Question
Rationalize the denominator : $\frac{1}{\sqrt{3}-\sqrt{2}}$
✓
Answer
$\frac{1}{\sqrt{3}-\sqrt{2}}=\frac{1}{(\sqrt{3}-\sqrt{2})} \times \frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})}$
...[Multiplying the numerator and denominator by $(\sqrt{3}+\sqrt{2})]$
$
\begin{aligned}
& =\frac{1 \times(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})} \\
= & \frac{\sqrt{3}+\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} \\
& \ldots\left[\because(a+b)(a-b)=a^2-b^2\right] \\
= & \frac{\sqrt{3}+\sqrt{2}}{3-2}=\frac{\sqrt{3}+\sqrt{2}}{1} \\
\therefore \quad \frac{1}{\sqrt{3}-\sqrt{2}}= & \sqrt{3}+\sqrt{2}
\end{aligned}
$
...[Multiplying the numerator and denominator by $(\sqrt{3}+\sqrt{2})]$
$
\begin{aligned}
& =\frac{1 \times(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})} \\
= & \frac{\sqrt{3}+\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} \\
& \ldots\left[\because(a+b)(a-b)=a^2-b^2\right] \\
= & \frac{\sqrt{3}+\sqrt{2}}{3-2}=\frac{\sqrt{3}+\sqrt{2}}{1} \\
\therefore \quad \frac{1}{\sqrt{3}-\sqrt{2}}= & \sqrt{3}+\sqrt{2}
\end{aligned}
$
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