Question
Rationalize the denominator : $\frac{1}{\sqrt{7}+\sqrt{2}}$
✓
Answer
$\quad \frac{1}{\sqrt{7}+\sqrt{2}}=\frac{1}{(\sqrt{7}+\sqrt{2})} \times \frac{(\sqrt{7}-\sqrt{2})}{(\sqrt{7}-\sqrt{2})}$
...[Multiplying the numerator and
denominator by $(\sqrt{7}-\sqrt{2})]$
$\begin{aligned}
= & \frac{\sqrt{7}-\sqrt{2}}{(\sqrt{7})^2-(\sqrt{2})^2} \\
& \ldots\left[\because(a-b)(a+b)=a^2-b^2\right] \\
= & \frac{\sqrt{7}-\sqrt{2}}{7-2} \\
\therefore \quad \frac{1}{\sqrt{7}+\sqrt{2}} & =\frac{\sqrt{7}-\sqrt{2}}{5}
\end{aligned}$
...[Multiplying the numerator and
denominator by $(\sqrt{7}-\sqrt{2})]$
$\begin{aligned}
= & \frac{\sqrt{7}-\sqrt{2}}{(\sqrt{7})^2-(\sqrt{2})^2} \\
& \ldots\left[\because(a-b)(a+b)=a^2-b^2\right] \\
= & \frac{\sqrt{7}-\sqrt{2}}{7-2} \\
\therefore \quad \frac{1}{\sqrt{7}+\sqrt{2}} & =\frac{\sqrt{7}-\sqrt{2}}{5}
\end{aligned}$
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