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Answer 1

Answer:-

We have to prove:-

$\sf \: \frac{1}{ \sqrt{2} - \sqrt{3} - \sqrt{5} } + \frac{1}{ \sqrt{2} + \sqrt{3} - 5 } = \frac{1}{ \sqrt{2} } \\ \\ \\ \implies \sf \: \frac{1}{ \sqrt{2} - \sqrt{5} - \sqrt{3} } + \frac{1}{ \sqrt{2} - \sqrt{5} + 3 } = \frac{1}{ \sqrt{2} }$

Taking LCM in LHS we get,

$\implies \sf \: \dfrac{ \sqrt{2} - \sqrt{5} + \sqrt{3} + \sqrt{2} - \sqrt{5} - \sqrt{3} }{( \sqrt{2} - \sqrt{5} - \sqrt{3})( \sqrt{2} - \sqrt{5} + 3)} = \dfrac{1}{ \sqrt{2} }$

Using (a + b)(a - b) = a² - b² in LHS we get,

$\implies \sf \: \dfrac{2 \sqrt{2} - 2 \sqrt{5} }{( \sqrt{2} - \sqrt{5}) ^{2} - {( \sqrt{3}) }^{2} } = \dfrac{1}{ \sqrt{2} }$

Using (a - b)² = a² + b² - 2ab in LHS we get,

$\implies \sf \: \frac{2( \sqrt{2} - \sqrt{5} )}{ {( \sqrt{2}) }^{2} + {( \sqrt{5}) }^{2} - 2 \times \sqrt{2} \times \sqrt{5} - 3 } = \frac{1}{ \sqrt{2} } \\ \\ \\ \implies \sf \: \frac{2( \sqrt{2} - \sqrt{5}) }{2 + 5 - 3 - 2 \times \sqrt{2 } \times \sqrt{5} } = \frac{1}{ \sqrt{2} } \\ \\ \\ \implies \sf \: \frac{2( \sqrt{2} - \sqrt{5} )}{4 - 2 \times \sqrt{2} \times \sqrt{5} } = \frac{1}{ \sqrt{2} } \\ \\ \\ \implies \sf \: \frac{2( \sqrt{2} - \sqrt{5} )}{2 \times \sqrt{2} \times \sqrt{2} - 2 \times \sqrt{2} \times \sqrt{5} } = \frac{1}{ \sqrt{2} } \\ \\ \\ \implies \sf \: \frac{ \cancel{2( \sqrt{2} - \sqrt{5} )}}{ \cancel2 \times \sqrt{2} \cancel{ ( \sqrt{2} - \sqrt{5}) }} = \frac{1}{ \sqrt{2} } \\ \\ \\ \implies \sf \: \frac{1}{ \sqrt{2} } = \frac{1}{ \sqrt{2} }$

Hence, Proved.