Soal Merasionalkan Penyebut Pecahan Bentuk Akar

Kumpulan latihan soal matematika, soal beserta jawabannya.
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Soal Merasionalkan Penyebut Pecahan Bentuk Akar

Post by opan » Wed Jan 30, 2019 10:25 am

Soal
Bentuk sederhana dari $\frac{(\sqrt{3}+\sqrt{7})(\sqrt{3}-\sqrt{7})}{2\sqrt{5}-4\sqrt{2}}$ adalah ...

Jawaban
\begin{align} \frac{(\sqrt{3}+\sqrt{7})(\sqrt{3}-\sqrt{7})}{2\sqrt{5}-4\sqrt{2}} &= \frac{3-7}{2\sqrt{5}-4\sqrt{2}}\times\frac{2\sqrt{5}+4\sqrt{2}}{2\sqrt{5}+4\sqrt{2}}\\ &=\frac{-4(2\sqrt{5}+4\sqrt{2})}{20-32}\\ &=\frac{-8(\sqrt{5}+2\sqrt{2})}{-12}\\ &=-\frac{2}{3}(\sqrt{5}+2\sqrt{2}) \end{align}

Soal
Hasil dari $\frac{(1+\sqrt7)(1-\sqrt7)}{2+\sqrt5}$ adalah...

A. $12+6\sqrt5$
B. $12+\sqrt5$
C. $-12+6\sqrt5$
D. $12-6\sqrt5$
E. $-12+\sqrt5$

Jawaban: D

Proses penyelesaian
\begin{align*}
\frac{(1+\sqrt7)(1-\sqrt7)}{2+\sqrt5}&=\frac{1-7}{2+\sqrt5}\\
&=\frac{-6}{2+\sqrt5}\times\frac{2+\sqrt5}{2+\sqrt5}\\
&=\frac{-12+6\sqrt5}{4-5}\\
&=12-6\sqrt5
\end{align*}

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