Express each one of the following with rational denominator:$ \frac{\sqrt{3}+1}{2 \sqrt{2}-\sqrt{3}} $

Given:

\( \frac{\sqrt{3}+1}{2 \sqrt{2}-\sqrt{3}} \)

To do: 

We have to express the given fraction with rational denominator.

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.

Therefore,

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

$=\frac{\sqrt{3} \times 2 \sqrt{2}+\sqrt{3} \times \sqrt{3}+2 \sqrt{2}+\sqrt{3}}{(2 \sqrt{2})^{2}-(\sqrt{3})^{2}}$                            [Since $(a+b)(a-b)=a^2-b^2$]

$=\frac{2 \sqrt{6}+3+2 \sqrt{2}+\sqrt{3}}{8-3}$

$=\frac{2 \sqrt{6}+2 \sqrt{2}+\sqrt{3}+3}{5}$

Hence, $\frac{\sqrt{3}+1}{2 \sqrt{2}-\sqrt{3}}=\frac{2 \sqrt{6}+2 \sqrt{2}+\sqrt{3}+3}{5}$.