Question

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Answer 1

$\frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+3\sqrt{7}}-2\sqrt{2}}$

Try to make this a perfect square :

$\implies \frac{\sqrt{2(8-2\sqrt{7})}}{\sqrt{2(16+6\sqrt{7}})-2\sqrt{2}}$

$\implies \frac{\sqrt{2(7-2\sqrt{7}+1)}}{\sqrt{2(9+6\sqrt{7}+7}-2\sqrt{2}}$

$\implies \frac{\sqrt{2((\sqrt{7})^2-2\times1\times\sqrt{7}+1)}}{\sqrt{2(3^2+2\times3\times\sqrt{7}+\sqrt{7})}-2\sqrt{2}}$

[ We know that a² + b² + 2ab = (a+b)²

$\implies \frac{\sqrt{2(\sqrt{7}-1)^2}}{\sqrt{2(3+\sqrt{7})}^2-2\sqrt{2}}$ ]

$\implies \frac{\sqrt{2}(\sqrt{7}-1)}{\sqrt{2}(3+\sqrt{7}-2\sqrt{2}}$

$\implies \frac{\sqrt{2}(\sqrt{7}-1)}{\sqrt{2}(3+\sqrt{2}-2}$

$\implies \frac{\sqrt{7}-1}{\sqrt{7}+1}$

$\implies \frac{\sqrt{7}-1}{\sqrt{7}+1}\times \frac{\sqrt{7}-1}{\sqrt{7}-1}$

$\implies \frac{(\sqrt{7}-1)^2}{\sqrt{7}^2-1^2}$

[[We know that  (a+b)² = a² + b² + 2ab

$\implies \frac{\sqrt{7}^2+1^2-2\sqrt{7}}{7-1}$ ]

$\implies \frac{7+1-2\sqrt{7}}{6}$

$\implies \frac{8-2\sqrt{7}}{6}$

$\implies \frac{2(4-\sqrt{7})}{6}$

$\implies \frac{4-\sqrt{7}}{3}$

The value of the given equation is coming :

$\boxed{\frac{4-\sqrt{7}}{3}}$

It is not matching the options sorry :(

But I am trying wait don't delete my answer moderator I will try it !

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Answer 2

HEY MATE HERE IS YOUR ANSWER

$<font color ="blue">$
Que :- Simplify :
$\huge \frac{ \sqrt{4 - \sqrt{7} } }{ \sqrt{8 + 3 \sqrt{7} - 2 \sqrt{2} } }$
A) 1
B) 2
C) -2
D) 3

$<font color ="aqua ">$

Answer :- OPTION (A) = 1

By using identity

$= > ( {a}^{2} + {b}^{2} ) = {a}^{2} + {b}^{2} + 2ab$
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Hope it will help you

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