Question

$\frac{1}{2 + \sqrt{3} } + \frac{2}{ \sqrt{5} - \sqrt{3} } + \frac{1}{2 - \sqrt{5} }$
please solve fast!!​

Answer 1

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

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

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

$= \frac{(2-\sqrt{3})}{4-3} + \frac{2(\sqrt{5}+\sqrt{3})}{ 5-3} + \frac{(2+\sqrt{5})}{4-5}$

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

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

$= 0$

Therefore.,

$\red{\frac{1}{2 + \sqrt{3} } + \frac{2}{ \sqrt{5} - \sqrt{3} } + \frac{1}{2 - \sqrt{5} } }\green{= 0}$

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