Rationalize the denominators of:

(i) $\dfrac{2\sqrt{3}}{\sqrt{5}}$

(ii) $\dfrac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}$

(i) Given,

$\dfrac{2\sqrt{3}}{\sqrt{5}}$

Let us rationalise the denominator,

$\Rightarrow \dfrac{2\sqrt{3} \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} \\[1em] \Rightarrow \dfrac{2\sqrt{15}}{(\sqrt{5})^2}\\[1em] \Rightarrow \dfrac{2\sqrt{15}}{5}$

Hence,$\dfrac{2\sqrt{3}}{\sqrt{5}} = \dfrac{2\sqrt{15}}{5}$

(ii) Given, $\dfrac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}$

Let us rationalise the denominator,

$\Rightarrow \dfrac{(\sqrt{6} - \sqrt{5}) \times (\sqrt{6} - \sqrt{5})}{(\sqrt{6} + \sqrt{5})\times (\sqrt{6} - \sqrt{5})}\\[1em] = \dfrac{(\sqrt{6} - \sqrt{5})^2}{(\sqrt{6})^2 - (\sqrt{5})^2}\\[1em] = \dfrac{(\sqrt{6})^2 + (\sqrt{5})^2 - 2 \times \sqrt{6} \times \sqrt{5}}{6 - 5}\\[1em] = \dfrac{6 + 5 - 2 \times \sqrt{30}}{1}\\[1em] = 11 - 2\sqrt{30}$

Hence, $\dfrac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}} = 11 - 2\sqrt{30}$.