \frac{2ab}{a^2+b^2} \\ cos (x+y) = cos 2p &= \frac{1-tan^2 \, p}{1+tan^2 \, p} \\ &= \frac{1-(\frac{a}{b})^2}{1+(\frac{a}{b})^2} \\ &= \frac{\frac{b^2-a^2}{b^2}}{\frac{b^2+a^2}{b^2}} \\ &= \frac{b^2-a^2}{a^2+b^2} \\ \end{align} </math> </div></div> <ol start=64> <li>Buktikan bahwa <math>\frac{sin \, x}{1-cos \, x} = cot \, \frac{y}{4}</math> jika <math>cot \, x = \frac{1+cos \, y}{sin \, 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} cot \, x &= \frac{2 cos^2 \, \frac{y}{2}}{2 sin \, \frac{y}{2} \, cos \, \frac{y}{2}} \\ &= \frac{cos \frac{y}{2}}{sin \, \frac{y}{2}} \\ &= cot \, \frac{y}{2} \\ x &=\frac{y}{2} \\ \frac{sin \, x}{1-cos \, x} &= \frac{sin \, \frac{y}{2}}{1-cos \, \frac{y}{2}} \\ &= \frac{2 sin \, \frac{y}{4} \, cos \, \frac{y}{4}}{2 sin^2 \, \frac{y}{4}} \\ &= \frac{cos \, \frac{y}{4}}{sin \, \frac{y}{4}} \\ &= cot \, \frac{y}{4} \end{align} </math> </div></div> <ol start=64> <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=65> <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=66> <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=67> <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=68> <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=69> <li>Berapakah nilai dari <math>x^5-3x^4-x^3+9x^2-11x+3</math> jika <math>x+\frac{1}{x}=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} *cara 1 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2-3x+1 &= 0 \\ (x^3-2x+3)(x^2-3x+1 &= 0(x^3-2x+3) \\ x^5-3x^4-x^3+9x^2-11x+3 &= 0 \\ *cara 2 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2-3x+1 &= 0 \\ x^5-3x^4-x^3+9x^2-11x+3 &= x^5-3x^4+x^3-2x^3+6x^2+3x^2-2x-9x+3 \\ &= x^5-3x^4+x^3-2x^3+6x^2-2x+3x^2-9x+3 \\ &= x^3(x^2-3x+1)-2x(x^2-3x+1)+3(x^2-3x+1) \\ &= x^3(0)-2x(0)+3(0) \\ &= 0 \\ *cara 3 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2 &= 3x-1 \\ x^2-3x+1 &= 0 \\ x^3 &= x \cdot x^2 \\ &= x (3x-1) \\ &= 3x^2-x x^4 &= x \cdot x^3 \\ &= x(3x^2-x) \\ &= 3x^3-x^2 \\ x^5 &= x \cdot x^4 \\ &= x(3x^3-x^2) \\ &= 3x^4-x^3 \\ x^5-3x^4-x^3+9x^2-11x+3 &= 3x^4-x^3-3(3x^3-x^2)-(3x^2-x)+9(3x-1)-11x+3 \\ \end{align} </math> </div></div> <ol start=70> <li>Berapakah nilai dari <math>\sqrt{x-7}</math> jika x=<math>16+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 &= 16+4\sqrt{5} \\ x-7 &= 16-7+4\sqrt{5} \\ &= 9+4\sqrt{5} \\ &= (2+\sqrt{5})^2 \\ \sqrt{x-7} &= 2+\sqrt{5} \\ \end{align} </math> </div></div> <ol start=71> <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=72> <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=73> <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=74> <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=75> <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=76> <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=77> <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=78> <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=79> <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=80> <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=81> <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=82> <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=83> <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=84> <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=85> <li>Jika <math>f(x) \cdot f(y) = f(x+y)</math> dan f(18)= 5 maka tentukan nilai f(72)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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) \cdot f(y) &= f(x+y) \\ f(18) \cdot f(18) &= f(18+18) \\ 5 \cdot 5 &= f(36) \\ f(36) &= 25 \\ f(36) \cdot f(36) &= f(36+36) \\ 25 \cdot 25 &= f(72) \\ f(72) &= 625 \\ \end{align} </math> </div></div> <ol start=86> <li>Jika <math>f(x) \cdot f(y) = f(xy)</math> dan f(4)= 16 maka tentukan nilai f(17)-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) \cdot f(y) &= f(xy) \\ f(4) \cdot f(4) &= f(4 \cdot 4) \\ 16 \cdot 16 &= f(16) \\ 16^2 &= f(16) \\ f(x) &= x^2 \\ f(17) &= 17^2 = 289 \\ f(17)-f(16) &= 289-256 = 33 \\ \end{align} </math> </div></div> <ol start=87> <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=88> <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=89> <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=90> <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=91> <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=92> <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=93> <li>Berapakah nilai xyz dan <math>\frac{xy+xz}{yz}</math> dari x+y=14, x+z=13 dan y+z=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} \text{ketiga persamaan tersebut dijumlahkan menjadi } \\ 2x+2y+2z &= 30 \\ x+y+z &= 15 \\ \text{untuk mencari x, y dan z masing-masing } \\ z &= 1 \\ y &= 2 \\ x &= 12 \\ xyz &= 24 \\ \frac{xy+xz}{yz} &= \frac{1 \cdot 2 + 1 \cdot 12}{2 \cdot 12} \\ &= \frac{14}{24} \\ &= \frac{7}{12} \\ \end{align} </math> </div></div> <ol start=94> <li>Tentukan nilai dari x+y+z dan yz(xy-xz) merupakan bilangan bulat positif jika xy = 2, xz = 6 dan yz = 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} xy \cdot yz \cdot xz &= 2 \cdot 3 \cdot 6 \\ (xyz)^2 &= 36 \\ xyz &= \pm 6 \\ xyz &= 6 \\ \frac{xyz}{xy} &= z = \frac{6}{2} = 3 \\ \frac{xyz}{yz} &= x = \frac{6}{3} = 2 \\ \frac{xyz}{xz} &= y = \frac{6}{6} = 1 \\ x+y+z &= 6 \\ yz(xy-xz) &= 3(2-6) = 3(-4) = -12 \\ \end{align} </math> </div></div> <ol start=95> <li>Tentukan nilai dari (x-z)^y jika <math>\frac{xy}{x+y} = \frac{1}{3}</math>, <math>\frac{yz}{y+z} = \frac{1}{4}</math> dan <math>\frac{xz}{x+z} = \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{xy}{x+y} &= \frac{1}{3} \\ \frac{x+y}{xy} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{y} + \frac{1}{x} &= 3 \\ \frac{yz}{y+z} &= \frac{1}{4} \\ \frac{y+z}{yz} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{z} + \frac{1}{y} &= 4 \\ \frac{xz}{x+z} &= \frac{1}{9} \\ \frac{x+z}{xz} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{z} + \frac{1}{x} &= 9 \\ \text{Misalkan 1/x = a, 1/y = b dan 1/z = c} \\ a+b &= 3 \\ b+c &= 4 \\ a+c &= 9 \\ a+b &= 3 \\ b+c &= 4 \\ a-c &= -1 \\ a-c &= -1 \\ a+c &= 9 \\ 2a &= 8 \\ a &= 4 \\ a-c &= -1 \\ 4-c &= -1 \\ c &= 5 \\ a+b &= 3 \\ 4+b &= 3 \\ b &= -1 \\ \frac{1}{x} &= 4 \\ x &= \frac{1}{4} \\ \frac{1}{y} &= -1 \\ y &= -1 \\ \frac{1}{z} &= 5 \\ z &= \frac{1}{5} \\ (x-z)^y &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=96> <li>Tentukan nilai dari (y+z)^x jika <math>\frac{x+y+z}{2} = \sqrt{x-2}+\sqrt{y-1}+\sqrt{z}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="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+z}{2} &= \sqrt{x-2}+\sqrt{y-1}+\sqrt{z} \\ x+y+z &= 2(\sqrt{x-2}+\sqrt{y-1}+\sqrt{z}) \\ x-2\sqrt{x-2}+y-2\sqrt{y-1}+z-2\sqrt{z} &= 0 \\ x-2-2\sqrt{x-2}+1+y-1-2\sqrt{y-1}+1+z-2\sqrt{z}+1 &= 0 \\ (\sqrt{x-2}-1)^2+(\sqrt{y-1}-1)^2+(\sqrt{z}-1)^2 &= 0 \\ (\sqrt{x-2}-1)^2 &= 0 \\ \sqrt{x-2}-1 &= 0 \\ \sqrt{x-2} &= 1 \\ x-2 &= 1 \\ x &= 3 \\ (\sqrt{y-1}-1)^2 &= 0 \\ \sqrt{y-1}-1 &= 0 \\ \sqrt{y-1} &= 1 \\ y-1 &= 1 \\ y &= 2 \\ (\sqrt{z}-1)^2 &= 0 \\ \sqrt{z}-1 &= 0 \\ \sqrt{z} &= 1 \\ z &= 1 \\ (y+z)^x &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=97> <li>Tentukan nilai dari x, y dan z jika <math>\frac{x+y}{2}=\frac{x+z}{4}=\frac{y+z}{5}</math> dan x+2y+3z=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{x+y}{2}=\frac{x+z}{4}=\frac{y+z}{5} &= k \\ x+y &= 2k \\ x+z &= 4k \\ y+z &= 5k \\ 2x+y+z &= 6k \\ 2x+5k &= 6k \\ k &= 2x \\ x &= \frac{k}{2} \\ y &= \frac{3k}{2} \\ z &= \frac{7k}{2} \\ x+2y+3z &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ x &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ y &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ z &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=98> <li>Tentukan nilai dari 2^x jika <math>2^{(x+y)}=40, 2^{(y+z)}=20, 2^{(x+z)}=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} *cara 1 \\ 2^{(x+y)} \cdot 2^{(y+z)} \cdot 2^{(x+z)} &= 40 \cdot 20 \cdot 10 \\ 2^{2(x+y+z)} &= 8000 \\ 2^{(x+y+z)} &= 40\sqrt{5} \\ 2^x \cdot 2^{(y+z)} &= 40\sqrt{5} \\ 2^x \cdot 20 &= 40\sqrt{5} \\ 2^x &= 2\sqrt{5} \\ *cara 2 \\ \frac{2^{(x+y)}}{2^{(y+z)}} &= \frac{40}{20} \\ 2^{(x-z)} &= 2 \\ 2^{(x+z)} \cdot 2^{(x-z)} &= 10 \cdot 2 \\ 2^{2x} &= 20 \\ 2^{x} &= 2\sqrt{5} \\ *cara 3 \\ \frac{2^{(x+y)} \cdot 2^{(x+z)}}{2^{(y+z)}} &= \frac{40(10)}{20} \\ 2^{2x} &= 20 \\ 2^x &= 2\sqrt{5} \\ *cara 4 \\ 2^{(x+y)}=40 &\text{ menjadi } x+y = ^2log 40 \\ 2^{(y+z)}=20 &\text{ menjadi } y+z = ^2log 20 \\ 2^{(x+z)}=10 &\text{ menjadi } x+z = ^2log 10 \\ x+y+x+z-(y+z) &= ^2log 40+^2log 10-^2log 20 \\ 2x &= ^2log 20 \\ 2^{2x} &= 20 \\ 2^x &= 2\sqrt{5} \\ \end{align} </math> </div></div> <ol start=98> <li>Tentukan nilai dari <math>\frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z}</math> jika <math>\frac{x^2}{by+cz} = \frac{y^2}{cz+ax} = \frac{z^2}{ax+by} = 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{x^2}{by+cz} = \frac{y^2}{cz+ax} = \frac{z^2}{ax+by} &= 1 \\ \frac{x^2}{by+cz} &= 1 \\ x^2 &= by+cz \\ \frac{y^2}{cz+ax} &= 1 \\ y^2 &= cz+ax \\ \frac{z^2}{ax+by} &= 1 \\ z^2 &= ax+by \\ \frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z} &= \frac{ax}{(a+x)x} + \frac{by}{(b+y)y} + \frac{cz}{(c+z)z} \\ &= \frac{ax}{ax+x^2} + \frac{by}{by+y^2} + \frac{cz}{cz+z^2} \\ &= \frac{ax}{ax+by+cz} + \frac{by}{by+cz+ax} + \frac{cz}{cz+ax+by} \\ &= \frac{ax+by+cz}{ax+by+cz} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=99> <li>Tentukan nilai dari <math>\frac{1}{xy+z-1} + \frac{1}{yz+x-1} + \frac{1}{xz+y-1}</math> jika x+y+z=2, <math>x^2+y^2+z^2=6</math> serta xyz = 4?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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 &= 2 \\ x &= 2-y-z \\ y &= 2-x-z \\ z &= 2-x-y \\ \frac{1}{xy+z-1} + \frac{1}{yz+x-1} + \frac{1}{xz+y-1} &= \frac{1}{xy+2-x-y-1} + \frac{1}{yz+2-y-z-1} + \frac{1}{xz+2-x-z-1} \\ &= \frac{1}{xy-x-y+1} + \frac{1}{yz-y-z+1} + \frac{1}{xz-x-z+1} \\ &= \frac{1}{x(y-1)-(y-1)} + \frac{1}{y(z-1)-(z-1)} + \frac{1}{x(z-1)-(z-1)} \\ &= \frac{1}{(x-1)(y-1)} + \frac{1}{(y-1)(z-1)} + \frac{1}{(x-1)(z-1)} \\ &= \frac{z-1+x-1+y-1}{(x-1)(y-1)(z-1)} \\ &= \frac{x+y+z-3}{(x-1)(y-1)(z-1)} \\ &= \frac{x+y+z-3}{xyz-(xy+yz+xz)+(x+y+z)-1} \\ (x+y+z)^2 &= x^2+y^2+z^2+2(xy+yz+xz) \\ 2^2 &= 6+2(xy+yz+xz) \\ 4 &= 6+2(xy+yz+xz) \\ -2 &= 2(xy+yz+xz) \\ xy+yz+xz &= -1 \\ \frac{x+y+z-3}{xyz-(xy+yz+xz)+(x+y+z)-1} &= \frac{2-3}{4-(-1)+2-1} \\ &= \frac{-1}{6} \\ &= -\frac{1}{6} \\ \end{align} </math> </div></div> <ol start=100> <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=101> <li>x, y dan z merupakan bilangan tak nol. Jika <math>x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}</math> maka berapa nilai 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} x+\frac{1}{y} &= y+\frac{1}{z} \\ x-y &= \frac{1}{z}-\frac{1}{y} \\ &= \frac{y-z}{yz} \\ y+\frac{1}{z} &= z+\frac{1}{x} \\ y-z &= \frac{1}{x}-\frac{1}{z} \\ &= \frac{z-x}{xz} \\ z+\frac{1}{x} &= x+\frac{1}{y} \\ z-x &= \frac{1}{y}-\frac{1}{x} \\ &= \frac{x-y}{xy} \\ (x-y)(y-z)(z-x) &= (\frac{y-z}{yz}) \cdot (\frac{z-x}{xz}) \cdot (\frac{x-y}{xy}) \\ x^2y^2z^2 &= 1 \\ (xyz)^2 &= 1 \\ xyz &= \pm 1 \\ \end{align} </math> </div></div> <ol start=102> <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=103> <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=104> <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=105> <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=106> <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=107> <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=108> <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=109> <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=110> <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=111> <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=112> <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=113> <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=114> <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=115> <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=116> <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=117> <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=118> <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=119> <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=120> <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=121> <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=122> <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=123> <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=124> <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=125> <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=126> <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. [[Kategori:Soal-Soal Matematika]] ack3d957pfftgwecjpw3tipfcpfm2vt 121435 121434 2026-10-23T02:41:24Z Akuindo 8654 121435 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>\frac{2x-4}{3}+2x-6 = \frac{3}{2x-4}+\frac{1}{2x-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{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> <ol start=7> <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} = y \\ y^5(1+y+y^2) &= 5^5 \cdot 31 \\ y &= 5 \\ x^{10} &= 5 \\ x &= ^5log \, 10 \\ \end{align} </math> </div></div> <ol start=8> <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=9> <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=10> <li>Berapa nilai x dari <math>\frac{\sqrt{(^4log \, x)^2+ ^2log \, x-4}}{16- ^2log \, x- (^4log \, x)^2} = \frac{1}{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} ^4 log \, x &= \frac{^2log \, x}{2} \\ \frac{\sqrt{(\frac{^2log \, x}{2})^2+ ^2log \, x-4}}{16- ^2log \, x- (\frac{^2log \, x}{2})^2} &= \frac{1}{4} \\ \text{misalkan } y &= ^2log \, x \\ \frac{\sqrt{(\frac{y}{2})^2+y-4}}{16-y-(\frac{y}{2})^2} &= \frac{1}{4} \\ \frac{\sqrt{(\frac{y}{2})^2+y-4}}{12-(-4+y+(\frac{y}{2})^2)} &= \frac{1}{4} \\ \text{misalkan } a &= (\frac{y}{2})^2+y-4 \\ \frac{\sqrt{a}}{12-a} &= \frac{1}{4}, a \ge 0 \\ \text{misalkan } \sqrt{a} &= b, b \ge 0 \\ \frac{b}{12-b^2} &= \frac{1}{4} \\ 4b &= 12-b^2 \\ b^2+4b-12 &= 0 \\ (b-2)(b+6) &= 0 \\ b = 2 &\text{ atau } b = -6 (TM) \\ \sqrt{a} &= 2 \\ a &= 4 \\ (\frac{y}{2})^2+y-4 &= 4 \\ \frac{y^2}{4}+y-8 &= 0 \\ y^2+4y-32 &= 0 \\ (y-4)(y+8) &= 0 \\ y = 4 &\text{ atau } y=-8 \\ ^2 log \, x &= 4 \\ x &= 16 \\ ^2 log \, x = -8 \\ x &= \frac{1}{256} \\ \end{align} </math> </div></div> <ol start=11> <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=12> <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=13> <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=14> <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=15> <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=16> <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=17> <li>Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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> <ol start=18> <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=19> <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=20> <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=21> <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=22> <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=23> <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=24> <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=25> <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=26> <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=27> <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=28> <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=29> <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=30> <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=31> <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=32> <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=33> <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=34> <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=35> <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=36> <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=37> <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=38> <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=39> <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=40> <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=41> <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=42> <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=43> <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=44> <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=45> <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=46> <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=47> <li>Berapakah nilai dari <math>5x+\frac{1}{4x}</math> jika <math>25x^2+\frac{1}{16x^2}=\frac{13}{2}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (5x+\frac{1}{4x})^2 &= 25x^2+\frac{5}{2}+\frac{16}{x^2} \\ &= 25x^2+\frac{16}{x^2}+\frac{5}{2} \\ &= \frac{13}{2}+\frac{5}{2} \\ &= 9 \\ &= \sqrt{9} \\ &= \pm 3 \\ \end{align} </math> </div></div> <ol start=48> <li>Berapakah nilai dari <math>x^{28}+x^{25}+x^{23}+x^{20}</math> jika <math>x+\frac{1}{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} x+\frac{1}{x} &= 1 \\ x-1+\frac{1}{x} &= 0 \\ x^2-x+1 &= 0 \\ (x+1)(x^2-x+1) &= (x+1)0 \\ x^3+1 &= 0 \\ x^3 &= -1 \\ x &= -1 \\ x^{28}+x^{25}+x^{23}+x^{20} &= (-1)^{28}+(-1)^{25}+(-1)^{23}+(-1)^{20} \\ &= 1-1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> <ol start=49> <li>Berapakah nilai dari <math>x^{28}-x^{25}+x^{23}-x^{20}</math> jika <math>x+\frac{1}{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} x+\frac{1}{x} &= -1 \\ x+1+\frac{1}{x} &= 0 \\ x^2-x+1 &= 0 \\ (x-1)(x^2-x+1) &= (x-1)0 \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{28}-x^{25}+x^{23}-x^{20} &= x^{25}(x^{3}-1)+x^{20}(x^{3}-1) &= x^{25}(0)+x^{20}(0) \\ &= 0 \\ \end{align} </math> </div></div> <ol start=50> <li>Berapakah nilai dari <math>\frac{3}{2x-3}+\frac{4}{2y-4}+\frac{5}{2z-5}</math> jika <math>\frac{2x}{2x-3}+\frac{2y}{2y-4}+\frac{2z}{2z-5}=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} \frac{2x}{2x-3}+\frac{2y}{2y-4}+\frac{2z}{2z-5} &= 7 \\ \frac{2x-3+3}{2x-3}+\frac{2y-4+4}{2y-4}+\frac{2z-5+5}{2z-5} &= 7 \\ 1+\frac{3}{2x-3}+1+\frac{4}{2y-4}+1+\frac{5}{2z-5} &= 7 \\ \frac{3}{2x-3}+\frac{4}{2y-4}+\frac{5}{2z-5} &= 4 \\ \end{align} </math> </div></div> <ol start=51> <li>Berapakah nilai dari <math>\frac{4x}{4x-7}+\frac{5y}{5y-7}+\frac{6z}{6z-7}</math> jika <math>\frac{7}{4x-7}+\frac{7}{5y-7}+\frac{7}{6z-7}=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{7}{4x-7}+\frac{7}{5y-7}+\frac{7}{6z-7} &= 10 \\ \frac{7+4x-4x}{4x-7}+\frac{7+5y-5y}{5y-7}+\frac{7+6z-6z}{6z-7} &= 10 \\ \frac{4x-(4x-7)}{4x-7}+\frac{5y-(5y-7)}{5y-7}+\frac{6z-(6z-7)}{6z-7} &= 10 \\ \frac{4x}{4x-7}-1+\frac{5y}{5y-7}-1+\frac{6z}{6z-7}-1 &= 10 \\ \frac{4x}{4x-7}+\frac{5y}{5y-7}+\frac{6z}{6z-7} &= 13 \\ \end{align} </math> </div></div> <ol start=52> <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=53> <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=54> <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=55> <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=56> <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=57> <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=58> <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=59> <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-3 sin \, x csc \, x(sin \, x-csc \, x) \\ 8^3 &= sin^3 \, x-csc^3 \, x-3 sin \, 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=60> <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{2 sin \, 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=61> <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=62> <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=63> <li>Berapa nilsi dari sin (x+y) dan cos (x+y) jika sin x + sin y = a serta cos x + cos y = 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} sin \, x+sin \, y &= 2 \, sin \, (\frac{x+y}{2}) \, cos \, (\frac{x-y}{2}) \\ cos \, x+cos \, y &= 2 \, cos \, (\frac{x+y}{2}) \, cos \, (\frac{x-y}{2}) \\ \frac{2 \, sin \, (\frac{x+y}{2}) \, cos \, (\frac{x-y}{2})}{2 \, cos \, (\frac{x+y}{2}) \, cos \, (\frac{x-y}{2})} &= \frac{sin \, (\frac{x+y}{2})}{cos (\frac{x+y}{2})} \\ &= tan \, (\frac{x+y}{2}) \\ \text{misalkan } p &= \frac{x+y}{2} \text{ maka } 2p = x+y \\ tan \, (\frac{x+y}{2}) &= tan \, p \\ &= \frac{a}{b} \\ sin (x+y) = sin 2p &= \frac{2 \, tan \, p}{1+tan^2 \, p} \\ &= \frac{2 \, (\frac{a}{b})}{1+(\frac{a}{b})^2} \\ &= \frac{2 \, (\frac{a}{b})}{\frac{b^2+a^2}{b^2}} \\ &= \frac{2 \, (\frac{a}{b})}{\frac{b^2+a^2}{b^2}} \\ &= \frac{2ab}{a^2+b^2} \\ cos (x+y) = cos 2p &= \frac{1-tan^2 \, p}{1+tan^2 \, p} \\ &= \frac{1-(\frac{a}{b})^2}{1+(\frac{a}{b})^2} \\ &= \frac{\frac{b^2-a^2}{b^2}}{\frac{b^2+a^2}{b^2}} \\ &= \frac{b^2-a^2}{a^2+b^2} \\ \end{align} </math> </div></div> <ol start=64> <li>Buktikan bahwa <math>\frac{sin \, x}{1-cos \, x} = cot \, \frac{y}{4}</math> jika <math>cot \, x = \frac{1+cos \, y}{sin \, 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} cot \, x &= \frac{2 cos^2 \, \frac{y}{2}}{2 sin \, \frac{y}{2} \, cos \, \frac{y}{2}} \\ &= \frac{cos \frac{y}{2}}{sin \, \frac{y}{2}} \\ &= cot \, \frac{y}{2} \\ x &=\frac{y}{2} \\ \frac{sin \, x}{1-cos \, x} &= \frac{sin \, \frac{y}{2}}{1-cos \, \frac{y}{2}} \\ &= \frac{2 sin \, \frac{y}{4} \, cos \, \frac{y}{4}}{2 sin^2 \, \frac{y}{4}} \\ &= \frac{cos \, \frac{y}{4}}{sin \, \frac{y}{4}} \\ &= cot \, \frac{y}{4} \end{align} </math> </div></div> <ol start=65> <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=66> <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=67> <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=68> <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=69> <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=70> <li>Berapakah nilai dari <math>x^5-3x^4-x^3+9x^2-11x+3</math> jika <math>x+\frac{1}{x}=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} *cara 1 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2-3x+1 &= 0 \\ (x^3-2x+3)(x^2-3x+1 &= 0(x^3-2x+3) \\ x^5-3x^4-x^3+9x^2-11x+3 &= 0 \\ *cara 2 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2-3x+1 &= 0 \\ x^5-3x^4-x^3+9x^2-11x+3 &= x^5-3x^4+x^3-2x^3+6x^2+3x^2-2x-9x+3 \\ &= x^5-3x^4+x^3-2x^3+6x^2-2x+3x^2-9x+3 \\ &= x^3(x^2-3x+1)-2x(x^2-3x+1)+3(x^2-3x+1) \\ &= x^3(0)-2x(0)+3(0) \\ &= 0 \\ *cara 3 \\ x+\frac{1}{x} &= 3 \\ x^2+1 &= 3x \\ x^2 &= 3x-1 \\ x^2-3x+1 &= 0 \\ x^3 &= x \cdot x^2 \\ &= x (3x-1) \\ &= 3x^2-x x^4 &= x \cdot x^3 \\ &= x(3x^2-x) \\ &= 3x^3-x^2 \\ x^5 &= x \cdot x^4 \\ &= x(3x^3-x^2) \\ &= 3x^4-x^3 \\ x^5-3x^4-x^3+9x^2-11x+3 &= 3x^4-x^3-3(3x^3-x^2)-(3x^2-x)+9(3x-1)-11x+3 \\ \end{align} </math> </div></div> <ol start=71> <li>Berapakah nilai dari <math>\sqrt{x-7}</math> jika x=<math>16+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 &= 16+4\sqrt{5} \\ x-7 &= 16-7+4\sqrt{5} \\ &= 9+4\sqrt{5} \\ &= (2+\sqrt{5})^2 \\ \sqrt{x-7} &= 2+\sqrt{5} \\ \end{align} </math> </div></div> <ol start=72> <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=73> <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=74> <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=75> <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=76> <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=77> <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=78> <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=79> <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=80> <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=81> <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=82> <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=83> <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=84> <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=85> <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=86> <li>Jika <math>f(x) \cdot f(y) = f(x+y)</math> dan f(18)= 5 maka tentukan nilai f(72)!</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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) \cdot f(y) &= f(x+y) \\ f(18) \cdot f(18) &= f(18+18) \\ 5 \cdot 5 &= f(36) \\ f(36) &= 25 \\ f(36) \cdot f(36) &= f(36+36) \\ 25 \cdot 25 &= f(72) \\ f(72) &= 625 \\ \end{align} </math> </div></div> <ol start=87> <li>Jika <math>f(x) \cdot f(y) = f(xy)</math> dan f(4)= 16 maka tentukan nilai f(17)-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) \cdot f(y) &= f(xy) \\ f(4) \cdot f(4) &= f(4 \cdot 4) \\ 16 \cdot 16 &= f(16) \\ 16^2 &= f(16) \\ f(x) &= x^2 \\ f(17) &= 17^2 = 289 \\ f(17)-f(16) &= 289-256 = 33 \\ \end{align} </math> </div></div> <ol start=88> <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=89> <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=90> <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=91> <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=92> <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=93> <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=94> <li>Berapakah nilai xyz dan <math>\frac{xy+xz}{yz}</math> dari x+y=14, x+z=13 dan y+z=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} \text{ketiga persamaan tersebut dijumlahkan menjadi } \\ 2x+2y+2z &= 30 \\ x+y+z &= 15 \\ \text{untuk mencari x, y dan z masing-masing } \\ z &= 1 \\ y &= 2 \\ x &= 12 \\ xyz &= 24 \\ \frac{xy+xz}{yz} &= \frac{1 \cdot 2 + 1 \cdot 12}{2 \cdot 12} \\ &= \frac{14}{24} \\ &= \frac{7}{12} \\ \end{align} </math> </div></div> <ol start=95> <li>Tentukan nilai dari x+y+z dan yz(xy-xz) merupakan bilangan bulat positif jika xy = 2, xz = 6 dan yz = 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} xy \cdot yz \cdot xz &= 2 \cdot 3 \cdot 6 \\ (xyz)^2 &= 36 \\ xyz &= \pm 6 \\ xyz &= 6 \\ \frac{xyz}{xy} &= z = \frac{6}{2} = 3 \\ \frac{xyz}{yz} &= x = \frac{6}{3} = 2 \\ \frac{xyz}{xz} &= y = \frac{6}{6} = 1 \\ x+y+z &= 6 \\ yz(xy-xz) &= 3(2-6) = 3(-4) = -12 \\ \end{align} </math> </div></div> <ol start=96> <li>Tentukan nilai dari (x-z)^y jika <math>\frac{xy}{x+y} = \frac{1}{3}</math>, <math>\frac{yz}{y+z} = \frac{1}{4}</math> dan <math>\frac{xz}{x+z} = \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{xy}{x+y} &= \frac{1}{3} \\ \frac{x+y}{xy} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{y} + \frac{1}{x} &= 3 \\ \frac{yz}{y+z} &= \frac{1}{4} \\ \frac{y+z}{yz} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{z} + \frac{1}{y} &= 4 \\ \frac{xz}{x+z} &= \frac{1}{9} \\ \frac{x+z}{xz} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{z} + \frac{1}{x} &= 9 \\ \text{Misalkan 1/x = a, 1/y = b dan 1/z = c} \\ a+b &= 3 \\ b+c &= 4 \\ a+c &= 9 \\ a+b &= 3 \\ b+c &= 4 \\ a-c &= -1 \\ a-c &= -1 \\ a+c &= 9 \\ 2a &= 8 \\ a &= 4 \\ a-c &= -1 \\ 4-c &= -1 \\ c &= 5 \\ a+b &= 3 \\ 4+b &= 3 \\ b &= -1 \\ \frac{1}{x} &= 4 \\ x &= \frac{1}{4} \\ \frac{1}{y} &= -1 \\ y &= -1 \\ \frac{1}{z} &= 5 \\ z &= \frac{1}{5} \\ (x-z)^y &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> <ol start=97> <li>Tentukan nilai dari (y+z)^x jika <math>\frac{x+y+z}{2} = \sqrt{x-2}+\sqrt{y-1}+\sqrt{z}</math>?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" style="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+z}{2} &= \sqrt{x-2}+\sqrt{y-1}+\sqrt{z} \\ x+y+z &= 2(\sqrt{x-2}+\sqrt{y-1}+\sqrt{z}) \\ x-2\sqrt{x-2}+y-2\sqrt{y-1}+z-2\sqrt{z} &= 0 \\ x-2-2\sqrt{x-2}+1+y-1-2\sqrt{y-1}+1+z-2\sqrt{z}+1 &= 0 \\ (\sqrt{x-2}-1)^2+(\sqrt{y-1}-1)^2+(\sqrt{z}-1)^2 &= 0 \\ (\sqrt{x-2}-1)^2 &= 0 \\ \sqrt{x-2}-1 &= 0 \\ \sqrt{x-2} &= 1 \\ x-2 &= 1 \\ x &= 3 \\ (\sqrt{y-1}-1)^2 &= 0 \\ \sqrt{y-1}-1 &= 0 \\ \sqrt{y-1} &= 1 \\ y-1 &= 1 \\ y &= 2 \\ (\sqrt{z}-1)^2 &= 0 \\ \sqrt{z}-1 &= 0 \\ \sqrt{z} &= 1 \\ z &= 1 \\ (y+z)^x &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> <ol start=98> <li>Tentukan nilai dari x, y dan z jika <math>\frac{x+y}{2}=\frac{x+z}{4}=\frac{y+z}{5}</math> dan x+2y+3z=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{x+y}{2}=\frac{x+z}{4}=\frac{y+z}{5} &= k \\ x+y &= 2k \\ x+z &= 4k \\ y+z &= 5k \\ 2x+y+z &= 6k \\ 2x+5k &= 6k \\ k &= 2x \\ x &= \frac{k}{2} \\ y &= \frac{3k}{2} \\ z &= \frac{7k}{2} \\ x+2y+3z &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ x &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ y &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ z &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> <ol start=99> <li>Tentukan nilai dari 2^x jika <math>2^{(x+y)}=40, 2^{(y+z)}=20, 2^{(x+z)}=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} *cara 1 \\ 2^{(x+y)} \cdot 2^{(y+z)} \cdot 2^{(x+z)} &= 40 \cdot 20 \cdot 10 \\ 2^{2(x+y+z)} &= 8000 \\ 2^{(x+y+z)} &= 40\sqrt{5} \\ 2^x \cdot 2^{(y+z)} &= 40\sqrt{5} \\ 2^x \cdot 20 &= 40\sqrt{5} \\ 2^x &= 2\sqrt{5} \\ *cara 2 \\ \frac{2^{(x+y)}}{2^{(y+z)}} &= \frac{40}{20} \\ 2^{(x-z)} &= 2 \\ 2^{(x+z)} \cdot 2^{(x-z)} &= 10 \cdot 2 \\ 2^{2x} &= 20 \\ 2^{x} &= 2\sqrt{5} \\ *cara 3 \\ \frac{2^{(x+y)} \cdot 2^{(x+z)}}{2^{(y+z)}} &= \frac{40(10)}{20} \\ 2^{2x} &= 20 \\ 2^x &= 2\sqrt{5} \\ *cara 4 \\ 2^{(x+y)}=40 &\text{ menjadi } x+y = ^2log 40 \\ 2^{(y+z)}=20 &\text{ menjadi } y+z = ^2log 20 \\ 2^{(x+z)}=10 &\text{ menjadi } x+z = ^2log 10 \\ x+y+x+z-(y+z) &= ^2log 40+^2log 10-^2log 20 \\ 2x &= ^2log 20 \\ 2^{2x} &= 20 \\ 2^x &= 2\sqrt{5} \\ \end{align} </math> </div></div> <ol start=100> <li>Tentukan nilai dari <math>\frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z}</math> jika <math>\frac{x^2}{by+cz} = \frac{y^2}{cz+ax} = \frac{z^2}{ax+by} = 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{x^2}{by+cz} = \frac{y^2}{cz+ax} = \frac{z^2}{ax+by} &= 1 \\ \frac{x^2}{by+cz} &= 1 \\ x^2 &= by+cz \\ \frac{y^2}{cz+ax} &= 1 \\ y^2 &= cz+ax \\ \frac{z^2}{ax+by} &= 1 \\ z^2 &= ax+by \\ \frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z} &= \frac{ax}{(a+x)x} + \frac{by}{(b+y)y} + \frac{cz}{(c+z)z} \\ &= \frac{ax}{ax+x^2} + \frac{by}{by+y^2} + \frac{cz}{cz+z^2} \\ &= \frac{ax}{ax+by+cz} + \frac{by}{by+cz+ax} + \frac{cz}{cz+ax+by} \\ &= \frac{ax+by+cz}{ax+by+cz} \\ &= 1 \\ \end{align} </math> </div></div> <ol start=101> <li>Tentukan nilai dari <math>\frac{1}{xy+z-1} + \frac{1}{yz+x-1} + \frac{1}{xz+y-1}</math> jika x+y+z=2, <math>x^2+y^2+z^2=6</math> serta xyz = 4?</li></ol> <div class="toccolours mw-collapsible mw-collapsed" 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 &= 2 \\ x &= 2-y-z \\ y &= 2-x-z \\ z &= 2-x-y \\ \frac{1}{xy+z-1} + \frac{1}{yz+x-1} + \frac{1}{xz+y-1} &= \frac{1}{xy+2-x-y-1} + \frac{1}{yz+2-y-z-1} + \frac{1}{xz+2-x-z-1} \\ &= \frac{1}{xy-x-y+1} + \frac{1}{yz-y-z+1} + \frac{1}{xz-x-z+1} \\ &= \frac{1}{x(y-1)-(y-1)} + \frac{1}{y(z-1)-(z-1)} + \frac{1}{x(z-1)-(z-1)} \\ &= \frac{1}{(x-1)(y-1)} + \frac{1}{(y-1)(z-1)} + \frac{1}{(x-1)(z-1)} \\ &= \frac{z-1+x-1+y-1}{(x-1)(y-1)(z-1)} \\ &= \frac{x+y+z-3}{(x-1)(y-1)(z-1)} \\ &= \frac{x+y+z-3}{xyz-(xy+yz+xz)+(x+y+z)-1} \\ (x+y+z)^2 &= x^2+y^2+z^2+2(xy+yz+xz) \\ 2^2 &= 6+2(xy+yz+xz) \\ 4 &= 6+2(xy+yz+xz) \\ -2 &= 2(xy+yz+xz) \\ xy+yz+xz &= -1 \\ \frac{x+y+z-3}{xyz-(xy+yz+xz)+(x+y+z)-1} &= \frac{2-3}{4-(-1)+2-1} \\ &= \frac{-1}{6} \\ &= -\frac{1}{6} \\ \end{align} </math> </div></div> <ol start=102> <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=103> <li>x, y dan z merupakan bilangan tak nol. Jika <math>x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}</math> maka berapa nilai 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} x+\frac{1}{y} &= y+\frac{1}{z} \\ x-y &= \frac{1}{z}-\frac{1}{y} \\ &= \frac{y-z}{yz} \\ y+\frac{1}{z} &= z+\frac{1}{x} \\ y-z &= \frac{1}{x}-\frac{1}{z} \\ &= \frac{z-x}{xz} \\ z+\frac{1}{x} &= x+\frac{1}{y} \\ z-x &= \frac{1}{y}-\frac{1}{x} \\ &= \frac{x-y}{xy} \\ (x-y)(y-z)(z-x) &= (\frac{y-z}{yz}) \cdot (\frac{z-x}{xz}) \cdot (\frac{x-y}{xy}) \\ x^2y^2z^2 &= 1 \\ (xyz)^2 &= 1 \\ xyz &= \pm 1 \\ \end{align} </math> </div></div> <ol start=104> <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=105> <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=106> <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=107> <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=108> <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=109> <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=110> <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=111> <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=112> <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=113> <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=114> <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=115> <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=116> <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=117> <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=118> <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=119> <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=120> <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=121> <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=122> <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=123> <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=124> <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=125> <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=126> <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=127> <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=128> <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. [[Kategori:Soal-Soal Matematika]] 53rncnooc0quh5mnwj44ssv8i93s1e9 0 27301 121439 120192 2026-10-23T06:13:23Z Bennylin 425 /* Referensi */ 121439 wikitext text/x-wiki '''Suwe Ora Jamu''' adalah judul salah satu lagu daerah Jawa. Sejumlah sumber menyebut lagu ini diciptakan R.C Hardjosoebroto, seniman bidang seni karawitan dan tembang dari Kulon Progo, Daerah Istimewa Yogyakarta, pada era 1970.<ref>https://repositori.kemenbud.go.id/id/eprint/2741/ kemenbud 26 september 2026</ref> {{Wikipedia}} ==Lirik lagu suwe ora jamu== ==== Lirik Lagu Suwe Ora Jamu oleh R C Hardjosubroto ==== ''Suwé ora jamu, Jamu Godhong téla'' ''Suwé ora ketemu, ketemu pisan gawé gela'' ==== Lirik lagu Suwe ora jamu Versi Waldjinah ==== ''Suwé ora jamu ya mas, Jamu godhong téla'' ''Suwé ora ketemu, temu pisan aja gela'' ''Suwé ora jamu ya mas, Jamu godhong meniran'' ''Wis suwé ora ketemu, Temu pisan dadi pikiran'' ''Wis suwé ya mas ora jamu, Jamu godhong sligi'' ''Wis suwé ora ketemu, Temu pisan dadi ati'' ''Abang-abang ora kurang, Sing ijo di undhuhi'' ''Sing legan ora kurang, Sing duwé bojo kok di rusuhi'' ''Ireng-ireng montoré kanjeng ya mas, Ganen rene tak setirané'' ''Yen padha seneng aja mung mandeng, Golèkana ngendi omahé '' ''BWis suwé ya mas ora jamu, Jamu rempah rempah'' ''Wis suwé ora ketemu, Temu manèh karo mbak Wadljinah'' == terjemah lagu suwe ora jamu == ==== Arti Lagu Suwe Ora Jamu oleh R C Hardjosubroto ==== Lama tidak minum jamu, jamu daun ketela, Lama tidak bertemu, ketemu sekali membuat kecewa.[https://www.detik.com/jateng/budaya/d-6181676/lirik-lagu-suwe-ora-jamu-lengkap-dengan-makna-dan-kisahnya] ==== Arti lagu Suwe ora jamu Versi Waldjinah ==== Lama tidak minum jamu Mas, jamu daun ketela Lama tidak bertemu, bertemu pertama kali jangan kecewa Lama tidak minum jamu Mas, jamu daun Meniran Sudah lama tidak bertemu, bertemu pertama jadi pikiran Sudah lama ya Mas, tidak minum jamu, jamu daun Sligi, Sudah lama tidak bertemu, bertemu sekali jadi hati.[https://www.detik.com/jateng/budaya/d-6181676/lirik-lagu-suwe-ora-jamu-lengkap-dengan-makna-dan-kisahnya] == Referensi == * https://www.detik.com/jateng/budaya/d-6181676/lirik-lagu-suwe-ora-jamu-lengkap-dengan-makna-dan-kisahnya [[Kategori:Lagu daerah]] lf291w16n6m3os2rc4mjgbzdsq41loo 0 27659 121440 116768 2026-10-23T06:13:41Z Bennylin 425 Bennylin memindahkan halaman [[PUISI UNTUK AYAH]] ke [[Puisi Untuk Ayah]] 116768 wikitext text/x-wiki Puisi berjudul Puisi untuk Ayah adalah karya karangan Pramoedya Ananta Toer.<ref>https://tegartes.wordpress.com/2014/06/06/puisi-pramoedya-ananta-toer/</ref> '''''PUISI UNTUK AYAH''''' ''Tidak, Bapak, aku tak akan kembali ke kampung. Aku mau pergi yang jauh (Gadis Pantai. hal. 269)'' Sebenarnya, aku ingin kembali, Ayah Pulang ke teduh matamu Berenang di kolam yang kau beri nama rindu Aku, ingin kembali Pulang menghitung buah mangga yang ranum di halaman Memetik tomat di belakang rumah nenek. Tapi jalanan yang jauh, cita-cita yang panjang tak mengizinkanku, Mereka selalu mengetuk daun pintu saat aku tertidur Menggaruk-garuk bantal saat aku bermimpi Aku ingin kembali ke rumah, Ayah Tapi nasib memanggilku Seekor kuda sembrani datang, menculikku dari alam mimpi Membawaku terbang melintasi waktu dan dimensi kata-kata Aku menyebut pulang, tapi ia selalu menolaknya Aku menyebut rumah, tapi ia bilang tak pernah ada rumah Aku sebut kampung halaman, ia bilang kampung halaman tak pernah ada ''Maka aku menungganginya'' ''Maka aku menungganginya'' Menyusuri hutan-hutan jati Melihat rumput-rumput yang terbakar di bawahnya Menyaksikan sepur-sepur yang batuk membelah tanah Jawa Arwah-arwah pekerja bergentayangan menuju ibu kota, Mencipta banjir dari genangan air mata Arwah-arwah buruh menggiring hujan air mata, mata mereka menyeret banjir Kota yang tua telah lelah menggigil, sudah lupa bagaimana bermimpi dan bangun pagi Hujan ingin bercerai dengan banjir Tapi kota yang pikun membuatnya bagai cinta sejati dua anak manusia Aku tak bisa pulang lagi, Ayah, kuda ini telah menambatkan hatiku di pelananya Orang-orang datang ke pasar malam, satu persatu, seperti katamu Berjudi dengan nasib, menunggu peruntungan menjadi kaya raya Tapi seperti rambu lalu lintas yang setia, sedih dan derita selalu berpelukan dengan setia Aku tak bisa pulang lagi, Ayah, kuda ini telah menambatkan hatiku di pelananya Orang bilang, apa yang ada di depan manusia hanya jarak. Dan batasnya adalah ufuk. Begitu jarak ditempuh sang ufuk menjauh. Yang tertinggal jarak itu juga-abadi. Di depan sana ufuk yang itu juga-abadi. Tak ada romantika cukup kuat untuk dapat menaklukan dan menggenggamnya dengan tangan-jarak dan ufuk abadi itu '''Pramoedya Ananta Toer''', ''Anak Semua Bangsa'' == Referensi == <references /> j6imsa31xv4f4eokgolhbv58oq91wnq 121442 121440 2026-10-23T06:29:34Z Bennylin 425 added [[Category:Pramoedya Ananta Toer]] using [[Help:Gadget-HotCat|HotCat]] 121442 wikitext text/x-wiki Puisi berjudul Puisi untuk Ayah adalah karya karangan Pramoedya Ananta Toer.<ref>https://tegartes.wordpress.com/2014/06/06/puisi-pramoedya-ananta-toer/</ref> '''''PUISI UNTUK AYAH''''' ''Tidak, Bapak, aku tak akan kembali ke kampung. Aku mau pergi yang jauh (Gadis Pantai. hal. 269)'' Sebenarnya, aku ingin kembali, Ayah Pulang ke teduh matamu Berenang di kolam yang kau beri nama rindu Aku, ingin kembali Pulang menghitung buah mangga yang ranum di halaman Memetik tomat di belakang rumah nenek. Tapi jalanan yang jauh, cita-cita yang panjang tak mengizinkanku, Mereka selalu mengetuk daun pintu saat aku tertidur Menggaruk-garuk bantal saat aku bermimpi Aku ingin kembali ke rumah, Ayah Tapi nasib memanggilku Seekor kuda sembrani datang, menculikku dari alam mimpi Membawaku terbang melintasi waktu dan dimensi kata-kata Aku menyebut pulang, tapi ia selalu menolaknya Aku menyebut rumah, tapi ia bilang tak pernah ada rumah Aku sebut kampung halaman, ia bilang kampung halaman tak pernah ada ''Maka aku menungganginya'' ''Maka aku menungganginya'' Menyusuri hutan-hutan jati Melihat rumput-rumput yang terbakar di bawahnya Menyaksikan sepur-sepur yang batuk membelah tanah Jawa Arwah-arwah pekerja bergentayangan menuju ibu kota, Mencipta banjir dari genangan air mata Arwah-arwah buruh menggiring hujan air mata, mata mereka menyeret banjir Kota yang tua telah lelah menggigil, sudah lupa bagaimana bermimpi dan bangun pagi Hujan ingin bercerai dengan banjir Tapi kota yang pikun membuatnya bagai cinta sejati dua anak manusia Aku tak bisa pulang lagi, Ayah, kuda ini telah menambatkan hatiku di pelananya Orang-orang datang ke pasar malam, satu persatu, seperti katamu Berjudi dengan nasib, menunggu peruntungan menjadi kaya raya Tapi seperti rambu lalu lintas yang setia, sedih dan derita selalu berpelukan dengan setia Aku tak bisa pulang lagi, Ayah, kuda ini telah menambatkan hatiku di pelananya Orang bilang, apa yang ada di depan manusia hanya jarak. Dan batasnya adalah ufuk. Begitu jarak ditempuh sang ufuk menjauh. Yang tertinggal jarak itu juga-abadi. Di depan sana ufuk yang itu juga-abadi. Tak ada romantika cukup kuat untuk dapat menaklukan dan menggenggamnya dengan tangan-jarak dan ufuk abadi itu '''Pramoedya Ananta Toer''', ''Anak Semua Bangsa'' == Referensi == <references /> [[Kategori:Pramoedya Ananta Toer]] 2adk0cb3b8raj3hg73w7ygwykn3czgc 0 28687 121436 120749 2026-10-23T03:58:57Z Abuhusainalbanjari 43686 /* Deskripsi */ 121436 wikitext text/x-wiki == Deskripsi == * Fungsi: *# Pemutar game *# Pemutar film *# Pemutar musik *# Penampil foto *# Pembaca ebook * Game yang didukung beserta ekstensi berkasnya: *# GB (gb) *# GBC (gbc) *# MD (bin) *# FC (nes) *# GBA (gba) *# MAME (kg) *# SFC (sfc) * Support 3 save state tiap game * Support micro-sd card * Kelengkapan: *# Kardus *# Konsol portabel X7 *# Kabel Audio/Video (A/V) *# Kabel charger *# Head phones *# Manual * Harga: Sekitar Rp345.000 == Menu == === Game di Menu Utama === # Knights of Valor (KOV) # Oriental Legend # Warriors of Fate (WOF) # Dynasty War # Final Fight # Street FIghter 2 # Captain Commando # Cadillacs and Dinosaurs # The Punisher # Raiden Trad # UN Squadron # Magic Sword === Layar 1 === # Permainan ## Disk lokal ### Classic ### GB ### GBC ### MD ### FC ### GBA ### MAME ### SFC ## Recent run list ## Disk yang dapat dilepas # Bioskop # Musik # Gambar === Layar 2 === # E-book # Browser # Pengaturan # Tema (ada 3 tema) == Kekurangan == * Tidak bisa untuk 2 pemain * Baterai drop setelah sekian lama dipakai * Tidak seperti Konsol Portabel X6, Konsol Portabel X7 ini tidak support game PS1 * Tidak ada pengaturan tombol, kecuali jika ada pada menu options game (pengaturan tombol lewat menu options game) [[Kategori:Panduan Konsol|{{SUBPAGENAME}}]] fdsg99jyd0tatj2crott4ni76ykvis6 3 28876 121437 121396 2026-10-23T06:11:04Z Bennylin 425 /* Halo */ Balas 121437 wikitext text/x-wiki ==Halo== Terkait dengan konten-konten berbahasa Banjar, apa tidak sebaiknya di Wikibuku Bahasa Banjar? <span style="text-shadow: 5px 3px 1px gray; font-color:black;">[[Pengguna:Bennylin|Bennylin]]</span> ([[Pembicaraan Pengguna:Bennylin|Bicara]]) 20 Oktober 2026 04.57 (UTC) :Halo, karena saat ini Wikibuku bahasa Banjar belum ada, saya dan yang lain berencana untuk meletakkannya terlebih dahulu di Wikibuku bahasa Indonesia, sambil menyiapkan inkubator Wikibuku bahasa Banjar. Begitu inkubator sudah berjalan, konten akan kami pindahkan kesana. [[Pengguna:Wadaihangit|Wadaihangit]] ([[Pembicaraan Pengguna:Wadaihangit|bicara]]) 20 Oktober 2026 05.02 (UTC) ::Oh begitu. Baiklah. Semoga sukses proyeknya! <span style="text-shadow: 5px 3px 1px gray; font-color:black;">[[Pengguna:Bennylin|Bennylin]]</span> ([[Pembicaraan Pengguna:Bennylin|Bicara]]) 23 Oktober 2026 06.11 (UTC) h5nbwzds6fkden94w0a1ec459pii5xv 0 28878 121441 2026-10-23T06:13:41Z Bennylin 425 Bennylin memindahkan halaman [[PUISI UNTUK AYAH]] ke [[Puisi Untuk Ayah]] 121441 wikitext text/x-wiki #ALIH [[Puisi Untuk Ayah]] s1qbb485at8crb3ct30ibc26ew59b1o