Rationalise the denominators of the following:
$(i)$ $\frac{1}{\sqrt{7}}$
$(ii)$ $\frac{1}{\sqrt{7}-\sqrt{6}}$
$(iii)$ $\frac{1}{\sqrt{5}+\sqrt{2}}$
$(iv)$ $\frac{1}{\sqrt{7}-2}$

(N/A) $(i)$ $\frac{1}{\sqrt{7}} = \frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{\sqrt{7}}{7}$
$(ii)$ $\frac{1}{\sqrt{7}-\sqrt{6}} = \frac{1 \times (\sqrt{7}+\sqrt{6})}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})} = \frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} = \frac{\sqrt{7}+\sqrt{6}}{7-6} = \sqrt{7}+\sqrt{6}$
$(iii)$ $\frac{1}{\sqrt{5}+\sqrt{2}} = \frac{1 \times (\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})} = \frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^2 - (\sqrt{2})^2} = \frac{\sqrt{5}-\sqrt{2}}{5-2} = \frac{\sqrt{5}-\sqrt{2}}{3}$
$(iv)$ $\frac{1}{\sqrt{7}-2} = \frac{1 \times (\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)} = \frac{\sqrt{7}+2}{(\sqrt{7})^2 - (2)^2} = \frac{\sqrt{7}+2}{7-4} = \frac{\sqrt{7}+2}{3}$