Lanjutan 2 Fungsi Eksponen

 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

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

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

$\begin{array}{rl}& \\ 1.& {\displaystyle \frac{c}{a+\sqrt{b}}=\frac{c}{a+\sqrt{b}}.\frac{a-\sqrt{b}}{a-\sqrt{b}}=\frac{c(a-\sqrt{b})}{a^2-b}}\\ 2.& {\displaystyle \frac{c}{a-\sqrt{b}}=\frac{c}{a-\sqrt{b}}.\frac{a+\sqrt{b}}{a+\sqrt{b}}=\frac{c(a+\sqrt{b})}{a^2-b}}\\ 3.& {\displaystyle \frac{c}{\sqrt{a}+\sqrt{b}}=\frac{c}{\sqrt{a}+\sqrt{b}}.\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{c(\sqrt{a}-\sqrt{b})}{a-b}}\end{array}$

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 $\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}$.

$\text{ CONTOH SOAL}$.

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

$\begin{array}{l}\\ 2.& \text{Rasionalkanlah penyebut pecahan berikut}\\ & \text{dan serderhankanlah hasilnya}\\ & \text{a}.{\displaystyle \frac{3}{6-\sqrt{5}}\text{f}.{\displaystyle \frac{3}{\sqrt{6}+\sqrt{5}}}}\\ & \text{b}.{\displaystyle \frac{3}{6+\sqrt{5}}\text{g}.\frac{\sqrt{3}}{\sqrt{6}-\sqrt{5}}}\\ & \text{c}.{\displaystyle \frac{\sqrt{3}}{6-\sqrt{5}}\text{h}.\frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}}\\ & \text{d}.{\displaystyle \frac{\sqrt{3}}{6+\sqrt{5}}\text{i}.\frac{\sqrt{3}}{\sqrt{6-2\sqrt{5}}}}\\ & \text{e}.{\displaystyle \frac{3}{\sqrt{6}-\sqrt{5}}\text{j}.\frac{\sqrt{3}}{\sqrt{6+2\sqrt{5}}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& {\displaystyle \frac{3}{6-\sqrt{5}}={\displaystyle \frac{3}{6-\sqrt{5}}\times \frac{6+\sqrt{5}}{6+\sqrt{5}}={\displaystyle \frac{3(6+\sqrt{5})}{6^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{18+3\sqrt{5}}{36-5}=\frac{18+3\sqrt{5}}{31}={\displaystyle \frac{1}{31}(18+3\sqrt{5})}}\\ \text{b}.& {\displaystyle \frac{3}{6+\sqrt{5}}={\displaystyle \frac{3}{6+\sqrt{5}}\times \frac{6-\sqrt{5}}{6-\sqrt{5}}={\displaystyle \frac{3(6-\sqrt{5})}{6^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{18-3\sqrt{5}}{36-5}=\frac{18-3\sqrt{5}}{31}={\displaystyle \frac{1}{31}(18-3\sqrt{5})}}\\ \text{c}.& {\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}(6+\sqrt{5})}{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}(6\sqrt{3}+\sqrt{15})}}\\ \text{d}.& {\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}(6-\sqrt{5})}{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}(6\sqrt{3}-\sqrt{15})}}\\ \text{e}.& {\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(\sqrt{6}+\sqrt{5})}{{\sqrt{6}}^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{3(\sqrt{6}+\sqrt{5})}{6-5}=\frac{3(\sqrt{6}+\sqrt{5})}{1}=3(\sqrt{6}+\sqrt{5})}\\ \text{f}.& {\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(\sqrt{6}-\sqrt{5})}{{\sqrt{6}}^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{3(\sqrt{6}-\sqrt{5})}{6-5}=\frac{3(\sqrt{6}-\sqrt{5})}{1}=3(\sqrt{6}-\sqrt{5})}\\ \text{g}.& {\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}(\sqrt{6}+\sqrt{5})}{{\sqrt{6}}^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{\sqrt{18}+\sqrt{15}}{6-5}=\frac{\sqrt{9.2}+\sqrt{15}}{1}=(3\sqrt{2}+\sqrt{15})}\\ \text{h}.& {\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}(\sqrt{6}-\sqrt{5})}{{\sqrt{6}}^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{\sqrt{18}-\sqrt{15}}{6-5}=\frac{\sqrt{9.2}-\sqrt{15}}{1}=(3\sqrt{2}-\sqrt{15})}\\ \text{i}.& \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}(\sqrt{15}+\sqrt{3})}\\ \text{j}.& \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}(\sqrt{15}-\sqrt{3})}\end{array}\end{array}$.

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