Rationalise

$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$

$=\frac{\sqrt{3}-\sqrt{2}}{3-2}$

$=\frac{\sqrt{3}-\sqrt{2}}{1}$

$=\sqrt{3}-\sqrt{2}$

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