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.