Question

rationalize the denominator $\frac{1}{\sqrt{6}+\sqrt{5}- \sqrt{11} }$

Answer 1

$\huge{\star}{\underline{\boxed{\red{\sf{Answer :}}}}}{\star}$

Given :-

$\sf{ \frac{1}{ \sqrt{ 6} + \sqrt{5} - \sqrt{11} } }$

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Solution :-

We know that,

When there is root in the denominator then we factorise it.

$\hookrightarrow\sf{ \frac{1}{ \sqrt{ 6} + \sqrt{5} - \sqrt{11} } } \\ \\ \sf{ \hookrightarrow \frac{1}{ \sqrt{6} + \sqrt{5} - \sqrt{11} } \: \: \: \times \: \: \: \frac{ \sqrt{6} + \sqrt{5} + \sqrt{11} }{ \sqrt{6} + \sqrt{5} + \sqrt{11} } } \\ \\ \bf{here \: \sqrt{6} + \sqrt{5} \: is \: a \: and \: \sqrt{11} \: is \: b \: }$

$\LARGE{\boxed{\blue{\sf{(a + b) (a - b) = a^2 - b^2 }}}}$

$\Large \sf{ \hookrightarrow \frac{ \sqrt{6} + \sqrt{5} + \sqrt{11} }{( { \sqrt{6} + \sqrt{5} )}^{2} - ( { \sqrt{11)} }^{2} } }$

$\LARGE{\boxed{\green{\sf{(a + b)^2 = a^2 + b^2 + 2ab}}}}$

$\sf{ \hookrightarrow \frac{ \sqrt{6} + \sqrt{5} + \sqrt{11} }{( \sqrt{6})^{2} + ({ \sqrt{5}) }^{2} + 2( \sqrt{6)} ( \sqrt{5) } - 11 } } \\ \\ \sf{ \hookrightarrow \frac{ \sqrt{6} + \sqrt{5} + \sqrt{11} }{6 + 5 - 11 + 2 \sqrt{30} } } \\ \\ \sf{ \hookrightarrow \frac{ \sqrt{6} + \sqrt{5} + \sqrt{11} }{ \cancel{11} - { \cancel11} + 2 \sqrt{30} } } \\ \\ \sf{ \hookrightarrow \frac{ \sqrt{6} + \sqrt{5} + \sqrt{11} }{2 \sqrt{30} } }$

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$\huge{\boxed{\orange{\sf{\frac{ \sqrt{6} + \sqrt{5} + \sqrt{11} }{2 \sqrt{30}}}}}}$