Rationalise the denominator of the following :

$\begin{matrix} \text{(i)} & \dfrac{3}{4\sqrt{5}} \\[1.5em] \text{(ii)} & \dfrac{5\sqrt{7}}{\sqrt{3}} \\[1.5em] \text{(iii)} & \dfrac{3}{4 - \sqrt{7}} \\[1.5em] \text{(iv)} & \dfrac{17}{3\sqrt{2} + 1} \\[1.5em] \text{(v)} & \dfrac{16}{\sqrt{41}-5} \\[1.5em] \text{(vi)} & \dfrac{1}{\sqrt{7} - \sqrt{6}} \\[1.5em] \text{(vii)} & \dfrac{1}{\sqrt{5} + \sqrt{2}} \\[1.5em] \text{(viii)} & \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \\[1.5em] \end{matrix}$

$\text{(i)}$

$\dfrac{3}{4\sqrt{5}}$

Let us rationalise the denominator,

Then,

$\dfrac{3}{4\sqrt{5}} = \dfrac{3×\sqrt{5}}{4\sqrt{5} × \sqrt{5}} \\[1.5em] \Rightarrow\dfrac{3\sqrt{5}}{4 × 5} \\[1.5em] \bold{\Rightarrow\dfrac{3\sqrt{5}}{20}}$

$\text{(ii)}$

$\dfrac{5\sqrt{7}}{\sqrt{3}}$

Let us rationalise the denominator,

Then,

$\dfrac{5\sqrt{7}}{\sqrt{3}} = \dfrac{5\sqrt{7}×\sqrt{3}}{\sqrt{3} × \sqrt{3}} \\[1.5em] \bold{\Rightarrow\dfrac{5\sqrt{21}}{3}}$

$\text{(iii)}$

$\dfrac{3}{4 - \sqrt{7}}$

Let us rationalise the denominator,

Then,

$\dfrac{3}{4 - \sqrt{7}} = \dfrac{3}{4 - \sqrt{7}} × \dfrac{4 + \sqrt{7}}{4 + \sqrt{7}} \\[1.5em] \Rightarrow\dfrac{3(4 + \sqrt{7})}{(4)^2 - (\sqrt{7})^2} \\[1.5em] \Rightarrow\dfrac{3(4 + \sqrt{7})}{(16) - (7)} \\[1.5em] \Rightarrow\dfrac{3(4 + \sqrt{7})}{9} \\[1.5em] \bold{\Rightarrow\dfrac{(4 + \sqrt{7})}{3}} \\[1.5em]$

$\text{(iv)}$

$\dfrac{17}{3\sqrt{2}+1}$

Let us rationalise the denominator,

Then,

$\dfrac{17}{3\sqrt{2} + 1} = \dfrac{17}{3\sqrt{2} + 1}×\dfrac{3\sqrt{2} - 1}{3\sqrt{2} - 1} \\[1.5em] \Rightarrow\dfrac{17(3\sqrt{2} - 1)}{(3\sqrt{2})^2 - 1} \\[1.5em] \Rightarrow\dfrac{17(3\sqrt{2} - 1)}{17} \\[1.5em] \bold{\Rightarrow(3\sqrt{2} - 1)} \\[1.5em]$

$\text{(v)}$

$\dfrac{16}{\sqrt{41}-5}$

Let us rationalise the denominator,

Then,

$\dfrac{16}{\sqrt{41} - 5} = \dfrac{16}{\sqrt{41} - 5}×\dfrac{\sqrt{41} + 5}{\sqrt{41} + 5} \\[1.5em] \Rightarrow\dfrac{16({\sqrt{41} + 5})}{(\sqrt{41})^2 - 5^2} \\[1.5em] \Rightarrow\dfrac{16({\sqrt{41} + 5})}{41 - 25} \\[1.5em] {\Rightarrow\dfrac{16({\sqrt{41} + 5})}{16}} \\[1.5em] \bold{\sqrt{41} + 5}$

$\text{(vi)}$

$\dfrac{1}{\sqrt{7} - \sqrt{6}}$

Let us rationalise the denominator,

Then,

$\dfrac{1}{\sqrt{7} - \sqrt{6}} = \dfrac{1}{\sqrt{7} - \sqrt{6}} × \dfrac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} \\[1.5em] \Rightarrow\dfrac{({\sqrt{7} + \sqrt{6}})}{(\sqrt{7})^2 - (\sqrt{6})^2} \\[1.5em] \Rightarrow\dfrac{\sqrt{7} + \sqrt{6}}{7 - 6 } \\[1.5em] \bold{{\Rightarrow\sqrt{7} + \sqrt{6}}} \\[1.5em]$

$\text{(vii)}$

$\dfrac{1}{\sqrt{5}+\sqrt{2}}$

Let us rationalise the denominator,

Then,

$\dfrac{1}{\sqrt{5} + \sqrt{2}} = \dfrac{1}{\sqrt{5} +\sqrt{2}} × \dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{5} - \sqrt{2}}}{(\sqrt{5})^2 - (\sqrt{2})^2} \\[1.5em] \bold{\Rightarrow\dfrac{{\sqrt{5} - \sqrt{2}}}{3}} \\[1.5em]$

$\text{(viii)}$

$\dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}}$

Let us rationalise the denominator,

Then,

$\dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} × \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}} \\[1.5em] = \dfrac{{(\sqrt{2} + \sqrt{3})^2}}{(\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \dfrac{{(\sqrt{2})^2 + (\sqrt{3})^2} + 2 × \sqrt{2} × \sqrt{3}}{(\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \dfrac{2 + 3 + 2\sqrt{2}\sqrt{3}}{2 - 3} \\[1.5em] = \bold{-(5 + 2\sqrt{6})} \\[1.5em]$