EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 7)

 $\begin{array}{ll}\\ 31.&\textrm{Nilai dari}\: \: \displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2^{\displaystyle 2}}}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&4\\ \textrm{b}.&6\\ \textrm{c}.&8\\ \textrm{d}.&10\\ \textrm{e}.&12 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 1}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 2}}=\displaystyle 2^{\displaystyle 4}=16\\ &\textrm{Sedangkan untuk}\\ &\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2}}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 1}}=\displaystyle 2^{\displaystyle 2}=4\\ &\textrm{Selanjutnya untuk}\\ &\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2^{\displaystyle 2}}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 4}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}=2^{\displaystyle 1}=2\\ &\textrm{Sehingga}\\ &\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2^{\displaystyle 2}}}}=16-4-2=10 \end{aligned}  \end{array}$.

EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 6)

$\begin{array}{ll}\\ 26.&\textrm{Nilai dari}\: \: \displaystyle \sqrt[4]{4}-\sqrt{2}=....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \sqrt[4]{4}-\sqrt{2}=\displaystyle \sqrt[2]{\sqrt[2]{2^{2}}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 27.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sqrt[\displaystyle 3]{64}}{\sqrt[\displaystyle 4]{64}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{b}.&\sqrt{2}\\ \textrm{c}.&2\sqrt{2}\\ \textrm{d}.&3\\ \textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \frac{\sqrt[\displaystyle 3]{64}}{\sqrt[\displaystyle 4]{64}}=\displaystyle \frac{\sqrt[\displaystyle 3]{4^{\displaystyle 3}}}{\sqrt[\displaystyle 4]{8^{\displaystyle 2}}}=\displaystyle \frac{\sqrt[\displaystyle 3]{4^{\displaystyle 3}}}{\sqrt[\displaystyle 4]{\sqrt{8}^{\displaystyle 4}}}=\frac{4}{\sqrt{8}}\\ &=\frac{4}{\sqrt{8}}\times \frac{\sqrt{8}}{\sqrt{8}}=\frac{4}{8}\left( \sqrt{8} \right)\\ &=\frac{1}{2}\left( 2\sqrt{2} \right)=\sqrt{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 28.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sqrt{\sqrt{25}}}{\sqrt{5}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&5\\ \textrm{b}.&\sqrt{5}\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{5}\\ \textrm{d}.&1\\ \textrm{e}.&\displaystyle \frac{1}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \frac{\sqrt{\sqrt{25}}}{\sqrt{5}}=\frac{\sqrt{\sqrt[\displaystyle 2]{5^{\displaystyle 2}}}}{\sqrt{5}}=\frac{\sqrt{5}}{\sqrt{5}}=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 29.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sqrt{49\sqrt{7}}}{\sqrt{7\sqrt{49}}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&7\\ \textrm{b}.&7\sqrt{7}\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{7}\\ \textrm{d}.&\sqrt{7}\\ \textrm{e}.&\displaystyle \sqrt[\displaystyle 4]{7} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \frac{\sqrt{49\sqrt{7}}}{\sqrt{7\sqrt{49}}}=\displaystyle \frac{7.\sqrt[\displaystyle 4]{7}}{\sqrt{7}.\sqrt[\displaystyle 4]{49}}=\displaystyle \frac{7.\sqrt[\displaystyle 4]{7}}{\sqrt{7}.\sqrt[\displaystyle 4]{7^{\displaystyle 2}}}\\ &=\displaystyle \frac{7.\sqrt[\displaystyle 4]{7}}{\sqrt{7}.\sqrt{7}}=\displaystyle \frac{7.\sqrt[\displaystyle 4]{7}}{7}=\sqrt[\displaystyle 4]{7} \end{aligned} \end{array}$.

 $\begin{array}{ll}\\ 30.&\textrm{Nilai dari}\: \: \displaystyle \left( \frac{1}{2} \right)^{\displaystyle \frac{1}{2}}-\left( \frac{1}{4} \right)^{\displaystyle \frac{1}{4}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \left( \frac{1}{2} \right)^{4}=\left( \frac{1}{4} \right)^{2}\Leftrightarrow \displaystyle \left( \frac{1}{2} \right)^{\displaystyle \frac{4}{1}}=\left( \frac{1}{4} \right)^{\displaystyle \frac{2}{1}}\\ &\Leftrightarrow \left( \frac{1}{2} \right)^{\displaystyle \frac{1}{2}}=\left( \frac{1}{4} \right)^{\displaystyle \frac{1}{4}}\\ &\Leftrightarrow \left( \frac{1}{2} \right)^{\displaystyle \frac{1}{2}}-\left( \frac{1}{4} \right)^{\displaystyle \frac{1}{4}}=0 \end{aligned} \end{array}$.

EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 5)

 $\begin{array}{ll}\\ 21.&(\textbf{UM UGM 05})\textrm{Hasil dari}\\ &\sqrt{0,3+\sqrt{0,08}}=\sqrt{a}+\sqrt{b}\: ,\: \textrm{maka}\: \: \displaystyle \frac{1}{a}+\frac{1}{b}=....\\ &\begin{array}{llll}\\ \textrm{a}.&25\\ \textrm{b}.&20\\ \textrm{c}.&15\\ \textrm{d}.&10\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\sqrt{0,3+\sqrt{0,08}}&=\sqrt{0,2+0,1+\sqrt{4\times 0,2\times 0,1}}\\ &=\sqrt{0,2+0,1+2\sqrt{\times 0,2\times 0,1}}\\ &=\sqrt{0,2}+\sqrt{0,1}\\ \textrm{maka},\: \: a=0,2&,\: \: b=0,1\\ \textrm{sehingga}\: \displaystyle \frac{1}{a}+\frac{1}{b}&=\displaystyle \frac{1}{0,2}+\frac{1}{0,1}=5+10=15\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&(\textbf{SPMB 06})\textrm{Jika bilangan bulat}\: \: a\: \: \: \textrm{dan}\: \: b\: \: \textrm{memenuhi}\\ &\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=a+b\sqrt{30}\: ,\: \textrm{maka}\: \: ab=....\\ &\begin{array}{llll}\\ \textrm{a}.&-22\\ \textrm{b}.&-11\\ \textrm{c}.&-9\\ \textrm{d}.&2\\ \textrm{e}.&13 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}&=\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}\times \displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}\\ &=\displaystyle \frac{5-2\sqrt{30}+6}{5-6}\\ &=\displaystyle \frac{11-2\sqrt{30}}{-1}\\ &=-11+2\sqrt{30}\\ \textrm{sehingga}&\: \: \: a=-11,\: \: b=2,\: \: \textrm{maka}\\ ab&=(-11)\times 2\\ &=-22 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 23.&(\textbf{OSK 2013})\textrm{Misal}\: \: a\: \: \textrm{dan}\: \: b\: \: \textrm{bilangan asli}\\ &\textrm{dengan}\: \: a>b.\: \: \textrm{Jika} \: \: \sqrt{94+2\sqrt{2013}}=\sqrt{a}+\sqrt{b}\\ &\textrm{maka nilai} \: \: a-b\: \: \textrm{adalah... .}\\\\ &\textrm{Jawab}:\\ &\begin{aligned} \sqrt{94+2\sqrt{2013}}&=\sqrt{61+33+2\sqrt{61\times 33}}\\ &=\sqrt{61}+\sqrt{33}\\ &=\sqrt{a}+\sqrt{b}\\ \textrm{Sehingga}\: \: a&=61,\: \: b=33,\: \: \textrm{maka}\\ a-b&=61-33\\ &=28 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 24.&\textrm{Daerah hasil dari fungsi eksponen}\: \: y\: =x^{- \frac{2}{3}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&y< 0\\ \textrm{b}.&y> 0\\ \textrm{c}.&y\geq 0\\ \textrm{d}.&y\leq 0\\ \textrm{e}.&\textrm{Semua bilangan real} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikanlah gambar berikut} \end{array}$

$.\quad\: \, \begin{aligned}\textrm{diketahui}&\\ y\: &=x^{-\displaystyle \frac{2}{3}}\\ y^{3}\: &=x^{-2}\\ y^{3}\: &=\displaystyle \frac{1}{x^{2}},\: \textrm{atau}\\ y^{3}\times x^{2}\: &=1,\\ \textrm{sehingga}&\: \: y\: \: \textrm{tidak mungkin berharga}\: \: 0 \end{aligned}$

$\begin{array}{ll}\\ 25.&\textrm{Jika}\: \: f(x)=b^{x},\: \: \textrm{di mana konstan positif},\\\\ &\displaystyle \frac{f\left ( x^{2}+x \right )}{f(x+1)}= ....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&f\left ( x^{2} \right )&&&\\ \textrm{b}.&f(x+1)f(x-1)\\ \textrm{c}.&f(x+1)+f(x-1)\\ \textrm{d}.&f(x+1)-f(x-1)\\ \textrm{e}.&f\left ( x^{2}-1 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\displaystyle \frac{f\left ( x^{2}+x \right )}{f(x+1)}&=\frac{b^{x^{2}+x}}{b^{x+1}}\\ &=b^{x^{2}+x-(x+1)}\\ &=b^{x^{2}-1}\\ &=f\left ( x^{2}-1 \right ) \end{aligned} \end{array}$

EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 4)

$\begin{array}{ll}\\ 16.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x^{\displaystyle x^{\displaystyle x^{\displaystyle x^{\displaystyle x^{\cdots }}}}}=2026 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\sqrt{2026}\\ \textrm{b}.&\sqrt[\displaystyle 2026]{2026}\\ \textrm{c}.&2026^{\sqrt{2026}}\\ \textrm{d}.&\sqrt{2026}^{\sqrt{2026}}\\ \textrm{e}.&\sqrt{2026\sqrt{2026}} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}x^{\displaystyle x^{\displaystyle x^{\displaystyle x^{\displaystyle x^{\cdots }}}}}&=2026\\ x^{2026}&=2026\\ x&=\sqrt[\displaystyle 2026]{2026} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ & \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}=3 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&3\\ \textrm{b}.&6\\ \textrm{c}.&7\\ \textrm{d}.&8\\ \textrm{e}.&9 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Misalkan}\quad A&=\sqrt{+\sqrt{x+\sqrt{+\cdots }}}\\ \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}&=3\\ \textrm{dikuadratkan}&\\ x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}&=9\\ x+3&=9\\ x&=9-3\\ x&=6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&7\sqrt[7]{7}\\ \textrm{b}.&7\\ \textrm{c}.&14\\ \textrm{d}.&49\\ \textrm{e}.&\sqrt[3]{81} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}x&=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^{3}&=49\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^{3}&=49x\\ x^{2}&=49\\ x&=\sqrt{49}\\ &=7 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Nilai dari}\\ &\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\\ &\textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&\sqrt[3]{2}+1\\ \textrm{c}.&\sqrt[3]{2}-1\\ \textrm{d}.&\sqrt[3]{4}+1\\ \textrm{e}.&\sqrt[3]{4}-1 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\times \frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1}\\ &=\displaystyle \frac{\left ( \sqrt[3]{2} \right )^{2}-1}{\sqrt[3]{2}+\sqrt[3]{4}+\sqrt[3]{8}-1-\sqrt[3]{2}-\sqrt[3]{4}}\\ &=\displaystyle \frac{\sqrt[3]{4}-1}{\sqrt[3]{8}-1}\\ &=\displaystyle \frac{\sqrt[3]{4}-1}{2-1}\\ &=\sqrt[3]{4}-1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Nilai dari}\\ &\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\\ &\textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\sqrt{2}-1\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{d}.&\sqrt{\displaystyle \frac{5}{3}}\\ \textrm{e}.&\sqrt{\displaystyle \frac{2}{5}} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\times \frac{\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}} -\sqrt{3-2\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}}{\sqrt{5}+1}-\left ( \sqrt{2}-1 \right )\\ &=\displaystyle \frac{\left ( \displaystyle \frac{3+\sqrt{5}}{\sqrt{2}} \right )+\left ( \displaystyle \frac{\sqrt{5}-1}{\sqrt{2}} \right )}{\sqrt{5}+1}+1-\sqrt{2}\\ &=\displaystyle \frac{\displaystyle \frac{2+2\sqrt{5}}{\sqrt{2}}}{1+\sqrt{5}}+1-\sqrt{2}\\ &=\displaystyle \frac{2}{\sqrt{2}}+1-\sqrt{2}\\ &=\sqrt{2}+1-\sqrt{2}\\ &=1 \end{aligned} \end{array}$

EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 3)

$\begin{array}{l}\\ 11.&\textrm{Nilai dari}\\ & \displaystyle \frac{2^{2026}+2^{2027}}{3}-2^{2026}\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-2&&&\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\displaystyle \frac{2^{2026}+2^{2027}}{3}-2^{2026}\\ &=\displaystyle \frac{2^{2026}+2^{2027}-3.2^{2026}}{3}\\ &=\displaystyle \frac{1.2^{2026}+2^{1}.2^{2026}-3.2^{2026}}{3}\\ &=\displaystyle \frac{(3-3).2^{2026}}{3}\\ &=0 \end{aligned} \end{array}$.

$\begin{array}{l}\\ 12.&\textrm{Nilai dari}\\ &\displaystyle \frac{2^{2026}+2^{2027}+2^{2028}}{2^{2029}+2^{2030}+2^{2031}}\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&0,125&&&\\ \textrm{b}.&0,\overline{333}\\ \textrm{c}.&0,45\\ \textrm{d}.&0,5\\ \textrm{e}.&0,\overline{666} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\displaystyle \frac{2^{2026}+2^{2027}+2^{2028}}{2^{2029}+2^{2030}+2^{2031}}\\ &=\displaystyle \frac{1.2^{2026}+2^{1}.2^{2026}+2^{2}.2^{2026}}{2^{3}.2^{2026}+2^{4}.2^{2026}+2^{5}.2^{2026}}\\ &=\displaystyle \frac{(1+2+4).2^{2026}}{(8+16+32).2^{2026}}\\ &=\displaystyle \frac{7}{56}=\frac{1}{8}=0,125 \end{aligned} \end{array}$.

$\begin{array}{l}\\ 13.&\textrm{Jika nilai dari}\\ &a^{\displaystyle a}=3\: ,\: \textrm{maka nilai}\quad a^{\displaystyle a^{\displaystyle a+1}}\quad \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&9&&&\\ \textrm{b}.&18\\ \textrm{c}.&27\\ \textrm{d}.&81\\ \textrm{e}.&243 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\displaystyle a^{\displaystyle a^{\displaystyle a+1}}=a^{\displaystyle a^{\displaystyle a}.a}=a^{\displaystyle 3.a}=\left( a^{\displaystyle a} \right)^{3}=3^{3}=27 \end{aligned} \end{array}$.

 $\begin{array}{ll}\\ 14.&\textrm{Hitunglah}\:\\\\ &\quad\quad\qquad \displaystyle \sqrt[8]{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{...}}}}}\\\\ &\textrm{nyatakan jawabannya dalam bentuk }\: \displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textrm{dengan a, b, c, dan d bilangan-bilangan bulat}\\ \end{array}$

Pembahasan:

$\begin{aligned}x^{8}&=2207-\displaystyle \underset{x^{8}}{\underbrace{\displaystyle \frac{1}{2207-\frac{1}{2207-\frac{1}{2207-...}}}}}\\ x^{8}&=2207-\displaystyle \frac{1}{x^{8}}\\ x^{8}+\displaystyle \frac{1}{x^{8}}&=2207\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )^{2}&=2207+2\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )&=\sqrt{2209}=47 \end{aligned}$

$\begin{aligned}x^{4}+\displaystyle \frac{1}{x^{4}}&=47\\ \left ( x^{2}+\displaystyle \frac{1}{x^{2}} \right )^{2}&=47+2\\ x^{2}+\displaystyle \frac{1}{x^{2}}&=\sqrt{49}=7\\ \left ( x+\displaystyle \frac{1}{x} \right )^{2}&=7+2\\ x+\displaystyle \frac{1}{x}&=\sqrt{9}=3\\ x^{2}-3x+1&=0,\\ &\textrm{persamaan kuadrat dalam x,}\\ & \textbf{gunakan rumus abc}\\ x_{1,2}=&\displaystyle \frac{3\pm \sqrt{5}}{2}=\displaystyle \frac{3\pm 1\sqrt{5}}{2}=\displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textbf{Sehingga},\quad \begin{cases} & a=3 \\ & b=1 \\ & c=5 \\ & d=2 \end{cases} \end{aligned}$.

$\begin{array}{ll}\\ 15.&\textrm{Diketahui}\\ &x=\displaystyle \frac{1+p+p^{2}+p^{3}+\cdots +p^{n-1}}{1+p+p^{2}+p^{3}+\cdots +p^{n-2}+p^{n-1}+p^{n}} \\ &y=\displaystyle \frac{1+q+q^{2}+q^{3}+\cdots +q^{n-1}}{1+q+q^{2}+q^{3}+\cdots +q^{n-2}+q^{n-1}+q^{n}}\\\\ &\textrm{dan}\: \: p>q>0\\\\ &\textrm{Tunjukkan bahwa}\: \: x<y \\\\\\ &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Perhatikan bahwa}:\: \: p>q>0\\ &\textrm{sehingga}\\ &\displaystyle \frac{1}{p}< \frac{1}{q},\: \: \displaystyle \frac{1}{p^{2}}< \frac{1}{q^{2}},\cdots , \displaystyle \frac{1}{p^{n}}< \frac{1}{q^{n}}\\ &\textrm{Jika bentuk di atas dijumlahkan, maka}\\ &\displaystyle \frac{1}{p}+\frac{1}{p^{2}}+\cdots +\frac{1}{p^{n}}< \frac{1}{q}+\frac{1}{q^{2}}+\cdots +\frac{1}{q^{n}}\\ &\Leftrightarrow \displaystyle \frac{p^{n-1}+\cdots +p^{2}+p+1}{p^{n}}< \displaystyle \frac{q^{n-1}+\cdots +q^{2}+q+1}{q^{n}}\\ &\Leftrightarrow \displaystyle \frac{p^{n}}{1+p+p^{2}+\cdots +p^{n-1}}>\displaystyle \frac{q^{n}}{1+q+q^{2}+\cdots +q^{n-1}}\\ &\Leftrightarrow \displaystyle \frac{p^{n}}{1+p+p^{2}+\cdots +p^{n-1}}+1>\displaystyle \frac{q^{n}}{1+q+q^{2}+\cdots +q^{n-1}}+1\\ &\Leftrightarrow \displaystyle \frac{1+p+p^{2}+\cdots +p^{n-1}+p^{n}}{1+p+p^{2}+\cdots +p^{n-1}}>\displaystyle \frac{1+q+q^{2}+\cdots +q^{n-1}+q^{n}}{1+q+q^{2}+\cdots +q^{n-1}}\\ &\Leftrightarrow \displaystyle \frac{1+p+p^{2}+\cdots +p^{n-1}}{1+p+p^{2}+\cdots +p^{n-1}+p^{n}}<\displaystyle \frac{1+q+q^{2}+\cdots +q^{n-1}}{1+q+q^{2}+\cdots +q^{n-1}+q^{n}}\\ &\Leftrightarrow x<y\qquad \blacksquare  \end{aligned}  \end{array}$.

EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 2)

 $\begin{array}{l}\\ 06.&\textrm{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-2&&&\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}5^{x+1}+5^{2-x}-30&=0\\ \left (5^{x} \right ).5^{1}+\displaystyle \frac{5^{2}}{5^{x}}-30&=0\\ 5\left ( 5^{x} \right )^{2}+25-30\left ( 5^{x} \right )&=0\\ \textrm{Persamaan kuadrat}&\: \textrm{dalam}\: \: 5^{x},\: \textrm{maka}\\ 5(5^{x})^{2}-30(5^{x})+25&=0\begin{cases} a & =5 \\ b & =-30 \\ c & =25 \end{cases}\\ (5^{x_{1}}).\left ( 5^{x_{2}} \right )&=\displaystyle \frac{c}{a}\\ 5^{x_{1}+x_{2}}&=\displaystyle \frac{25}{5}=5\\ 5^{x_{1}+x_{2}}&=5^{1}\\ x_{1}+x_{2}&=1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 07.&\textrm{Jumlah akar-akar persamaan}\\ &2020^{x^{2}-7x+7}=2021^{x^{2}-7x+7}\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-7\\ \textrm{b}.&-5\\ \textrm{c}.&-3\\ \textrm{d}.&5\\ \textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}2020^{x^{2}-7x+7}&=2021^{x^{2}-7x+7}\\ \textrm{Karena basis}&\: \textrm{tidak sama},\\ \textrm{maka harusl}&\textrm{ah pangkatnya}=0,\\ x^{2}-7x+7&=0\\ \textrm{dan jumlah}\: &\textrm{akar-akarnya adalah}:\\ x_{1}+x_{2}&=-\displaystyle \frac{b}{a}, \: \: \textrm{dari persamaan}\\ x^{2}-7x+7&=0\begin{cases} a &=1 \\ b &=-7 \\ c &=7 \end{cases}\\ \textrm{maka}\: \: x_{1}+x_{2}&=-\displaystyle \frac{b}{a}=-\frac{-7}{1}=7 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 08.&\textrm{Nilai dari}\: \: \displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&5\\ \textrm{c}.&10\\ \textrm{d}.&20\\ \textrm{e}.&40 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}&=\displaystyle \frac{2^{4}.2^{2016}+2^{2}.2^{2016}}{2^{2}.2^{2018}+2^{2016}}\\ &=\displaystyle \frac{2^{2016}\left ( 2^{4}+2^{2} \right )}{2^{2016}\left ( 2^{2}+1 \right )}\\ &=\displaystyle \frac{16+4}{4+1}\\ &=\displaystyle \frac{20}{5}\\ &=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 09.&(\textbf{UM IPB})\textrm{Jika}\: \: ab=a^{b} \: \: \textrm{dan}\: \: \displaystyle \frac{a}{b}=a^{3b}\\ &\textrm{maka nilai}\: \: a\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&0,5\\ \textrm{c}.&1\\ \textrm{d}.&0,25\\ \textrm{e}.&0,75 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Diketahui}&\\ ab&=a^{b}\\ b&=\displaystyle \frac{a^{b}}{a}=a^{b-1}.....\textbf{1}\\ \textrm{maka}&\\ \displaystyle \frac{a}{b}&=a^{3b}...............\textbf{2}\\ \textbf{1}&\: \: ke\: \: \textbf{2}\\ \displaystyle \frac{a}{a^{b-1}}&=a^{3b}\\ a^{2-b}&=a^{3b}\\ 2-b&=3b\\ -4b&=-2\\ b&=\displaystyle \frac{1}{2}................\textbf{3}\\ \textbf{3}&\: \: ke\: \: \textbf{1}\\ a\left ( \displaystyle \frac{1}{2} \right )&=a^{\frac{1}{2}}\\ \displaystyle \frac{1}{4}a^{2}&=a\\ a^{2}-4a&=0\\ a(a-4)&=0\\ a=0\: \: &\textrm{atau}\: \: a=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: \displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}=4\: ,\: \textrm{maka}\: \: x=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 1,1\\ \textrm{b}.&\displaystyle 1,2\\ \textrm{c}.&1,3\\ \textrm{d}.&\displaystyle 1,4\\ \textrm{e}.&1,5\\\\ &&&(\textbf{SAT Test Math Level 2})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}&=4\\ \left (3.x^{^{^{^{ \frac{3}{2}}}}} \right )^{2}&=4^{2}\\ 3^{2}.x^{3}&=4^{2}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\times \frac{3}{3}\\ x^{3}&\leq \displaystyle \frac{4^{2}}{3^{2}}\times \frac{4}{3}\\ x^{3}&\leq \left ( \displaystyle \frac{4^{3}}{3^{3}} \right )\\ x^{3}&\leq \left ( \displaystyle \frac{4}{3} \right )^{3}\\ x&\leq \displaystyle \frac{4}{3}\\ x&\leq 1,\overline{333}\\ x&\approx 1,3 \end{aligned} \end{array}$

EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 1)

$\begin{array}{ll}\\ 01.&\textrm{Jika bentuk}\: \: \displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}\\ & \textrm{dinyatakan dalam pangkat positif}=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{a^{2}}{a-b}&&&\\\\ \textrm{b}.&\displaystyle \frac{a^{2}}{a-1}\\\\ \textrm{c}.&\displaystyle \frac{b-a}{ab}\\\\ \textrm{d}.&\displaystyle \frac{a^{2}}{b-a}\\\\ \textrm{e}.&\displaystyle \frac{1}{a-b}\\\\ &&&&(\textbf{SAT Test Math Level 2})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}&=\displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}\times \frac{b}{b}\\ &=\displaystyle \frac{a}{a^{-1}b-1}\\ &=\displaystyle \frac{a}{a^{-1}b-1}\times \frac{a}{a}\\ &=\displaystyle \frac{a^{2}}{b-a} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 02.&\textrm{Jika terdapat hubungan berikut}\\ &\textrm{a}.\quad 2^{p}=3^{q}=6^{r},\: \: \textrm{tunjukkan bahwa}\: \: pr+qr-pq=0\\ &\textrm{b}.\quad 2^{x}=3^{2y}=6^{z},\: \: \textrm{tunjukkan bahwa }\: \: 2xy-2yz-xz=0\\ &\textrm{c}.\quad 3^{15a}=5^{5b}=15^{3c},\: \: \textrm{tunjukkan bahwa }\: \: 5ab-bc-3ac=0\\\\ &\textrm{Bukti}\\ &\textrm{Yang akan ditunjukkan adalah no. 02 yang poin c, yaitu:}\\ &\begin{aligned}3^{15a}=5^{5b}=15^{3c}&\begin{cases} 3=5^{\frac{5b}{15a}} & \\ 3^{\frac{15a}{5b}}=b &\left ( a^{b}=c^{d}\rightarrow a=c^{\frac{d}{b}}\: \: \textrm{atau}\: \: a^{\frac{b}{d}}=c \right ) \end{cases}\\ 3^{15a}&=15^{3c}\\ 3^{15a}&=(3\times 5)^{3c}\\ 3^{15a}&=(3\times 3^{\frac{15a}{5b}})^{3c}\\ 3^{15a}&=3^{3c+\frac{9c}{b}}\\ a^{f(x)}&=a^{g(x)}\\ f(x)&=g(x)\\ 15a&=3c+\frac{9ac}{b}\\ 15ab&=3bc+9ac\\ 5ab&=bc+3ac\\ 5ab-bc-3ac&=0\quad \color{black}\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 03.&\textrm{Diketahui bahwa}\quad 5^{p}=9^{q}=2025, \: \textrm{nilai}\: \: \displaystyle \frac{pq}{p+q}=....\\ &\textrm{Jawab}:\\  &\begin{aligned}&5^{p}=9^{q}=2025=45^{2} \quad \textrm{dengan}\quad\begin{cases} 5=9^{\frac{q}{p}} & \\ 5^{\frac{p}{q}}=9  \end{cases}\\ &\Leftrightarrow  5^{p}=(5\times 9)=5^{2}\times 9^{2}\\ &\Leftrightarrow 5^{p}=5^{2}\times \left(5^{\frac{p}{q}}  \right)^{2}\\ &\Leftrightarrow 5^{p}=5^{2+\displaystyle \frac{2p}{q}},\quad \textrm{ingat}\quad a^{f(x)}=a^{g(x)}\Rightarrow f(x)=g(x)\\ &\Leftrightarrow p=2+\displaystyle \frac{2p}{q}\\ &\Leftrightarrow pq=2q+2p=2(p+q)\\ &\Leftrightarrow \displaystyle \frac{pq}{p+q}=2   \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 04.&\textrm{Bentuk sederhana dari}\\ &\sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}=....\\\\ &\qquad\qquad\qquad(\textbf{SIMAK UI 2012 Mat IPA})\\ &\begin{array}{llllllll}\\ \textrm{a}.&2-\sqrt{2}\\ \textrm{b}.&8-\sqrt{2}\\ \textrm{c}.&-2+\sqrt{2}\\ \textrm{d}.&2+5\sqrt{2}\\ \textrm{e}.&8+5\sqrt{2} \end{array}\\\\ &\textrm{Jawab}:\qquad \textbf{b}\\ &\textrm{misalkan},\\ &\begin{aligned}x&=\sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}\\ &=\left ( \sqrt{33+20\sqrt{2}}-\sqrt{27-2.9\sqrt{2}}\: \right )\\ &=\sqrt{33+20\sqrt{2}}-\sqrt{27-18\sqrt{2}}\\ x^{2}&=33+20\sqrt{2}+27-18\sqrt{2}-2\sqrt{\left ( 33+20\sqrt{2} \right )\left ( 27-18\sqrt{2} \right )}\\ &=60+2\sqrt{2}-2\sqrt{33.27-33.18\sqrt{2}+27.20\sqrt{2}-20.18.2}\\ &=60+2\sqrt{2}-2\sqrt{891-720+540\sqrt{2}-594\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-54\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2.27\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2\sqrt{27.27}\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2\sqrt{162.9}}\\ &=60+2\sqrt{2}-2\sqrt{162+9-2\sqrt{162.9}}\\ &=60+2\sqrt{2}-2\left ( \sqrt{162}-\sqrt{9} \right )\\ &=60+2\sqrt{2}-2\left ( 9\sqrt{2}-3 \right )\\ x^{2}&=66-16\sqrt{2}\\ x&=\sqrt{66-2.8\sqrt{2}}\\ &=\sqrt{64+2-2\sqrt{64.2}}\\ &=\sqrt{64}-\sqrt{2}\\ &=8-\sqrt{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 05.&\textrm{Jika}\: \: \displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\displaystyle \frac{a\sqrt{2}+b\sqrt{3}+c\sqrt{5}}{12}\: ,\\\\ &\textrm{maka}\: \: a+b+c=....\\ &\qquad\qquad\qquad\qquad (\textbf{UM UGM 2016 Mat Das})\\ &\begin{array}{llllllll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\ \textrm{c}.&2\\ \textrm{d}.&3\\ \textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\qquad \textbf{e}\\ &\begin{aligned}\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}&=\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\times \displaystyle \frac{\left ( \sqrt{2}+\sqrt{3}-\sqrt{5} \right )}{\left ( \sqrt{2}+\sqrt{3}-\sqrt{5} \right )}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\left ( \sqrt{2}+\sqrt{3} \right )^{2}-\left ( \sqrt{5} \right )^{2}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2\sqrt{2.3}-5}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}\times \displaystyle \frac{\sqrt{6}}{\sqrt{6}}\\ &=\displaystyle \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}\\ &=\displaystyle \frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}\\ &=\displaystyle \frac{3\sqrt{2}+2\sqrt{3}-\sqrt{30}}{12}\\ &\quad \begin{cases} a &=3 \\ b &=2 \\ c &=-1 \end{cases}\\ a+b+c&=3+2+(-1)\\ &=4 \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Baskoro, B.D. 2012. Aljabar dan Trigonometri Cespleng Olimpiade Matematika. Yogyakarta: BERLIAN
2. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bnadung: SEWU.
3. Kanginan, M., Terzalgi, Y. 2013. Matematika untuk SMA-MA/SMK Kelas X Wajib. Bandung: SEWU.
4. Tung, Khoe Yao. 2012. Pintar Matematika SMA Kelas XII IPA untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.

EKSPONEN (LANJUTAN 3)

D. PERSAMAAN EKSPONEN

Berikut bentuk persamaan eksponen yang sering digunakan terangkum dalam tabel berikut beserta cara penyelesaiannya

$\begin{array}{|c|l|l|}\hline \textbf{No}&\textbf{Persamaan Eksponen}&\textbf{Penyelesaian}\\\hline 1&a^{f(x)}=1,\: \: a>0,a\neq 1&f(x)=0\\\hline 2&a^{f(x)}=a^{p},\: \: a>0,a\neq 1&f(x)=p\\\hline 3&a^{f(x)}=a^{g(x)},\: \: a>0,a\neq 1&f(x)=g(x)\\\hline 4&a^{f(x)}=b^{f(x)},\: \: a>0,a\neq 1&f(x)=0\\ &\qquad\quad \textrm{dan}\: \: b>0,\: b\neq 1&\\\hline 5&h(x)^{f(x)}=h(x)^{g(x)}&\begin{aligned}(1)\: &f(x)=g(x)\\ (2)\: &h(x)=1\\ (3)\: &h(x)=0\\ &\textrm{dengan syarat}\\ &f(x)> 0\: \: \textrm{dan}\\ &g(x)> 0\\ (4)\: &h(x)=-1\\ &\textrm{dengan syarat}\\ &f(x)\: \textrm{dan}\: g(x)\\ &\textrm{keduanya}\\ &\textrm{genap atau}\\ &\textrm{keduanya}\\ &\textrm{ganjil}\\ &\textrm{atau}\\ &\textrm{dapat juga}\\ &\textrm{ditunjukkan}\\ &(-1)^{f(x)}=(-1)^{g(x)} \end{aligned}\\\hline 6&g(x)^{f(x)}=h(x)^{f(x)}&\begin{aligned}(1)\: &g(x)=h(x)\\ (2)\: &f(x)=0\\ &\textrm{dengan syarat}\\ &g(x)\neq 0\: \: \textrm{dan}\\ &h(x)\neq 0\\ \end{aligned}\\\hline 7&f(x)^{g(x)}=1&\begin{aligned}(1)\: &f(x)=1\\ (2)\: &f(x)=-1\\ &\textrm{dengan syarat}\\ &g(x)\: \: \textrm{genap}\\ (3)\: &g(x)=0\\ &\textrm{dengan syarat}\\ &f(x)\neq 0 \end{aligned}\\\hline 8&A\left ( a^{f(x)} \right )^{2}+B\left ( a^{f(x)} \right )+C=0&\begin{aligned}&\textrm{ubah}\: \: a^{f(x)}=y\\ &\textrm{sehingga}\\ &Ay^{2}+By+C=0\\ &\textrm{selanjutnya}\\ &\textrm{substitusikan}\\ &\textrm{nilai}\: \: y\: \: \textrm{ke}\\ &\textrm{persamaan}\\ &a^{f(x)}=y \end{aligned}\\\hline \end{array}$.

$\LARGE{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukan himpunan penyelesaian dari}\\ &\textrm{a}.\quad 2^{2x-2021}=1\\ &\textrm{b}.\quad \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}=1\\ &\textrm{c}.\quad \sqrt{2}^{2x-2021}=1\\\\  &\textbf{Jawab}:\\ &\begin{array}{|c|c|c|}\hline \textrm{a}&\textrm{b}&\textrm{c}\\\hline \begin{aligned} 2^{2x-2021}&=1\\ 2^{2x-2021}&=2^{0}\\ 2x-2021&=0\\ 2x&=2021\\ x&=\displaystyle \frac{2021}{2}\\ & \end{aligned}&\begin{aligned} \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}&=1\\ \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}&=\left ( \frac{1}{2} \right )^{0}\\ 2x-2021&=0\\ 2x&=2021\\ x&=\displaystyle \frac{2021}{2} \end{aligned}&\begin{aligned} \sqrt{2}^{2x-2021}&=1\\ \sqrt{2}^{2x-2021}&=\sqrt{2}^{0}\\ 2x-2021&=0\\ 2x&=2021\\ x&=\displaystyle \frac{2021}{2}\\ & \end{aligned}\\\hline \textbf{HP}=\left \{ \displaystyle \frac{2021}{2} \right \}&\textbf{HP}=\left \{ \displaystyle \frac{2021}{2} \right \}&\textbf{HP}=\left \{ \displaystyle \frac{2021}{2} \right \}\\\hline \end{array}\\  \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukan himpunan penyelesaian dari}\\ &\textrm{a}.\quad 2^{2x-2021}=128\\ &\textrm{b}.\quad \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}=128\\ &\textrm{c}.\quad \sqrt{2}^{2x-2021}=128\\\\  &\textbf{Jawab}:\\ &\begin{array}{|c|c|c|}\hline \textrm{a}&\textrm{b}&\textrm{c}\\\hline \begin{aligned} 2^{2x-2021}&=128\\ 2^{2x-2021}&=2^{7}\\ 2x-2021&=7\\ 2x=7&+2021\\ x=\displaystyle \frac{2028}{2}&=1014\\ & \end{aligned}&\begin{aligned} \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}&=128\\ \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}&=\left ( \displaystyle \frac{1}{2} \right )^{-7}\\ 2x-2021&=-7\\ 2x=2021&-7\\ x=\displaystyle \frac{2014}{2}&=1007 \end{aligned}&\begin{aligned} \sqrt{2}^{2x-2021}&=128\\ \sqrt{2}^{2x-2021}&=\sqrt{2}^{256}\\ 2x-2021&=256\\ 2x=2021&+256\\ x&=\displaystyle \frac{2277}{2}\\ & \end{aligned}\\\hline \textbf{HP}=\left \{ 1014 \right \}&\textbf{HP}=\left \{ \displaystyle 1007 \right \}&\textbf{HP}=\left \{ \displaystyle \frac{2277}{2} \right \}\\\hline \end{array}\\  \end{array}$.

$\begin{array}{ll}\\ 3.&(\textbf{SPMB 04})\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{27}{3^{2x-1}}=81^{-0,125} \: \: \textrm{adalah... .}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\displaystyle \frac{27}{3^{2x-1}}&=81^{-0,125}\\ 3^{3-(2x-1)}&=3^{4(\frac{1}{8})}\\ 3-2x+1&=-\displaystyle \frac{1}{2}\\ -2x+4&=-\displaystyle \frac{1}{2}\\ -x+2&=-\displaystyle \frac{1}{4}\\ -x&=-2-\displaystyle \frac{1}{4}\\ -x&=-2\displaystyle \frac{1}{4}\\ x&=2\displaystyle \frac{1}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&(\textbf{UMPTN 00})\\ &\textrm{Bentuk}\: \: \left (\sqrt[3]{\displaystyle \frac{1}{243}} \right )^{3x}=\left ( \displaystyle \frac{3}{3^{x-2}} \right )^{2}\sqrt[3]{\displaystyle \frac{1}{9}}\\ &\textrm{Jika}\: \: x_{0}\: \: \textrm{memenuhi persamaan, maka nilai}\\ &1-\displaystyle \frac{3}{4}x_{0}=\: ....\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\left (\sqrt[3]{\displaystyle \frac{1}{243}} \right )^{3x}&=\left ( \displaystyle \frac{3}{3^{x-2}} \right )^{2}\sqrt[3]{\displaystyle \frac{1}{9}}\\ 3^{-5x}&=3^{2(1-(x-2))}.3^{-\frac{2}{3}}\\ -5x&=2(1-(x-2))+\left ( -\displaystyle \frac{2}{3} \right ),\: \: \textrm{dikali}\: \: 3\\ -15x&=6(3-x)+(-2)\\ -15x&=18-6x-2\\ 6x-15x&=16\\ -9x&=16\\ x&=\displaystyle \frac{16}{-9}\\ x_{0}&=-\displaystyle \frac{16}{9},\: \: \textrm{selanjutnya}\\ 1-\displaystyle \frac{3}{4}x_{0}&=1-\displaystyle \frac{3}{4}\times \left (-\frac{16}{9} \right )\\ &=1+\frac{4}{3}\\ &=1+1\displaystyle \frac{1}{3}\\ &=2\displaystyle \frac{1}{3} \end{aligned} \end{array}$.

$\begin{array}{l}\\ 5.&\textrm{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\begin{aligned}5^{x+1}+5^{2-x}-30&=0\\ \left (5^{x} \right ).5^{1}+\displaystyle \frac{5^{2}}{5^{x}}-30&=0\\ 5\left ( 5^{x} \right )^{2}+25-30\left ( 5^{x} \right )&=0\\ \textrm{Persamaan kuadrat}&\: \textrm{dalam}\: \: 5^{x},\: \textrm{maka}\\ 5(5^{x})^{2}-30(5^{x})+25&=0\begin{cases} a & =5 \\ b & =-30 \\ c & =25 \end{cases}\\ (5^{x_{1}}).\left ( 5^{x_{2}} \right )&=\displaystyle \frac{c}{a}\\ 5^{x_{1}+x_{2}}&=\displaystyle \frac{25}{5}=5\\ 5^{x_{1}+x_{2}}&=5^{1}\\ x_{1}+x_{2}&=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Jumlah akar-akar persamaan}\\ &2023^{x^{2}-7x+7}=2024^{x^{2}-7x+7}\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\begin{aligned}2023^{x^{2}-7x+7}&=2024^{x^{2}-7x+7}\\ \textrm{Karena basis}&\: \textrm{tidak sama},\\ \textrm{maka harusl}&\textrm{ah pangkatnya}=0,\\ x^{2}-7x+7&=0\\ \textrm{dan jumlah}\: &\textrm{akar-akarnya adalah}:\\ x_{1}+x_{2}&=-\displaystyle \frac{b}{a}, \: \: \textrm{dari persamaan}\\ x^{2}-7x+7&=0\begin{cases} a &=1 \\ b &=-7 \\ c &=7 \end{cases}\\ \textrm{maka}\: \: x_{1}+x_{2}&=-\displaystyle \frac{b}{a}=-\frac{-7}{1}=7 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Tentukan himpunan penyelesaian dari}\\ &(x-2)^{x^{2}-7x+6}=1\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat bentuk}\: \: \: f(x)^{g(x)}=1\begin{cases} f(x) & =x-2 \\  g(x) & =x^{2}-7x+5 \end{cases}\\ &\begin{array}{|l|l|l|}\hline f(x)=1&f(x)=-1&g(x)=0\\ &\textrm{Syarat}\: \: g(x)\: \: \textrm{genap}&\textrm{Syarat}\: \: f(x)\neq 0\\\hline \begin{aligned}x&-2=1\\ x&=3\\ & \end{aligned}&\begin{aligned}x&-2=-1\\ x&=2-1=1\\ & \end{aligned}&\begin{aligned}x^{2}&-7x+6=0\\ \Leftrightarrow &(x-1)(x-6)\\ \Leftrightarrow &\: x=1\: \: \textrm{atau}\: \: x=6 \end{aligned}\\\hline &\begin{aligned}\textrm{S}&\textrm{yaratnya}\: \: x\\ \textrm{u}&\textrm{ntuk}\: \: x=1\\ g&(1)=1^{2}-7+6\\ &=0\: \: (\textrm{memenuhi}) \end{aligned}&\begin{aligned}f(1)&=1-2=-1\neq 0\\ f(6)&=6-2=4\neq 0\\ &\\ & \end{aligned}\\ &\textbf{Catatan}:\: 0&\\ &\textrm{paritasnya genap}&\\\hline \end{array}\\ &\textbf{HP}=\left \{ 1,3,6 \right \} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Tentukan himpunan penyelesaian dari}\\ &(x^{2}-9x+19)^{2x+3}=(x^{2}-9x+19)^{x-1}\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat bentuk}\: \: \: h(x)^{f(x)}=h(x)^{g(x)}\begin{cases} h(x) & =x^{2}-9x+19 \\  f(x) & =2x+3\\ g(x)&=x-1 \end{cases}\\ &\begin{array}{|l|l|}\hline \begin{aligned}&\textrm{Syarat-syaratnya}\\ &\bullet \: \: f(x)=g(x)\\ &\Leftrightarrow 2x+3=x-1\\ &\Leftrightarrow x=-4\\ &\bullet \: \: h(x)=1\\ &\Leftrightarrow x^{2}-9x+19=1\\ &\Leftrightarrow x^{2}-9x+18=0\\ &\Leftrightarrow (x-3)(x-6)=0\\ &\Leftrightarrow x=3\: \: \textrm{atau}\: \: x=6\\ &\bullet \: \: h(x)=0\\ &\Leftrightarrow x^{2}-9x+19=0\\ &\Leftrightarrow x_{1,2}=\displaystyle \frac{9\pm \sqrt{5}}{2}\\ &\quad \textrm{gunakan rumus ABC}\\ &\textrm{Setelah diuji keduanya}\\ &\textrm{positif, maka}\\ &x=\displaystyle \frac{9\pm \sqrt{5}}{2}\: \: \textrm{merupakan}\\ &\textbf{penyelesaian} \end{aligned} &\begin{aligned}&\textrm{lanjutannya}\\ &\bullet \: \: h(x)=-1\\ &\Leftrightarrow x^{2}-9x+19=-1\\ &\Leftrightarrow x^{2}-9x+20=0\\ &\Leftrightarrow (x-4)(x-5)=0\\ &\Leftrightarrow x=4\: \: \textrm{atau}\: \: x=5\\ &\textrm{Uji nilanya}\\ &\textrm{untuk}\: \: x=4\\ &\blacklozenge \: \: f(4)=2(4)+3\: \: \textrm{ganjil}\\ &\blacklozenge \: \: g(4)=4-1\: \: \textrm{ganjil}\\ &\textrm{karena}\: f(4),g(4)\: \textrm{keduanya }\\ &\textrm{ganjil, maka}\: \: x=4\\ &\textrm{adalah}\: \textbf{penyelesaian} \\ &\textrm{untuk}\: \: x=5\\ &\blacklozenge \: \: f(5)=2(5)+3\: \: \textrm{ganjil}\\ &\blacklozenge \: \: g(5)=5-1\: \: \textrm{genapl}\\ &\textrm{karena}\: f(4)\neq g(4),\: \textrm{maka}\: \: x=5\\ &\textrm{adalah}\: \textbf{bukan penyelesaian}\\ &\\  \end{aligned}\\\hline \end{array}\\ &\textbf{HP}=\left \{ -4,3,4,6,\displaystyle \frac{9-\sqrt{5}}{2},\frac{9+\sqrt{5}}{2} \right \} \end{array}$.

DAFTAR PUSTAKA

1. Kurnia, N, dkk. 2016. Jelajah Matematika I SMA Kelas X Peminatan MIPA. Jakarta: YUDHISTIRA.

EKSPONEN (LANJUTAN 2)

 C. 2. 2  Merasionalkan penyebut

Jika suatu pecahan penyebutnya mengandung bilangan irasional atau bentuk akar, maka penyebut ini dapat dibuat menjadi bilangan rasional. Perhatikanlah langkah berikut

$\begin{aligned}1.\quad&\displaystyle \frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\times \frac{\sqrt{b}}{\sqrt{b}}=\frac{a\sqrt{b}}{\left ( \sqrt{b^{2}} \right )}=\frac{a}{b}\sqrt{b}\\ 2.\quad&\displaystyle \frac{a}{\sqrt[3]{b}}=\frac{a}{\sqrt[3]{b}}\times \frac{\sqrt[3]{b^{2}}}{\sqrt[3]{b^{2}}}=\frac{a\sqrt[3]{b^{2}}}{\left ( \sqrt[3]{b^{3}} \right )}=\frac{a}{b}\sqrt[3]{b^{2}}\\ 3.\quad&\displaystyle \frac{a}{\sqrt[5]{b^{3}}}=\displaystyle \frac{a}{\sqrt[5]{b^{3}}}\times \frac{\sqrt[5]{b^{2}}}{\sqrt[5]{b^{2}}}=\frac{a\sqrt[5]{b^{2}}}{\sqrt[5]{b^{5}}}=\frac{a}{b}\sqrt[5]{b^{2}} \end{aligned}$

Merasionalkan di atas adalah contoh bebrapa contoh model merasionalkan jika berjenis tunggal tetapi jika nanti jenisnya lebih dari itu, maka perhatikanlah simulasi contoh berikut

$\begin{aligned}&\\ 1.\quad&\displaystyle \frac{c}{a+\sqrt{b}}=\frac{c}{a+\sqrt{b}}.\frac{a-\sqrt{b}}{a-\sqrt{b}}=\frac{c\left ( a-\sqrt{b} \right )}{a^{2}-b}\\ 2.\quad&\displaystyle \frac{c}{a-\sqrt{b}}=\frac{c}{a-\sqrt{b}}.\frac{a+\sqrt{b}}{a+\sqrt{b}}=\frac{c\left ( a+\sqrt{b} \right )}{a^{2}-b}\\ 3.\quad&\displaystyle \frac{c}{\sqrt{a}+\sqrt{b}}=\frac{c}{\sqrt{a}+\sqrt{b}}.\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{c\left ( \sqrt{a}-\sqrt{b} \right )}{a-b}\\ \end{aligned}$

Perhatikanlah simulasi contoh di atas, bentuk $a+\sqrt{b}$ memiliki bentuk sekawan (irasional juga) $a-\sqrt{b}$, demikian juga bentuk $\sqrt{a}+\sqrt{b}$ memiliki sekawan $\sqrt{a}-\sqrt{b}$. Disamping itu ada bentuk khusus yatu bentuk  $\sqrt[3]{a}+\sqrt[3]{b}$ memiliki bentuk sekawan $\sqrt[3]{a^{2}}-\sqrt[3]{ab}+\sqrt[3]{b^{2}}$.

$\LARGE{ CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Rasionalkanlah penyebut pecahan berikut}\\ &\textrm{dan serderhankanlah hasilnya}\\ &\textrm{a}.\quad \displaystyle \frac{2}{\sqrt{5}}\qquad\qquad \textrm{d}.\quad \displaystyle \frac{\sqrt{2}}{\sqrt{5}}\\ &\textrm{b}.\quad \displaystyle \frac{2}{5\sqrt{2}}\: \: \: \quad\quad\quad \textrm{e}.\quad \displaystyle \frac{p}{\sqrt{q}}\\ &\textrm{c}.\quad \displaystyle \frac{6}{3\sqrt{5}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\displaystyle \frac{2}{\sqrt{5}}=\displaystyle \frac{2}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\displaystyle \frac{2\sqrt{5}}{\sqrt{25}}=\displaystyle \frac{2}{5}\sqrt{5}\\ \textrm{b}.\quad&\displaystyle \frac{2}{5\sqrt{2}}=\displaystyle \frac{2}{5\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\displaystyle \frac{2\sqrt{2}}{5\sqrt{4}}=\frac{2\sqrt{2}}{5.2}=\displaystyle \frac{1}{5}\sqrt{2} \\ \textrm{c}.\quad&\displaystyle \frac{6}{3\sqrt{5}}=\displaystyle \frac{6}{3\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\displaystyle \frac{6\sqrt{5}}{3\sqrt{25}}=\frac{6\sqrt{5}}{3.5}=\displaystyle \frac{2}{5}\sqrt{5}\\ \textrm{d}.\quad &\displaystyle \frac{\sqrt{2}}{\sqrt{5}}=\displaystyle \frac{\sqrt{2}}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{10}}{\sqrt{25}}=\frac{\sqrt{10}}{5} =\displaystyle \frac{1}{5}\sqrt{10}\\ \textrm{e}.\quad &\displaystyle \frac{p}{\sqrt{q}}=\displaystyle \frac{p}{\sqrt{q}}\times \frac{\sqrt{q}}{\sqrt{q}}=\displaystyle \frac{p\sqrt{q}}{\sqrt{q^{2}}}=\frac{p\sqrt{q}}{q}=\displaystyle \frac{p}{q}\sqrt{q} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Rasionalkanlah penyebut pecahan berikut}\\ &\textrm{dan serderhankanlah hasilnya}\\ &\textrm{a}.\quad \displaystyle \frac{3}{6-\sqrt{5}}\qquad\qquad \textrm{f}.\quad \displaystyle \frac{3}{\sqrt{6}+\sqrt{5}}\\ &\textrm{b}.\quad \displaystyle \frac{3}{6+\sqrt{5}}\quad \quad\quad\quad \textrm{g}.\quad \frac{\sqrt{3}}{\sqrt{6}-\sqrt{5}}\\ &\textrm{c}.\quad \displaystyle \frac{\sqrt{3}}{6-\sqrt{5}}\qquad\qquad\textrm{h}.\quad \frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}\\ &\textrm{d}.\quad \displaystyle \frac{\sqrt{3}}{6+\sqrt{5}}\qquad\qquad\textrm{i}.\quad \frac{\sqrt{3}}{\sqrt{6-2\sqrt{5}}}\\ &\textrm{e}.\quad \displaystyle \frac{3}{\sqrt{6}-\sqrt{5}}\: \qquad\quad\textrm{j}.\quad \frac{\sqrt{3}}{\sqrt{6+2\sqrt{5}}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\displaystyle \frac{3}{6-\sqrt{5}}=\displaystyle \frac{3}{6-\sqrt{5}}\times \frac{6+\sqrt{5}}{6+\sqrt{5}}=\displaystyle \frac{3\left ( 6+\sqrt{5} \right )}{6^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{18+3\sqrt{5}}{36-5}=\frac{18+3\sqrt{5}}{31}=\displaystyle \frac{1}{31}\left ( 18+3\sqrt{5} \right )\\ \textrm{b}.\quad & \displaystyle \frac{3}{6+\sqrt{5}}=\displaystyle \frac{3}{6+\sqrt{5}}\times \frac{6-\sqrt{5}}{6-\sqrt{5}}=\displaystyle \frac{3\left ( 6-\sqrt{5} \right )}{6^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{18-3\sqrt{5}}{36-5}=\frac{18-3\sqrt{5}}{31}=\displaystyle \frac{1}{31}\left ( 18-3\sqrt{5} \right )\\ \textrm{c}.\quad &\displaystyle \frac{\sqrt{3}}{6-\sqrt{5}}=\displaystyle \frac{\sqrt{3}}{6-\sqrt{5}}\times \frac{6+\sqrt{5}}{6+\sqrt{5}}=\displaystyle \frac{\sqrt{3}\left ( 6+\sqrt{5} \right )}{6^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{6\sqrt{3}+\sqrt{15}}{36-5}=\frac{6\sqrt{3}+\sqrt{15}}{31}=\displaystyle \frac{1}{31}\left ( 6\sqrt{3}+\sqrt{15} \right )\\ \textrm{d}.\quad &\displaystyle \frac{\sqrt{3}}{6+\sqrt{5}}=\displaystyle \frac{\sqrt{3}}{6+\sqrt{5}}\times \frac{6-\sqrt{5}}{6-\sqrt{5}}=\displaystyle \frac{\sqrt{3}\left ( 6-\sqrt{5} \right )}{6^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{6\sqrt{3}-\sqrt{15}}{36-5}=\frac{6\sqrt{3}-\sqrt{15}}{31}=\displaystyle \frac{1}{31}\left ( 6\sqrt{3}-\sqrt{15} \right )\\ \textrm{e}.\quad &\displaystyle \frac{3}{\sqrt{6}-\sqrt{5}}=\displaystyle \frac{3}{\sqrt{6}-\sqrt{5}}\times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\displaystyle \frac{3\left ( \sqrt{6}+\sqrt{5} \right )}{\sqrt{6}^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{3\left ( \sqrt{6}+\sqrt{5} \right )}{6-5}=\frac{3\left ( \sqrt{6}+\sqrt{5} \right )}{1}=3\left ( \sqrt{6}+\sqrt{5} \right ) \\ \textrm{f}.\quad &\displaystyle \frac{3}{\sqrt{6}+\sqrt{5}}=\displaystyle \frac{3}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}=\displaystyle \frac{3\left ( \sqrt{6}-\sqrt{5} \right )}{\sqrt{6}^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{3\left ( \sqrt{6}-\sqrt{5} \right )}{6-5}=\frac{3\left ( \sqrt{6}-\sqrt{5} \right )}{1}=3\left ( \sqrt{6}-\sqrt{5} \right )\\ \textrm{g}.\quad &\displaystyle \frac{\sqrt{3}}{\sqrt{6}-\sqrt{5}}=\displaystyle \frac{\sqrt{3}}{\sqrt{6}-\sqrt{5}}\times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\displaystyle \frac{\sqrt{3}\left ( \sqrt{6}+\sqrt{5} \right )}{\sqrt{6}^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{\sqrt{18}+\sqrt{15}}{6-5}=\frac{\sqrt{9.2}+\sqrt{15}}{1}=\left ( 3\sqrt{2}+\sqrt{15} \right )\\ \textrm{h}.\quad &\displaystyle \frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}=\displaystyle \frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}=\displaystyle \frac{\sqrt{3}\left ( \sqrt{6}-\sqrt{5} \right )}{\sqrt{6}^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{\sqrt{18}-\sqrt{15}}{6-5}=\frac{\sqrt{9.2}-\sqrt{15}}{1}=\left ( 3\sqrt{2}-\sqrt{15} \right )\\ \textrm{i}.\quad &\frac{\sqrt{3}}{\sqrt{6-2\sqrt{5}}}=\frac{\sqrt{3}}{\sqrt{5+1-2\sqrt{5.1}}}=\displaystyle \frac{\sqrt{3}}{\sqrt{5}-\sqrt{1}}=\frac{\sqrt{3}}{\sqrt{5}-1}\\ &=\frac{\sqrt{3}}{\sqrt{5}-1}\times \frac{\sqrt{5}+1}{\sqrt{5}+1}=\displaystyle \frac{\sqrt{3.5}+\sqrt{3.1}}{\sqrt{5}^{2}-1^{2}}=\frac{\sqrt{15}+\sqrt{3}}{5-1}\\ &=\displaystyle \frac{1}{4}\left ( \sqrt{15}+\sqrt{3} \right )\\ \textrm{j}.\quad &\frac{\sqrt{3}}{\sqrt{6+2\sqrt{5}}}=\frac{\sqrt{3}}{\sqrt{5+1+2\sqrt{5.1}}}=\displaystyle \frac{\sqrt{3}}{\sqrt{5}+\sqrt{1}}=\frac{\sqrt{3}}{\sqrt{5}+1}\\ &=\frac{\sqrt{3}}{\sqrt{5}+1}\times \frac{\sqrt{5}-1}{\sqrt{5}-1}=\displaystyle \frac{\sqrt{3.5}-\sqrt{3.1}}{\sqrt{5}^{2}-1^{2}}=\frac{\sqrt{15}-\sqrt{3}}{5-1}\\ &=\displaystyle \frac{1}{4}\left ( \sqrt{15}-\sqrt{3} \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3&\textrm{Rasionalkan penyebut dan sederhanakanlah}\\ &\textrm{a}.\quad \displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}\\ &\textrm{b}.\quad\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}\\ &=\displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}\times \displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{\sqrt{2}+\sqrt{5}-\sqrt{7}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{\left (\sqrt{2}+\sqrt{5} \right )^{2}-\left (\sqrt{7} \right )^{2}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{(2+2\sqrt{10}+5)-7}=\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{2\sqrt{10}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{2\sqrt{10}}\times \displaystyle \frac{\sqrt{10}}{\sqrt{10}}=\displaystyle \frac{\sqrt{20}+\sqrt{50}-\sqrt{70}}{2\times 10}\\ &=\displaystyle \frac{2\sqrt{5}+5\sqrt{2}+\sqrt{70}}{20} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}\\ &=\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}\times \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{\left (\sqrt{2}+\sqrt{3} \right )^{2}-\left (\sqrt{5} \right )^{2}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{(2+2\sqrt{6}+3)-5}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2\sqrt{6}}\\ &=\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}\\ &=\displaystyle \frac{\sqrt{12}+\sqrt{18}+\sqrt{30}}{2\times 6}\\ &=\displaystyle \frac{2\sqrt{3}+3\sqrt{2}+\sqrt{30}}{12} \end{aligned} \end{array}$

EKSPONEN (LANJUTAN 1)

 $\Large\textrm{C.2  Operasi Bilangan Bentuk Akar}$.

C. 2. 1  Sifat-sifat yang berlaku pada operasi bilangan bentuk akar

$\begin{aligned}&\\ 1.\quad&a\sqrt[n]{c}+b\sqrt[n]{c}=\left ( a+b \right )\sqrt[n]{c}\\ 2.\quad&a\sqrt[n]{c}-b\sqrt[n]{c}=\left ( a-b \right )\sqrt[n]{c}\\ 3.\quad&\sqrt[n]{a}.\sqrt[n]{b}=\sqrt[n]{ab}\\ 4.\quad&\sqrt[n]{a^{n}}=a\\ 5.\quad&a\sqrt[n]{c} x b\sqrt[n]{d} = ab\sqrt[n]{cd}\\ 6.\quad&\frac{a\sqrt[n]{c}}{b\sqrt[n]{d}}=\frac{a}{b}.\sqrt[n]{\frac{c}{d}}\\ 7.\quad&\sqrt{\left ( a+b \right )+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}\\ 8.\quad&\sqrt{\left ( a+b \right )-2\sqrt{ab}}=\sqrt{a}-\sqrt{b} \end{aligned}$.

$\LARGE{ CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Sederhanakanlah bentuk akar berikut}\\ &\begin{array}{lllllll} \textrm{a}.&\sqrt{8}&\textrm{f}.&\sqrt[3]{16}&\textrm{k}.&\sqrt{8x^{5}},\: \: x\geq 0\\ \textrm{b}.&\sqrt{12}&\textrm{g}.&\sqrt[3]{32}&\textrm{l}.&\sqrt{48x^{6}y^{11}},\: \: y\geq 0\\ \textrm{c}.&\sqrt{27}&\textrm{h}.&\sqrt[3]{54}&\textrm{m}.&2\sqrt{8}\times \sqrt{3}\\ \textrm{d}.&\sqrt{28}&\textrm{i}.&\sqrt[3]{81}&\textrm{n}.&3\sqrt{6}\times 2\sqrt{2}\\ \textrm{e}.&\sqrt{32}&\textrm{j}.&\sqrt[3]{625}&\textrm{o}.&2\sqrt[3]{6}\times 6\sqrt[3]{9} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{lllllll}\\ \textrm{a}.&\sqrt{8}=\sqrt{4\times 2}=\sqrt{2^{2}}\times \sqrt{2}=2\sqrt{2}\\ \textrm{b}.&\sqrt{12}=\sqrt{4\times 3}=\sqrt{2^{2}}\times \sqrt{3}=2\sqrt{3}\\ \textrm{c}.&\sqrt{27}=\sqrt{9\times 3}=\sqrt{3^{2}}\times \sqrt{3}=3\sqrt{3}\\ \textrm{d}.&\sqrt{28}=\sqrt{4\times 7}=\sqrt{2^{2}}\times \sqrt{7}=2\sqrt{7}\\ \textrm{e}.&\sqrt{32}=\sqrt{16\times 2}=\sqrt{4^{2}}\times \sqrt{2}=4\sqrt{2}\\ \textrm{f}.&\sqrt[3]{16}=\sqrt[3]{8\times 2}=\sqrt[3]{2^{3}}\times \sqrt[3]{2}=2\sqrt[3]{2}\\ \textrm{g}.&\sqrt[3]{32}=\sqrt[3]{8\times 4}=\sqrt[3]{2^{3}}\times \sqrt[3]{4}=2\sqrt[3]{4}\\ \textrm{h}.&\sqrt[3]{54}=\sqrt[3]{27\times 2}=\sqrt[3]{3^{3}}\times \sqrt[3]{2}=3\sqrt[3]{2}\\ \textrm{i}.&\sqrt[3]{81}=\sqrt[3]{27\times 3}=\sqrt[3]{3^{3}}\times \sqrt[3]{3}=3\sqrt[3]{3}\\ \textrm{j}.&\sqrt[3]{625}=\sqrt[3]{125\times 5}=\sqrt[3]{5^{3}}\times \sqrt[3]{5}=5\sqrt[3]{5}\\ \textrm{k}.&\sqrt{8x^{5}}=\sqrt{4.2.x^{4}.x^{1}}=\sqrt{2^{2}}\times \sqrt{2}\times \sqrt{x^{4}}\times \sqrt{x}\\ &\quad\quad \: \: \: =2.\sqrt{2}.x^{2}.\sqrt{x}=2x^{2}\sqrt{2x},\: \: \: x\geq 0\\ \textrm{l}.&\sqrt{48x^{6}y^{11}}=\sqrt{16.3.x^{6}.y^{10}.y^{1}}\\ &\quad\quad \: \: \: =\sqrt{4^{2}}\times \sqrt{3}\times \sqrt{x^{6}}\times \sqrt{y^{10}}\times \sqrt{y}\\ &\quad\quad \: \: \: =4\sqrt{3}.x^{3}.y^{5}.\sqrt{y}=4x^{3}y^{5}\sqrt{3y},\: \: \: \geq 0\\ \textrm{m}.&2\sqrt{8}\times \sqrt{3}=2\sqrt{4\times 2}\times \sqrt{3}\\ &\quad\quad \: \: \: =2\sqrt{2^{2}}\times \sqrt{2}\times \sqrt{3}=2.2.\sqrt{2.3}\\ &\quad\quad \: \: \: =4\sqrt{6}\\ \textrm{n}.&3\sqrt{6}\times 2\sqrt{2}=3\sqrt{2\times 3}\times 2\sqrt{2}\\ &\quad\quad \: \: \: =3\times 2\times \sqrt{2^{2}\times 3}=6\times \sqrt{2^{2}}\times \sqrt{3}\\ &\quad\quad \: \: \: =6\times 2\times \sqrt{3}=12\sqrt{3}\sqrt{6}\\ \textrm{o}.&2\sqrt[3]{6}\times 6\sqrt[3]{9}=2.6.\sqrt[3]{6\times 9}=12\times \sqrt[3]{2.3.3.3}\\ &\quad\quad \: \: \: =12\times \sqrt[3]{2.3^{3}}=12\times \sqrt[3]{2}\times \sqrt[3]{3^{3}}\\ &\quad\quad \: \: \: =12\times \sqrt[3]{2}\times 3\\ &\quad\quad \: \: \: =36\sqrt[3]{2} \end{array} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah pangkat rasional dari}\\ &\textrm{a}.\quad \sqrt{y\sqrt[3]{x^{2}y}}\\ &\textrm{b}.\quad \sqrt[3]{x^{3}\sqrt[5]{x^{3}\sqrt{x^{3}}}}\\ &\textrm{c}.\quad \sqrt[3]{x^{2}\sqrt{x\sqrt[5]{x^{2}}}}\\ &\textrm{d}.\quad xyz\sqrt[3]{\displaystyle \frac{xy}{z^{5}}}\sqrt[3]{\displaystyle \frac{xz}{y^{5}}}\sqrt[3]{\displaystyle \frac{yz}{x^{5}}}\\\\ &\textrm{Jawab}:\\ &\begin{array}{lllllll}\\ \textrm{a}.&\sqrt{y\sqrt[3]{x^{2}y}}=\sqrt{y\left ( x^{2}y \right )^{\frac{1}{3}}}=\left ( y\left ( x^{2}y \right )^{\frac{1}{3}} \right )^{\frac{1}{2}}\\ &\quad\quad \: \: \: =y^{.^{\frac{1}{2}}}.x^{.^{\frac{2}{3}.\frac{1}{2}}}.y^{.^{\frac{1}{3}.\frac{1}{2}}}=y^{.^{\frac{1}{2}+\frac{1}{6}}}x^{.^{\frac{1}{3}}}=x^{.^{\frac{1}{3}}}.y^{.^{\frac{4}{6}}}\\ &\quad\quad \: \: \: =x^{.^{\frac{1}{3}}}.y^{.^{\frac{2}{3}}}\\ \textrm{b}.&\sqrt[3]{x^{3}\sqrt[5]{x^{3}\sqrt{x^{3}}}}=\sqrt[3]{x^{3}\sqrt[5]{x^{3}.x^{.^{\frac{3}{2}}}}}=\sqrt[3]{x^{3}.x^{.^{\frac{3}{5}}}x^{.^{\frac{3}{2.5}}}}\\ &\quad\quad \: \: \: =x^{.^{\frac{3}{3}}}.x^{.^{\frac{3}{5.3}}}.x^{.^{\frac{3}{2.5.3}}}=x^{1}+x^{.^{\frac{1}{5}}}.x^{.^{\frac{1}{10}}}\\ &\quad\quad \: \: \: =x^{.^{1+\frac{1}{5}+\frac{1}{10}}}=x^{.^{\frac{10+2+1}{10}}}=x^{.^{\frac{13}{10}}}\\ \textrm{c}.&\sqrt[3]{x^{2}\sqrt{x\sqrt[5]{x^{2}}}}=\sqrt[3]{x^{2}\sqrt{x.x^{.^{\frac{2}{5}}}}}=\sqrt[3]{x^{2}.x^{.^{\frac{1}{2}}}.x^{.^{\frac{2}{5.2}}}}\\ &\quad\quad \: \: \: =x^{.^{\frac{2}{3}}}.x^{.^{\frac{1}{2.3}}}.x^{.^{\frac{2}{5.2.3}}}=x^{.^{\frac{2}{3}}}.x^{.^{\frac{1}{6}}}.x^{.^{\frac{1}{15}}}\\ &\quad\quad \: \: \: =x^{.^{\frac{20+5+2}{30}}}=x^{.^{\frac{27}{30}}}=x^{.^{\frac{9}{10}}}\\ \textrm{d}.&xyz\sqrt[3]{\displaystyle \frac{xy}{z^{5}}}\sqrt[3]{\displaystyle \frac{xz}{y^{5}}}\sqrt[3]{\displaystyle \frac{yz}{x^{5}}}=xyz\sqrt[3]{\displaystyle \frac{x^{2}y^{2}z^{2}}{x^{5}y^{5}z^{5}}}\\ &\quad\quad \: \: \: =xyz\sqrt[3]{\displaystyle \frac{1}{x^{(5-2)}y^{(5-2)}z^{(5-2)}}}\\ &\quad\quad \: \: \: =xyz\sqrt[3]{\displaystyle \frac{1}{x^{3}y^{3}z^{3}}}=xyz\sqrt[3]{\displaystyle \frac{1}{(xyz)^{3}}}\\ &\quad\quad \: \: \: =xyz.\displaystyle \frac{1}{xyz}=\displaystyle \frac{xyz}{xyz}=1 \end{array} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: a,\: b\: \: \textrm{bilangan positif dan}\\ &\sqrt{a^{2}b\sqrt[3]{ab^{2}\sqrt{ab}}}=a^{x}.b^{y},\: \: \textrm{tentukan nilai}\: \: x-y\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\sqrt{a^{2}b\sqrt[3]{ab^{2}\sqrt{ab}}}=a^{x}.b^{y}\\ &\textrm{perhatikan cara menguraikannya}\\ &\sqrt{a^{2}b\sqrt[3]{ab^{2}\sqrt{ab}}}=\sqrt{a^{2}b\sqrt[3]{ab^{2}.a^{.^{\frac{1}{2}}}b^{.^{\frac{1}{2}}}}}\\ &=\sqrt{a^{2}b\sqrt[3]{a^{.^{1+\frac{1}{2}}}b^{.^{2+\frac{1}{2}}}}}=\sqrt{a^{2}b\sqrt[3]{a^{.^{\frac{3}{2}}}b^{.^{\frac{5}{2}}}}}\\ &=\sqrt{a^{2}b.a^{.^{\frac{3}{2.3}}}b^{.^{\frac{5}{2.3}}}}=\sqrt{a^{2}b.a^{.^{\frac{1}{2}}}b^{.^{\frac{5}{6}}}}\\ &=\sqrt{a^{.^{2+\frac{1}{2}}}.b^{.^{1+\frac{5}{6}}}}=\sqrt{a^{.^{\frac{5}{2}}}b^{.^{\frac{11}{6}}}}=a^{.^{\frac{5}{2.2}}}b^{.^{\frac{11}{6.2}}}\\ &=a^{.^{\frac{5}{4}}}b^{.^{\frac{11}{12}}}\\ &=a^{x}b^{y}\\ &\quad \textrm{maka}\: \: x=\displaystyle \frac{5}{4},\: \: \textrm{dan}\: \: y=\frac{11}{12}\\ &\quad x-y=\displaystyle \frac{5}{4}-\frac{11}{12}=\displaystyle \frac{15-11}{12}=\frac{4}{12}=\displaystyle \frac{1}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&(\textbf{Matematika Dasar UM UGM 2008})\\ &\textrm{Bentuk sederhana dari}\\ &\qquad\qquad \displaystyle \frac{\sqrt[6]{x^{2}}\sqrt[3]{x^{2}\sqrt{x+1}}}{x\sqrt[6]{x+1}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sqrt[6]{x^{2}}\sqrt[3]{x^{2}\sqrt{x+1}}}{x\sqrt[6]{x+1}}\\ &=\displaystyle \frac{\sqrt[6]{x^{2}}.\sqrt[3.2]{\left (x^{2}.\sqrt{x+1} \right )^{2}}}{\sqrt[6]{x^{6}}.\sqrt[6]{x+1}}=\displaystyle \frac{\sqrt[6]{x^{2}}.\sqrt[6]{x^{4}.(x+1)}}{\sqrt[6]{x^{6}(x+1)}}\\ &=\displaystyle \frac{\sqrt[6]{x^{(2+4)}(x+1)}}{\sqrt[6]{x^{6}(x+1)}}=\displaystyle \frac{\sqrt[6]{x^{6}(x+1)}}{\sqrt[6]{x^{6}(x+1)}}\\ &=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Nilai dari}\: \: \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}=....\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\\ &\textrm{berikut uraiannya}\\ &\textrm{Misalkan}\\ &x^{2}=x^{2},\quad \textrm{maka}\: \: \: x^{2}=1+\left ( x^{2}-1 \right )\\ &x^{2}=1+(x-1)(x+1)\\ &x^{2}=1+(x-1)\sqrt{(x+1)^{2}}\\ &x^{2}=1+(x-1)\sqrt{1+((x+1)^{2}-1)}\\ &x^{2}=1+(x-1)\sqrt{1+(x+1-1)(x+1+1)}\\ &x^{2}=1+(x-1)\sqrt{1+x(x+2)}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{(x+2)^{2}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+((x+2)^{2}-1)}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+2-1)(x+2+1)}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)(x+3)}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{(x+3)^{2}}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+((x+3)^{2}-1)}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+3-1)(x+3+1)}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)(x+4)}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{(x+4)^{2}}}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+((x+4)^{2}-1)}}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+4-1)(x+4+1)}}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)(x+5)}}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\sqrt{...}}}}}\\ &x=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\cdots }}}}}\\ &\textrm{maka}\\ &\cdots \: =\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\\ &\textrm{Jelas tampak bahwa nilai}\: \: x\: \: \textrm{yang memenuhi adalah}\\ &x=3 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Sederhanakanlah bentuk}\\ &\textrm{a}.\quad \sqrt{3+2\sqrt{2}}\qquad\textrm{d}.\quad \sqrt{21-4\sqrt{5}}\\ &\textrm{b}.\quad \sqrt{6-\sqrt{32}}\qquad\textrm{e}.\quad \sqrt{6-2\sqrt{8}}\\ &\textrm{c}.\quad \sqrt{7+4\sqrt{3}}\qquad\textrm{f}.\quad \sqrt{5+\sqrt{24}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Inga}&\textrm{t bahwa}:\quad \sqrt{a+b\pm 2\sqrt{ab}}=\sqrt{a}\pm \sqrt{b},\quad a\geq b\\ \textrm{a}.\quad&\sqrt{3+2\sqrt{2}}=\sqrt{2+1+2\sqrt{2.1}}=\sqrt{2}+1\\ \textrm{b}.\quad&\sqrt{6-\sqrt{32}}=\sqrt{6-\sqrt{4.4.2}}=\sqrt{4+2-2\sqrt{4.2}}\\ &=\sqrt{4}-\sqrt{2}=2-\sqrt{2}\\ \textrm{c}.\quad&\sqrt{7+4\sqrt{3}}=\sqrt{4+3+2.2\sqrt{3}}=\sqrt{4+3+2\sqrt{4.3}}\\ &=\sqrt{4}+\sqrt{3}=2+\sqrt{3}\\ \textrm{d}.\quad&\sqrt{21-4\sqrt{5}}=\sqrt{20+1-2.2\sqrt{5}}=\sqrt{20+1-2\sqrt{4.5}}\\ &=\sqrt{20+1-2\sqrt{20.1}}=\sqrt{20}-1=\sqrt{4.5}-1=2\sqrt{5}-1\\ \textrm{e}.\quad&\sqrt{6-2\sqrt{8}}=\sqrt{4+2-2\sqrt{4.2}}=\sqrt{4}-\sqrt{2}=2-\sqrt{2}\\ \textrm{f}.\quad&\sqrt{5+\sqrt{24}}=\sqrt{3+2+\sqrt{4.3.2}}=\sqrt{3+2+2\sqrt{3.2}}\\ &=\sqrt{3}+\sqrt{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Sederhanakan bentuk berikut}\\ &\textrm{a}.\quad \sqrt{0,3+\sqrt{0,08}}\\ &\textrm{b}.\quad \sqrt{94+2\sqrt{2013}}\\ &\textrm{c}.\quad \sqrt{17+4\sqrt{15}}=a\sqrt{3}+b\sqrt{5},\: \: \textrm{tentukan}\: \: b-a\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Inga}&\textrm{t bahwa}:\quad \sqrt{p+q\pm 2\sqrt{pq}}=\sqrt{p}\pm \sqrt{q},\quad p\geq q\\ \textrm{a}.\quad&\sqrt{0,3+\sqrt{0,08}}=\sqrt{0,3+\sqrt{4.(0,02)}}=\sqrt{0,3+2\sqrt{0,02}}\\ &=\sqrt{0,2+0,1+2\sqrt{(0,2).(0,1)}}=\sqrt{0,2}+\sqrt{0,1}\\ \textrm{b}.\quad&\sqrt{94+2\sqrt{2013}}=\sqrt{61+33+2\sqrt{61.33}}=\sqrt{61}+\sqrt{33}\\ \textrm{c}.\quad&\sqrt{17+4\sqrt{15}}=\sqrt{17+2.2\sqrt{15}}=\sqrt{17+2\sqrt{4.15}}\\ &=\sqrt{17+2\sqrt{60}}=\sqrt{12+5+2\sqrt{12.5}}=\sqrt{12}+\sqrt{5}\\ &=\sqrt{4.3}+\sqrt{1.5}=2\sqrt{3}+1\sqrt{5}=a\sqrt{3}+b\sqrt{5}\quad\begin{cases} a & =2 \\ b & = 1 \end{cases}\\ &\textrm{maka}\quad b-a=1-2=-1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Bentuk paling sederhana dari}\\ &\qquad\qquad \sqrt[4]{49-20\sqrt{6}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\sqrt[4]{49-20\sqrt{6}}\\ &=\sqrt{\sqrt{49-2.10\sqrt{6}}}=\sqrt{\sqrt{49-2\sqrt{100.6}}}\\ &=\sqrt{\sqrt{49-2\sqrt{600}}}=\sqrt{\sqrt{25+24-2\sqrt{25.24}}}\\ &=\sqrt{\sqrt{25}-\sqrt{24}}=\sqrt{5-\sqrt{24}}=\sqrt{5-\sqrt{4.6}}\\ &=\sqrt{5-2\sqrt{6}}=\sqrt{3+2-2\sqrt{3.2}}=\sqrt{3}-\sqrt{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Bentuk paling sederhana dari}\\ &\qquad\qquad \left (\sqrt{52+6\sqrt{43}} \right )^{3}-\left (\sqrt{52-6\sqrt{43}} \right )^{3}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Inga}&\textrm{tlah bentuk}:\quad (A-B)^{3}=A^{3}-B^{3}-3AB(A-B)\\ &\qquad\qquad\qquad \Leftrightarrow A^{3}-B^{3}=(A-B)^{3}+3AB(A-B)\\ \bullet \: \: &\sqrt{52+6\sqrt{43}}=\sqrt{52+2.3\sqrt{43}}=\sqrt{52+2\sqrt{43.9}}\\ &=\sqrt{43+9+2\sqrt{43.9}}=\sqrt{43}+\sqrt{9}=\sqrt{43}+3\\ \bullet \: \: &\sqrt{52-6\sqrt{43}}=\sqrt{52-2.3\sqrt{43}}=\sqrt{52-2\sqrt{43.9}}\\ &=\sqrt{43+9-2\sqrt{43.9}}=\sqrt{43}-\sqrt{9}=\sqrt{43}-3\\ &\textrm{misalkan}\: \: \: \begin{cases} A & =\sqrt{43}+3 \\ B & =\sqrt{43}-3 \end{cases}\\ &A^{3}-B^{3}=(A-B)^{3}+3AB(A-B)\\ &=\left (\sqrt{43}+3 -\left (\sqrt{43}-3 \right ) \right )^{3}+3(\sqrt{43}+3)(\sqrt{43}-3)(\sqrt{43}+3-\left (\sqrt{43}-3 \right )) \\ &=\left ( 6 \right )^{3}+3\left ( \sqrt{43}^{2}-3^{2} \right )\left ( 6 \right )\\ &=216+18\left ( 43-9 \right )\\ &=216+18.34\\ &=216+612\\ &=828 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Sederhanakanlah bentuk akar berikut}\\ &\begin{array}{lllllll} \textrm{a}.&3\sqrt{2}+5\sqrt{8}-\sqrt{32}\\ \textrm{b}.&5\sqrt{3}+5\sqrt{27}-2\sqrt{75}\\ \textrm{c}.&3\sqrt{50}-4\sqrt{32}-\sqrt{2}\\ \textrm{d}.&\left ( 2+\sqrt{2} \right )\left ( 4-\sqrt{2} \right )\\ \textrm{e}.&\left ( 3\sqrt{2}+\sqrt{3} \right )\left ( \sqrt{2}-2\sqrt{3} \right )\\ \textrm{f}.&\left ( 4\sqrt{3}-3\sqrt{5} \right )\left ( 2\sqrt{3}+\sqrt{5} \right ) \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{lllllll}\\ \textrm{a}.&3\sqrt{2}+5\sqrt{8}-\sqrt{32}\\ &\quad\quad \: \: \: =3\sqrt{2}+5\sqrt{4.2}-\sqrt{16.2}\\ &\quad\quad \: \: \: =3\sqrt{2}+5.2\sqrt{2}-4\sqrt{2}\\ &\quad\quad \: \: \: =(3+10-4)\sqrt{2}=9\sqrt{2}\\ \textrm{b}.&5\sqrt{3}+5\sqrt{27}-2\sqrt{75}\\ &\quad\quad \: \: \: =5\sqrt{3}+5\sqrt{9.3}-2\sqrt{25.3}\\ &\quad\quad \: \: \: =5\sqrt{3}+5.3\sqrt{3}-2.5\sqrt{3}\\ &\quad\quad \: \: \: =(5+15-10)\sqrt{3}=10\sqrt{3}\\ \textrm{c}.&3\sqrt{50}-4\sqrt{32}-\sqrt{2}\\ &\quad\quad \: \: \: =3\sqrt{25.2}-4\sqrt{16.2}-\sqrt{1.2}\\ &\quad\quad \: \: \: =3.5\sqrt{2}-4.4\sqrt{2}-1\sqrt{2}\\ &\quad\quad \: \: \: =(15-16-1)\sqrt{2}=-2\sqrt{2}\\ \textrm{d}.&\left ( 2+\sqrt{2} \right )\left ( 4-\sqrt{2} \right )\\ &\quad\quad =2.4-2.\sqrt{2}+4.\sqrt{2}-\sqrt{2.2}\\ &\quad\quad =8+(4-2)\sqrt{2}-2\\ &\quad\quad =6+2\sqrt{2}\\ \textrm{e}.&\left ( 3\sqrt{2}+\sqrt{3} \right )\left ( \sqrt{2}-2\sqrt{3} \right )\\ &\quad\quad =3\sqrt{2.2}-3.2.\sqrt{2.3}+\sqrt{3.2}-2\sqrt{3.3}\\ &\quad\quad =3.2-6\sqrt{6}+1\sqrt{6}-3.2\\ &\quad\quad = 6-6+(1-6)\sqrt{6}=-5\sqrt{6}\\ \textrm{f}.&\left ( 4\sqrt{3}-3\sqrt{5} \right )\left ( 2\sqrt{3}+\sqrt{5} \right )\\ &\quad\quad =4.2.\sqrt{3.3}+4\sqrt{3.5}-3.2.\sqrt{5.3}-3\sqrt{5.5}\\ &\quad\quad =8.3+4\sqrt{15}-6\sqrt{15}-3.5\\ &\quad\quad =24-15+(4-6)\sqrt{15}=9-2\sqrt{15} \end{array} \end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
2. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.

EKSPONEN

A. EKSPONEN

Eksponen (bilangan berpangkat) adalah bentuk perkalian berulang dari suatu bilangan dengan dirinya sendiri. Secara notasi 

Misalkan diketahui bahwa $a$ adalah suatu bilangan tidak nol dan $n$ adalah bilangan asli, maka bilangan ekponen atau bilangan berpangkat didefinisikan dengan:

$\LARGE a^{n}=\underset{n}{\underbrace{a\times a\times \times a\times ...\times a}}$

$\begin{aligned}\textrm{Bilangan}&:\\ a&\: \: \textrm{disebut basis atau bilangan pokok}\\ n&\: \: \textrm{disebut sebagai bilangan pangkat/eksponen} \end{aligned}$.

$\LARGE{ CONTOH SOAL}$.

$(1).\quad 3^{4}=3\times 3\times 3\times 3=81$

$(2).\quad 5^{4}=5\times 5\times 5\times 5=625$

$(3).\quad 2^{6}=2\times 2\times 2\times 2\times 2\times 2=64$

$(4).\quad 6^{7}=6\times 6\times 6\times 6\times 6\times 6\times 6=279936$

$(5).\quad (-3)^{3}=(-3)\times (-3)\times (-3)=-27$

$(6).\quad (-2)^{4}=(-2)\times (-2)\times (-2)\times (-2)=16$

$(7).\quad \left ( \displaystyle \frac{1}{5} \right )^{3}=\left ( \displaystyle \frac{1}{5} \right )\times \left ( \displaystyle \frac{1}{5} \right )\times \left ( \displaystyle \frac{1}{5} \right )= \displaystyle \frac{1}{125}$

$(8).\quad \left ( -\displaystyle \frac{1}{2} \right )^{3}=\left ( -\displaystyle \frac{1}{2} \right )\times \left (- \displaystyle \frac{1}{2} \right )\times \left ( -\displaystyle \frac{1}{2} \right )=- \displaystyle \frac{1}{8}$

$\Large\textrm{B. Sifat-Sifat Bilangan Pangkat Positif}$

$\begin{aligned}\\ 1.\quad&a^{m}.a^{n}=a^{m+n}\\ 2.\quad&a^{m}:a^{n}=a^{m-n}\\ 3.\quad&\left ( a^{m} \right )^{n}=a^{m.n},\: \: \textrm{syarat}\: \: a\neq 0\\ 4.\quad&\left ( ab \right )^{n}=a^{n}.b^{n}\\ 5.\quad&\left ( \frac{a}{b} \right )^{n}=\frac{a^{n}}{b^{n}},\: \: \textrm{syarat}\: \: b\neq 0 \end{aligned}$

Beberpa hal yang perlu diketahui juga, yaitu

$\begin{aligned}\\ 1.\quad&(a+b)^{2}=a^{2}+2ab+b^2\\ 2.\quad&(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}\\ 3.\quad&\left ( a+\frac{1}{a} \right )^{2}=a^{2}+2+\displaystyle \frac{1}{a^{2}},\: \: \textrm{syarat}\: \: a\neq 0\\ \end{aligned}$

$\LARGE{ CONTOH SOAL}$

$\begin{aligned}\\ (1).\quad&2^{6} \times 2^{4} \times 2^{7} = 2^{6+4+7}=2^{17}\\ (2).\quad&2^{5} \times 3^{5} \times 7^{5} = \left ( 2 . 3 . 7 \right )^{5}=\left ( 42 \right )^{5}\\ (3).\quad&\displaystyle \frac{a^{3}.a^{7}.a^{6}}{a^{9}}=\displaystyle \frac{a^{3+7+6}}{a^{9}}=\frac{a^{16}}{a^{9}}=a^{16-9}=a^{7},\: \: \textrm{syarat}\: \: a\neq 0\\ \end{aligned}$

$\begin{aligned}(4).\quad\displaystyle \frac{3^{7}.7^{3}.2}{\left ( 42 \right )^{3}}&=\frac{2^{1}.3^{7}.7^{3}}{\left ( 2.3.7 \right )^{3}}=\frac{2^{1}.3^{7}.7^{3}}{2^{3}.3^{3}.7^{3}}\\ &=2^{1-3}.3^{7-3}.7^{3-3}=2^{-2}.3^{4}.7^{0}\\ &=\frac{1}{2^{2}}.3^{4}.1=\frac{3^{4}}{2^{2}} \end{aligned}$

$\begin{aligned}(5).\quad\displaystyle \frac{2^{2025}+2^{2026}+2^{2027}}{7}&=\displaystyle \frac{1.2^{2025}+2^{1}.2^{2025}+2^{2}.2^{2025}}{7}\\ &=\frac{\left ( 1+2+4 \right ).2^{2025}}{7}\\ &=\frac{7.2^{2025}}{7}\\ &=2^{2025} \end{aligned}$

$\begin{aligned}(6)\quad \displaystyle \frac{\left ( 2^{n+2} \right )^{2}-2^{2}.2^{2n}}{2^{n}.2^{n+2}}&=\displaystyle \frac{2^{2(n+2)}-2^{2}.2^{2n}}{2^{n}.2^{n}.2^{2}}\\ &=\displaystyle \frac{2^{2n}.2^{2.2}-2^{2}.2^{2n}}{2^{n+n}.2^{2}}\\ &=\displaystyle \frac{2^{2n}(2^{4}-2^{2})}{2^{2n}.2^{2}}\\ &=\displaystyle \frac{(2^{4}-2^{2})}{2^{2}}=\frac{16-4}{4}\\ &=\displaystyle \frac{12}{4}=3 \end{aligned}$

$\LARGE\textrm{C. Bentuk Akar}$

Bilangan bentuk akar di sini adalah kebalikan dari bilangan bentuk pangkat. Bilangan bentuk akar selanjutnya disebut bilangan irasional. Sebagai contoh $\sqrt{2}$, $\sqrt{3}$, $\sqrt{8}$, $\sqrt[3]{3}$, $\sqrt[3]{4}$, $\sqrt[3]{7}$ dan tapi ingat $\sqrt{4}$ dan  $\sqrt[3]{8}$ serta  $\sqrt[3]{27}$ adalah bukan bentuk akar, karena nantinya akan menghasilkan masing-masing 2 dan 3 serta 3.

$\begin{aligned}&\\ 1.\quad&a^{ \frac{1}{n}}=\sqrt[n]{a}\\ 2.\quad&a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\\ 3.\quad&a^{\frac{1}{2}}=\sqrt[2]{a^{1}}=\sqrt{a} \end{aligned}$.

$\textrm{Cara membaca}$.

$\begin{aligned}1.\quad&\sqrt[n]{p}\: \: \: \textbf{dibaca}\: \: \: \textrm{akar pangkat n dari p}\\ 2.\quad&\sqrt[n]{p^{2}}\: \: \: \textbf{dibaca}\: \: \: \textrm{akar pangkat n dari p kuadrat}\\ 3.\quad&\sqrt[n]{p^{3}}\: \: \: \textbf{dibaca}\: \: \: \textrm{akar pangkat n dari p pangkat tiga}\\ 4.\quad&\sqrt{p}\: \: \: \textbf{dibaca}\: \: \: \textrm{akar dari p}\: \: \: \textrm{atau}\\ &\qquad\qquad\qquad \textrm{akar kuadrat dari p}\\ &\qquad\qquad\qquad \textrm{ingat bahwa}:\: \: \sqrt{p}=\sqrt[2]{p} \end{aligned}$.

$\begin{aligned}\textrm{Defini}&\textrm{si}\\ \textrm{Jika}\: &\: a\: \: \textrm{dan}\: \: b\: \: \textrm{bilangan real dan}\\ &n\: \: \textrm{bilangan bulat positif, maka}:\\ &a^{n}=b\Leftrightarrow \sqrt[n]{b}=a\\ \textrm{keter}&\textrm{angan}:\\ \sqrt[n]{b}&\quad \textrm{disebut}\: \: \textbf{akar (radikal)}\\ b&\quad \textrm{disebut}\: \: \textbf{radikan}\\ &\quad \textrm{(bilangan pokok yang ditarik akarnya)}\\ n&\quad \textrm{disebut}\: \: \textbf{indeks}\\ &\quad (\textrm{pangkat akar}) \end{aligned}$.

$\Large\textrm{C.1  Bilangan Pangkat Pecahan}$.

Operasi Bilangan pangkat pecahan sama dengan operasi pangkat bilangan bulat.

$\LARGE{ CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&a^{.^{\frac{1}{2}}}\times a^{.^{\frac{1}{3}}}=a^{.^{\frac{1}{2}+\frac{1}{3}}}=a^{.^{\frac{5}{6}}}\\ 2.&a^{.^{\frac{1}{5}}}: a^{.^{\frac{1}{3}}}=a^{.^{\frac{1}{5}-\frac{1}{3}}}=a^{.^{-\frac{2}{15}}}\\ 3.&\left (a^{.^{\frac{2}{5}}} \right )^{\frac{4}{7}}=a^{.^{\frac{8}{35}}}\\ 4.&81^{.^{\frac{1}{2}}}=\left ( 9^{2} \right )^{.^{\frac{1}{2}}}=9^{1}=9\\ 5.&27^{.^{-\frac{2}{3}}}=\left ( 3^{3} \right )^{.^{-\frac{2}{3}}}=\left (3 \right )^{-2}=\displaystyle \frac{1}{3^{2}}=\frac{1}{9} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Sederhanakanlah bentuk berikut dan}\\ &\textrm{nyatakan hasilnya dalam pangkat positif}\\\\ &\textrm{a}.\quad \left ( 3p^{.^{\frac{5}{3}}}q^{.^{-\frac{3}{4}}} \right )\left ( 2p^{.^{-\frac{2}{3}}}q^{.^{\frac{5}{4}}} \right )\\\\ &\textrm{b}.\quad \displaystyle \frac{\left ( 8p^{.^{\frac{2}{3}}}q^{0}r^{.^{-\frac{1}{2}}} \right )}{\left ( 4p^{.^{-\frac{1}{2}}}q^{.^{-\frac{1}{3}}}r \right )}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\left ( 3p^{.^{\frac{5}{3}}}q^{.^{-\frac{3}{4}}} \right )\left ( 2p^{.^{-\frac{2}{3}}}q^{.^{\frac{5}{4}}} \right )\\ &=3.2.p^{.^{\frac{5}{3}+\left ( -\frac{2}{3} \right )}}.q^{.^{-\frac{3}{4}+\frac{5}{4}}}\\ &=6.p^{.^{\frac{3}{3}}}q^{.^{\frac{2}{4}}}\\ &=6pq^{.^{\frac{1}{2}}} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\displaystyle \frac{\left ( 8p^{.^{\frac{2}{3}}}q^{0}r^{.^{-\frac{1}{2}}} \right )}{\left ( 4p^{.^{-\frac{1}{2}}}q^{.^{-\frac{1}{3}}}r \right )}\\ &=2.p^{.^{\frac{2}{3}-\left ( -\frac{1}{2} \right )}}q^{.^{0}-\left ( -\frac{1}{3} \right )}r^{.^{-\frac{1}{2}-1}}\\ &=2p^{.^{\frac{2}{3}+\frac{1}{2}}}q^{.^{\frac{1}{3}}}r^{.^{-\frac{3}{2}}}\\ &=2p^{.^{\frac{4+3}{6}}}q^{.^{\frac{1}{3}}}.r^{.^{-\frac{3}{2}}}\\ &=2p^{.^{\frac{7}{6}}}q^{.^{\frac{1}{3}}}.r^{.^{-\frac{3}{2}}}\\ &=\displaystyle \frac{2p^{.^{\frac{7}{6}}}q^{.^{\frac{1}{3}}}}{r^{.^{\frac{3}{2}}}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Sederhanakanlah bentuk berikut dan}\\ &\textrm{nyatakan hasilnya dalam pangkat positif}\\ &\textrm{a}.\quad \left ( \displaystyle \frac{p^{3n+1}q^{n}}{p^{3n+4}q^{4n}} \right )^{\frac{1}{3}}\\ &\textrm{b}.\quad \left ( \displaystyle \frac{p^{-2}q^{3}}{p^{4}q^{-3}} \right )^{-\frac{1}{2}}\left ( \displaystyle \frac{p^{4}q^{-5}}{pq} \right )^{-\frac{1}{3}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\left ( \displaystyle \frac{p^{3n+1}q^{n}}{p^{3n+4}q^{4n}} \right )^{\frac{1}{3}}\\ &=\left ( p^{(3n+1)-(3n+4)}q^{n-4n} \right )^{\frac{1}{3}}\\ &=\left ( p^{-3}q^{-3n} \right )^{\frac{1}{3}}\\ &=p^{-3.\frac{1}{3}}q^{-3n.\frac{1}{3}}\\ &=p^{-1}q^{-n}\\ &=\displaystyle \frac{1}{pq^{n}} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\left ( \displaystyle \frac{p^{-2}q^{3}}{p^{4}q^{-3}} \right )^{-\frac{1}{2}}\left ( \displaystyle \frac{p^{4}q^{-5}}{pq} \right )^{-\frac{1}{3}}\\ &=\left ( \displaystyle \frac{p^{-2.(-\frac{1}{2})}q^{3.(-\frac{1}{2})}}{p^{4.(-\frac{1}{2})}q^{-3.(-\frac{1}{2})}} \right )\left ( \displaystyle \frac{p^{4.(-\frac{1}{3})}q^{-5.(-\frac{1}{3})}}{p^{.^{-\frac{1}{3}}}q^{.^{-\frac{1}{3}}}} \right )\\ &=\displaystyle \frac{p^{1}q^{.^{-\frac{3}{2}}}}{p^{-2}q^{.^{\frac{3}{2}}}}\times \frac{p^{.^{-\frac{4}{3}}}q^{.^{\frac{5}{3}}}}{p^{.^{-\frac{1}{3}}}q^{.^{-\frac{1}{3}}}}\\ &=p^{1-(-2)+(-\frac{4}{3})-(-\frac{1}{3})}q^{-\frac{3}{2}-\frac{3}{2}+\frac{5}{3}-(-\frac{1}{3})}\\ &=p^{3-\frac{3}{3}}q^{-\frac{6}{2}+\frac{6}{3}}\\ &=p^{3-1}q^{-3+2}\\ &=p^{2}q^{-1}\\ &=\displaystyle \frac{p^{2}}{q} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Jabarkanlah bentuk}\\ &\qquad\qquad\quad \left ( 2m^{.^{\frac{3}{2}}}+n^{.^{\frac{3}{4}}} \right )^{2}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\left ( 2m^{.^{\frac{3}{2}}}+n^{.^{\frac{3}{4}}} \right )^{2}\\ &=\left ( 2m^{.^{\frac{3}{2}}} \right )^{2}+2\left ( 2m^{.^{\frac{3}{2}}} \right )\left ( n^{.^{\frac{3}{4}}} \right )+\left ( n^{.^{\frac{3}{4}}} \right )^{2}\\ &\textrm{INGAT}\: :\: \: \color{black}\left ( A+B \right )^{2}=A^{2}+2AB+B^{2}\\ &=2^{2}m^{.^{\frac{3.2}{2}}}+2.2.m^{.^{\frac{3}{2}}}n^{.^{\frac{3}{4}}}+n^{.^{\frac{3.2}{4}}}\\ &=4m^{3}+4m^{.^{\frac{3}{2}}}n^{.^{\frac{3}{4}}}+n^{.^{\frac{3}{2}}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jabarkanlah bentuk}\\ &\qquad\qquad\quad \left ( 2m^{.^{\frac{3}{2}}}-n^{.^{\frac{3}{4}}} \right )^{3}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\left ( 2m^{.^{\frac{3}{2}}}-n^{.^{\frac{3}{4}}} \right )^{3}\\ &=\left ( 2m^{.^{\frac{3}{2}}} \right )^{3}-3\left ( 2m^{.^{\frac{3}{2}}} \right )^{2}\left ( n^{.^{\frac{3}{4}}} \right )+3\left ( 2m^{.^{\frac{3}{2}}} \right )\left ( n^{.^{\frac{3}{4}}} \right )^{2}-\left ( n^{.^{\frac{3}{4}}} \right )^{3}\\ &\textrm{INGAT}\: :\: \: \color{black}\left ( A-B \right )^{3}=A^{3}-3A^{2}B+3AB^{2}-B^{3}\\ &=2^{3}m^{.^{\frac{3.3}{2}}}-3.2^{2}.m^{.^{\frac{3.2}{2}}}n^{.^{\frac{3}{4}}}+3.2.m^{.^{\frac{3}{2}}}n^{.^{\frac{3.2}{4}}}-n^{.^{\frac{3.3}{4}}}\\ &=8m^{.^{\frac{9}{2}}}-12m^{3}n^{.^{\frac{3}{4}}}+6m^{.^{\frac{3}{2}}}n^{.^{\frac{3}{2}}}-n^{.^{\frac{9}{2}}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Jabarkanlah bentuk berikut}\\\\ &\textrm{a}.\quad \left ( 2p^{.^{\frac{1}{2}}}-3q^{.^{\frac{1}{2}}} \right )\left ( p^{.^{\frac{1}{2}}}+4q^{.^{\frac{1}{2}}} \right )\\\\ &\textrm{b}.\quad \left ( p^{.^{\frac{1}{3}}}-q^{.^{\frac{1}{3}}} \right )\left ( p^{.^{\frac{2}{3}}}+p^{.^{\frac{1}{3}}}q^{.^{\frac{1}{3}}}+q^{.^{\frac{2}{3}}} \right )\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\left ( 2p^{.^{\frac{1}{2}}}-3q^{.^{\frac{1}{2}}} \right )\left ( p^{.^{\frac{1}{2}}}+4q^{.^{\frac{1}{2}}} \right )\\ &=2\left (p^{.^{\frac{1}{2}}} \right )^{2}+2.4.p^{.^{\frac{1}{2}}}q^{.^{\frac{1}{2}}}-3q^{.^{\frac{1}{2}}}.p^{.^{\frac{1}{2}}}-3.4.\left (q^{.^{\frac{1}{2}}} \right )^{2}\\ &=2p^{1}+8q^{.^{\frac{1}{2}}}q^{.^{\frac{1}{2}}}-3p^{.^{\frac{1}{2}}}q^{.^{\frac{1}{2}}}-12.q^{1}\\ &=2p+5(pq)^{.^{\frac{1}{2}}}-12q \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\left ( p^{.^{\frac{1}{3}}}-q^{.^{\frac{1}{3}}} \right )\left ( p^{.^{\frac{2}{3}}}+p^{.^{\frac{1}{3}}}q^{.^{\frac{1}{3}}}+q^{.^{\frac{2}{3}}} \right )\\ &=p^{.^{\frac{1+2}{3}}}+\left (p^{.^{\frac{1}{3}}} \right )^{2}q^{.^{\frac{1}{3}}}+p^{.^{\frac{1}{3}}}q^{.^{\frac{2}{3}}}-p^{.^{\frac{2}{3}}}q^{.^{\frac{1}{3}}}-p^{.^{\frac{1}{3}}}\left (q^{.^{\frac{1}{3}}} \right )^{2}-q^{.^{\frac{1+2}{3}}}\\ &=p^{1}+0+0-q^{1}\\ &=p-q \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas X kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.

Fungsi Kuadrat

 $\begin{array}{|l|l|}\hline \textrm{Pengertian}&\begin{aligned}&\textrm{Suatu fungsi yang berbentuk}\\ &f(x)=ax^{2}+bx+c\\ & a,\: b,\: c,\: \in \mathbb{R},\: a\neq 0 \end{aligned}\\\hline \textrm{Grafik Fungsi}&\textrm{Keterangan}\\\hline \textrm{Titik potong sumbu x}&\textrm{Jika ada}\\\hline &\begin{aligned}&\textrm{untuk titik potong}\\ &\textrm{terhadap sumbu x }\\ &\textrm{Jika y = 0 maka }\\ &ax^{2}+bx+c=0\\ &\textrm{Selanjutnya tinggal}\\ &\textrm{menentukan nilai D}\\ &D=b^{2}-4ac\: \: \textrm{adalah}\\ &\: \: \: \: \: \: \: \: \: \textrm{nilai diskriminan}.\\ &\textrm{Jika} \: D>0\\ &\textrm{maka grafik}\\ &\textrm{memotong sumbu x}\\ &\textrm{di dua tempat berbeda}\\ &\textrm{yaitu di} \: (x_{1},0)\: \textrm{dan}\: (x_{2},0).\\ &\textrm{dan jika D = 0}\\ &\textrm{maka grafik}\\ &\textrm{ hanya menyinggung}\\ &\textrm{sumbu x di satu titik}\\ &\textrm{yaitu di }\: (x_{1},0)\\ &\textrm{dan jika}\: D<0 \\ &\textrm{maka grafik}\\ &\textrm{tidak memotong}\\ &\textrm{atau menyinggung sumbu x} \end{aligned}\\\hline \textrm{Titik potong sumbu y}&\begin{aligned}&\textrm{titik potong terhadap}\\ &\textrm{sumbu y, jika x = 0}\\ &y=f(x)=ax^{2}+bx+c\\ &y=f(0)=a(0)^{2}+b(0)+c\\ &y=c \end{aligned}\\\hline \textrm{Sumbu Simetri (SS)}&x=\displaystyle \frac{-b}{2a}\\\hline \textrm{Titik Puncak}&\left ( \displaystyle \frac{-b}{2a},\displaystyle \frac{D}{-4a} \right )\\\hline \textrm{Posisi grafik}&\textrm{Jika}\: a>0\: \textrm{maka}\\ &\textrm{grafik terbuka ke atas}\\ &\textrm{Dan jika nilai}\: a<0\: \textrm{maka}\\ &\textrm{grafik terbuka ke bawah}\\\hline \end{array}$.

Selanjutnya cara membuat grafik fungsi kudratnya adalah sebagai berikut:

$\begin{array}{|c|c|}\hline \textrm{Jika memotong sumbu}-\textrm{X}&\textrm{Jika menyinggung sumbu}-\textrm{X}\\ \textrm{di titik}\: \left ( x_{1},0 \right )\: \textrm{dan}\: \left ( x_{2},0 \right )&\textrm{di titik}\: \left ( x_{1},0 \right )\: \textrm{dan melalui}\\ \textrm{dan melalui sebuah titik lain}&\textrm{sebuah titik lain} \\\hline &\\ y=f(x)=a\left ( x-x_{1} \right )\left ( x-x_{2} \right )&y=f(x)=a\left ( x-x_{1} \right )^{2}\\ &\\\hline \textrm{Jika grafik fungsi itu melalui}&\textrm{Jika grafik fungsi itu melalui}\\\hline \textrm{Titik puncak}\: \: P\left ( x_{p},y_{p} \right )\: \textrm{dan}&\textrm{tiga buah titik yaitu}\: \left ( x_{1},y_{1} \right )\\ \textrm{sebuah titik lain}&\left ( x_{2},y_{2} \right )\: \: \textrm{dan}\: \: \left ( x_{3},y_{3} \right )\\\hline &\\ y=f(x)=a\left ( x-x_{p} \right )^{2}+y_{p}&y=f(x)=ax^{2}+bx+c\\ &\\\hline \end{array}$.

$\LARGE{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika}\: \: f\: \: \textrm{adalah fungsi linear dengan}\\ & f(2)-f(-2)=8,\\ & \textrm{maka nilai dari}\: \: f(4)-f(-2)\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\\ &f(x)=ax+b\\ &f(2)-f(-2)\\ &=\left (a(2)+b \right )-\left ( a(-2)+b \right )=8\\ &8=2a+2a\\ &8=4a\\ &2=a\\ &f(x)=2x+b,\quad \textrm{dengan}\: \: b\: \: \textrm{konstan}\\ &\textrm{Sehingga nilai}\quad\\ &f(4)-f(-2)=\left (2(4)+b \right )-\left (2(-2)+b \right )\\ &=8+b+4-b\\ &=12 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Ubahlah}\: \: 8-6x-x^{2}\: \: \textrm{ke dalam bentuk}\\ & a-(x+b)^{2},\: \textrm{selanjutnya tentukan}\\ & \textrm{daerah hasil dari}\: \: f(x)=8-6x-x^{2}\\ & \textrm{untuk}\: \: x\: \: \textrm{bilangan real}\\ &\qquad(\textit{NTU Entrance Examination AO-level})\\\\ &\textbf{Jawab}:\\ &\begin{array}{|c|l|}\hline 1.&\textrm{Diketahui}\\\hline &\begin{aligned}\textrm{Misal}\quad\qquad&\\ 8-6x-x^{2}&=f(x)\\ f(x)&=-x^{2}-6x+8\\ &=-\left ( x^{2}+6x-8 \right )\\ &=-\left ( x^{2}+6x+9-17 \right )\\ &=-\left ( (x+3)^{2}-17 \right )\\ &=-(x+3)^{2}+17\\ & \end{aligned}\\\hline 2.&\textrm{Mencari koordinat}\: \: \left ( x_{SS},y_{SS} \right )\\\hline &\begin{aligned}f(x)&=-x^{2}-6x+8\left\{\begin{matrix} a=-1\\ b=-6\\ c=\: \: 8\: \: \end{matrix}\right.\\ \textrm{Maka}&\\ x_{SS}&=\frac{-b}{2a}=\displaystyle \frac{-(-6)}{2(-1)}\\ &=-3\\ y_{SS}&=f(-3)=-\left ( -3+3 \right )^{2}+17=17\\ \therefore &\left ( x_{SS},y_{SS} \right )=(-3,17) \end{aligned}\\\hline 3.&\textrm{Nilai fungsi}\\\hline &\begin{aligned}\textrm{Karena}&\: \: a=-1<0\\ \textrm{maka f}&\textrm{ungsi menghadap}\\ \textbf{ke ba}&\textbf{wah},\: \: \textrm{sehingga}\\ \textrm{daerah}&\: \: \textrm{hasilnya}\: \: \left (R_{f} \right )\\ \textrm{adalah}&:\\ &\left \{ -\infty <y\leq 17 \right \}\\ &\\ &\textrm{Berikut ilustrasinya} \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{adalah akar-akar dari }\\ &\textrm{persamaan kuadrat}\: \: x^{2}+mx+m=0,\\ &\textrm{maka nilai}\: \: m\: \: \textrm{yang menyebabkan }\\ &\textrm{jumlah kuadrat akar-akar mencapai}\\ &\textrm{minimum adalah}\: ....\\ &\qquad \: \textbf{(UM UNDIP 2014 Mat Das)}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \: x^{2}+mx+m=0\\ & \textbf{persamaan kuadrat}\: \textrm{dalam}\: \: x,\\ & \textrm{maka}\\ &x^{2}+mx+m=x^{2}-(\alpha +\beta )x+(\alpha \beta )=0\\ &\begin{cases} \alpha +\beta &=-m \\ & \\ \alpha \beta &=m \end{cases}\\ &\textrm{Selanjutnya}\\ &\alpha ^{2}+\beta ^{2}=\left ( \alpha +\beta \right )^{2}-2\alpha \beta\\ &=(-m)^{2}-2m\: \: \textrm{dan dapat kita tuliskan sebagai}\\ &f(m)=m^{2}-2m\begin{cases} a &=1 \\ b &=-2 \\ c &=0 \end{cases} \\ &\textrm{fungsi kuadrat dalam}\: \: m,\\ &\textrm{sehingga kita perlu mencari titik}\: \: \left ( m_{SS},f\left ( m_{SS} \right ) \right ),\\ & \textrm{tetapi yang kita perlukan}\\ &\textrm{cuma}\: \: m-\textrm{nya saja, yaitu}:\: \: m=m_{SS},\\ &\textrm{dengan}\quad m_{SS}=\displaystyle \frac{-b}{2a}=\frac{-(-2)}{2.1}=1 \end{aligned} \end{array}$.

Link Materi

https://ahmadthohir1098.blogspot.com/2024/11/kumpulan-materi-matematika-ma-sma-kelas.html

Sistem Pertidaksamaan Linear

 B. Sistem Pertidaksamaan Linear

$\begin{array}{ll}\\ &\textbf{BENTUK UMUM}\\ &\begin{cases} ax+by<c \\ ax+by\leq c \\ ax+by>c \\ ax+by\geq c \end{cases}\\\\ &\textbf{LANGKAH-LANGKAH}\\ &\textrm{dalam membuat gambar grafik persamaan linear}\\ &\: \textrm{adalah sebagai berikut}:\\ &\bullet\quad \textrm{membuat gambar grafik}\: \: ax+by=c\\ &\quad \: \: \textrm{untuk batas wilayahnya}\\ &\bullet \quad \textrm{menyelidiki wilayah yang dimaksud di sekitar}\\ &\quad \: \: \textrm{garis} \: \: ax+by=c\\ &\bullet \quad \textrm{ambillah sebuah titik}\: \left ( x_{0},y_{0} \right )\: \textrm{sembarang}\\ &\: \: \quad \textrm{kemudian substitusikan ke pertidaksamaan}\\ &\quad \: \: ax+by\: ....\: c\\ &\bullet \quad \textrm{jika diperoleh nilai ketaksamaan yang benar},\\ &\: \: \quad \textrm{maka daerah di mana titik uji}\: \left ( x_{0},y_{0} \right )\\ &\: \: \quad \textrm{berada merupakan wilayah penyelesaiannya}\\ &\: \: \quad \textrm{demikian juga sebaliknya} \end{array}$

$\LARGE\fbox{CONTOH SOAL}$

$\begin{array}{l}\\ 1.&\textrm{Gambarlah himpunan penyelesaian (HP)}\\ &\textrm{dari pertidaksamaan linear berikut}\\ &\textrm{a}.\quad 3x+2y< 6\\ &\textrm{b}.\quad 3x+2y\leq 6\\ &\textrm{c}.\quad 3x+2y> 6\\ &\textrm{d}.\quad 3x+2y\geq 6\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Mula}-\textrm{mula kita gambar garis}\: \: 3x+2y=6\\\\ &\begin{array}{|c|c|c|}\hline \textrm{Komponen}&\textrm{pada}&\textrm{pada}\\ \textrm{titik}&\textrm{sumbu}-y&\textrm{sumbu}-x\\\hline x&0&2\\\hline y&3&0\\\hline (x,y)&(0,3)&(2,0)\\\hline \end{array}\\\\ &\textrm{Selanjutnya gambar grafiknya sebagai berikut}. \end{aligned} \end{array}$

Dan berikut untuk wilayah dan juga batas-batas untuk pertidalsamaan

$3x+2y<6$

Kita dapat menggunakan titik uji untuk memastikan kondisi gambar di atas, yaitu di antaranya

$\begin{array}{|c|c|c|}\hline \textrm{Titik}&\textrm{Pengujian}&\textrm{Keterangan}\\ &\textrm{Uji}&3x+2y<6\\\hline (0,0)&3(0)+2(0)=0<\textbf{6}&\textrm{Dalam wilayah}\\\hline (0,1)&3(0)+2(1)=2<\textbf{6}&\textrm{Dalam wilayah}\\\hline (1,0)&3(1)+2(0)=3<\textbf{6}&\textrm{Dalam wilayah}\\\hline (1,1)&3(1)+2(1)=5<\textbf{6}&\textrm{Dalam wilayah}\\\hline (0,2)&3(0)+2(2)=4<\textbf{6}&\textrm{Dalam wilayah}\\\hline (2,0)&3(2)+2(0)=6=\textbf{6}&\textrm{Di luar wilayah}\\\hline (2,2)&3(2)+2(2)=10>\textbf{6}&\textrm{Di luar wilayah}\\\hline (0,3)&3(0)+2(3)=6=\textbf{6}&\textrm{Di luar wilayah}\\\hline (3,0)&3(3)+2(0)=9>\textbf{6}&\textrm{Di luar wilayah}\\\hline (3,3)&3(3)+2(3)=15>\textbf{6}&\textrm{Di luar wilayah}\\\hline \vdots &\vdots&\vdots \\\hline \end{array}$

Dan berikut untuk wilayah yang memenuhi  $"3x+2y\leq 6$

$\begin{array}{ll}\\ 2.&\textrm{Selesaikanlah pertidaksamaan berikut}\\ &\textrm{a}.\quad 12x+2>4x+6\\ &\textrm{b}.\quad 2-3x<6-x\\ &\textrm{c}.\quad 6x+1\geq 2\\ &\textrm{d}.\quad \displaystyle \frac{2-3x}{2}<\frac{3-x}{3}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{a}.\: \: 12x&+2>4x+6\\ 12x&-4x>6-2\\ 8x&>4\\ x&>\displaystyle \frac{1}{2} \end{aligned}\\ &\begin{aligned}\color{black}\textrm{b}.\: \: &2-3x<6-x\\ &-3x+x<6-2\\ &-2x<4\: \: \textrm{dikali}\: \: (-1)\\ &2x>-4\: \: (\textrm{tanda berubah})\\ &x>-2 \end{aligned}\\ &\begin{aligned}\textrm{c}.\: \: 6x&+1\geq 2\\ 6x&\geq 2-1\\ x&\geq \displaystyle \frac{1}{6} \end{aligned}\\ &\begin{aligned}\textrm{d}.\: \: \: \: \displaystyle \frac{2-3x}{2}&<\frac{3-x}{3}\\ 3(2-3x)&<2(3-x)\\ 6-9x&<6-2x\\ -9x+2x&<6-6\\ -7x&<0\: \: \textrm{di kali}\: \: (-1)\\ 7x&>0\: \: (\textrm{tanda berubah})\\ x&>\displaystyle \frac{0}{7}\\ x&>0 \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Heryadi, D. 2007. Modul Matematikauntuk SMK Kelas X. Bogor: YUDHISTIRA.
2. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo. PT. TIGA SERANGKAI PUSTAKA MANDIRI.