Question

Rationalise $\frac{1}{\sqrt{3}+\sqrt{2}}$.

Answer 1

$$\begin{gathered} \frac{1}{\sqrt{3}+\sqrt{2}}=\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ =\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{\sqrt{3}-\sqrt{2}}{3-2} \\ =\frac{\sqrt{3}-\sqrt{2}}{1} \\ =\sqrt{3}-\sqrt{2} \end{gathered}$$

Hence, the rationalised form is $\sqrt{3}-\sqrt{2}$.