Rationalise the denominator of the following and hence evaluate by taking $\sqrt{2}$ = 1.414 and $\sqrt{3}$ = 1.732 , upto three places of decimal:

$\begin{matrix} \text{(i)} & \dfrac{\sqrt2}{2+\sqrt2} \\[1.5em] \text{(ii)} & \dfrac{1}{\sqrt3+\sqrt2} \\[1.5em] \end{matrix}$

(i) Rationalise the denominator ,

$\dfrac{\sqrt{2}}{2 + \sqrt{2}} = \dfrac{\sqrt{2}}{2+ \sqrt{2}} × \dfrac{2 - \sqrt{2}}{2 - \sqrt{2}} \\[1.5em] \Rightarrow \sqrt{2} × \dfrac{(2 - \sqrt{2})}{2^2 - (\sqrt{2})^2} \\[1.5em] \Rightarrow \dfrac{2{\sqrt{2}} - 2}{4 - 2} \\[1.5em] \Rightarrow (\sqrt{2} - 1) \\[1.5em]$

Since, $\sqrt{2}$ = 1.414

$\Rightarrow {1.414 - 1} \\[1.5em] \Rightarrow\bold{ 0.414} \\[1.5em]$

$(\text{ii})$ Rationalise the denominator ,

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

Since, $\sqrt{3}$ = 1.732

$\Rightarrow {1.732 - 1.414} \\[1.5em] \Rightarrow\bold{0.318} \\[1.5em]$