# Simplify:$\frac{1}{2+\sqrt{3}}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}$

Given:

$\frac{1}{2+\sqrt{3}}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}$

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,

$\frac{1}{2+\sqrt{3}}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}$

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

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

$=\frac{2-\sqrt{3}}{4-3}$

$=\frac{2-\sqrt{3}}{1}$

$=2-\sqrt{3}$

$\frac{2}{\sqrt{5}-\sqrt{3}}=\frac{2(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}$

$=\frac{2(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}$

$=\frac{2(\sqrt{5}+\sqrt{3})}{5-3}$

$=\frac{2(\sqrt{5}+\sqrt{3})}{2}$

$=\sqrt{5}+\sqrt{3}$

$\frac{1}{2-\sqrt{5}}=\frac{1(2+\sqrt{5})}{(2-\sqrt{5})(2+\sqrt{5})}$

$=\frac{2+\sqrt{5}}{(2)^{2}-(\sqrt{5})^{2}}$

$=\frac{2+\sqrt{5}}{4-5}$

$=\frac{2+\sqrt{5}}{-1}$

$=-2-\sqrt{5}$

Therefore,

$\frac{1}{2+\sqrt{3}}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}=(2-\sqrt{3})+(\sqrt{5}+\sqrt{3})+(-2-\sqrt{5})$

$=2-\sqrt{3}+\sqrt{5}+\sqrt{3}-2-\sqrt{5}$

$=0$

Hence, $\frac{1}{2+\sqrt{3}}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}=0$.

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Updated on: 10-Oct-2022

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