Question:
Rationalise $\frac{1}{\sqrt{3}+\sqrt{2}}$.
Solution:
$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$
$=\frac{\sqrt{3}-\sqrt{2}}{3-2}$
$=\frac{\sqrt{3}-\sqrt{2}}{1}$
$=\sqrt{3}-\sqrt{2}$
Hence, the rationalised form is $\sqrt{3}-\sqrt{2}$.
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