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