Rationalise the denominators of the following :

$(i)$ $\frac{1}{\sqrt{7}}$

$(ii)$ $\frac{1}{\sqrt{7}-\sqrt{6}}$

$(iii)$ $\frac{1}{\sqrt{5}+\sqrt{2}}$

$(iv)$ $\frac{1}{\sqrt{7}-2}$

$(i)$ $\frac{1}{\sqrt{7}}$  $=\frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}}=\frac{\sqrt{7}}{7}$

$(ii)$ $\frac{1}{\sqrt{7}-\sqrt{6}}$  $=\frac{1 \times(\sqrt{7}+\sqrt{6})}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})}$                   $[\because$ R.F. of $(\sqrt{x}-\sqrt{y})=(\sqrt{x}+\sqrt{y})$

$=\frac{(\sqrt{7}+\sqrt{6})}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}=\frac{(\sqrt{7}+\sqrt{6})}{7-6}=\frac{(\sqrt{7}+\sqrt{6})}{1}=(\sqrt{7}+\sqrt{6})$

Thus,   $\frac{1}{\sqrt{7}-\sqrt{6}}=(\sqrt{7}+\sqrt{6})$

$(iii)$  $\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{1(\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}$ $[\because $ RF of $(\sqrt{x}+\sqrt{y})$ is $(\sqrt{x}-\sqrt{y})]$

$=\frac{(\sqrt{5}-\sqrt{2})}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}=\frac{(\sqrt{5}-\sqrt{2})}{5-2}=\frac{(\sqrt{5}-\sqrt{2})}{3}$

Thus,  $\frac{1}{(\sqrt{5}+\sqrt{2})}=\frac{\sqrt{5}-\sqrt{2}}{3}$

$(iv)$ $\frac{1}{\sqrt{7}-2}$  $=\frac{1 \times(\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)}=\frac{(\sqrt{7}+2)}{(\sqrt{7})^{2}-(2)^{2}}$            $[\because$ R.F. of $(\sqrt{7}-2)$ is $(\sqrt{7}+2)]$

$=\frac{(\sqrt{7}+2)}{7-4}=\frac{\sqrt{7}+2}{3}$

Thus,  $\frac{1}{(\sqrt{7}-2)}=\frac{(\sqrt{7}+2)}{3}$