Simplify:
$ \frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}+\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}} $

Given:

\( \frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}+\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}} \)

To do: 

We have to simplify the given expression.

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,

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

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

$=\frac{9 \times 2+4 \times 3-2 \times 3 \sqrt{2} \times 2 \sqrt{3}}{9 \times 2-4 \times 3}$

$=\frac{18+12-12 \sqrt{6}}{18-12}$

$=\frac{30-12 \sqrt{6}}{6}$

$=5-2 \sqrt{6}$

$\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}=\frac{(\sqrt{4 \times 3})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$

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

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

$=\frac{6+2 \sqrt{6}}{1}$

$=6+2 \sqrt{6}$

Therefore,

$\frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}+\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}=5-2 \sqrt{6}+6+2 \sqrt{6}$

$=11$

Hence, $\frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}+\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}=11$.

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Updated on: 10-Oct-2022

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