Question

rationalise 1/root 5+root 6-root 11

Answer 1

Here's your answer!!

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We have to rationalise ,

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

So,

$= > \frac{1}{( \sqrt{5} + \sqrt{6}) - \sqrt{11} } \times \frac{ (\sqrt{5} + \sqrt{6}) + \sqrt{11} }{ (\sqrt{5} + \sqrt{6} )+ \sqrt{11} }$



$= > \frac{ \sqrt{5} + \sqrt{6} + \sqrt{11} }{ {( \sqrt{5} + \sqrt{6} ) }^{2} - {( \sqrt{11} )}^{2} }$


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We know that,
$= > {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2 \times a \times b$
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$= > \frac{ \sqrt{5} + \sqrt{6} + \sqrt{11} }{5 + 6 + 2 \times \sqrt{5} \times \sqrt{6} - 11 }$


$= > \frac{ \sqrt{5} + \sqrt{6} + \sqrt{11} }{11 + 2 \sqrt{30} - 11}$


$= > \frac{ \sqrt{5} + \sqrt{6} + \sqrt{11} }{2 \sqrt{30} } (11 \: and \: - 11 \: got \: cancel)$


By rationalizing it's denominator,we get:-

$= > \frac{ \sqrt{5} + \sqrt{6} + \sqrt{11} }{2 \sqrt{30} } \times \frac{2 \sqrt{30} }{2 \sqrt{30} }$


$= > \frac{2 \sqrt{150} + 2 \sqrt{180} + 2 \sqrt{330} }{4 \times 30}$


$= > \frac{ 2\sqrt{150} + 2 \sqrt{180} + 2 \sqrt{330} }{120}$


$= > 2( \frac{150 + 180 + 330}{120} )$


$= > \frac{150 + 180 + 330}{60}$



We can write it as ,


$= > \frac{5 \sqrt{6} + 6 \sqrt{5} + \sqrt{330} }{60}$

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Hope it helps you!! :)

Answer 2

Here it's clear.hope it gonna help