Question

Hi answers fast....


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Answer 1

Here's your answer!!

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$= > \frac{4}{2 + \sqrt{3} + \sqrt{7} }$

We have to rationalise it ,

$= > \frac{4}{(2 + \sqrt{3} )+ \sqrt{7} } \times \frac{(2 + \sqrt{3} ) - \sqrt{7} }{(2 + \sqrt{3} ) - \sqrt{7} }$

$= > \frac{4((2 + \sqrt{3}) - \sqrt{7} ) }{ {(2 + \sqrt{3}) }^{2} - {(7)}^{2} }$

$= > \frac{8 + 4 \sqrt{3} - 4 \sqrt{7} }{4 + 3 + 4 \sqrt{3} - 7 }$

$= > \frac{8 + 4 \sqrt{3} - 4 \sqrt{7} }{7 - 7 + 4 \sqrt{3} }$

$7 \: and \: - 7 \: get \: cancelled$

$= > \frac{8 + 4 \sqrt{3} - 4 \sqrt{7} }{4 \sqrt{3} }$

We can see that 4 is common here,so we took 4 as common .

$= > \frac{4(2 + \sqrt{3} - \sqrt{7} )}{4( \sqrt{3}) }$

$4 \: and \: 4 \: get \: cancelled$

$= > \frac{2 + \sqrt{3} - \sqrt{7} }{ \sqrt{3} }$

By multiplying it with √3 ,we got ,

$= > \frac{2 + \sqrt{3} - \sqrt{7} }{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} }$

$= > \frac{2 \sqrt{3} + 3 - \sqrt{21} }{3}$

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Hope it helps you!! :)