Question

Solve No.3................................

Answer 1

hope it will help you.
please mark it as brainliest.

Answer 2

Hey friend,
Here is the answer you were looking for:
$\frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } - \frac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } + \frac{2 \sqrt{3} }{ \sqrt{6} + 2} \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } \times \frac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} + \sqrt{3} } - \frac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } \times \frac{ \sqrt{6} + \sqrt{2} }{ \sqrt{6} + \sqrt{2} } \\ + \frac{2 \sqrt{3} }{ \sqrt{6} + 2} \times \frac{ \sqrt{6} - 2}{ \sqrt{6} - 2} \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{3 \sqrt{2} \times \sqrt{6} + 3 \sqrt{2} \times \sqrt{3} }{ {( \sqrt{6} )}^{2} - {( \sqrt{3}) }^{2} } - \frac{4 \sqrt{3} \times \sqrt{6} + 4 \sqrt{3} \times \sqrt{2} }{ {( \sqrt{6} )}^{2} - {( \sqrt{2}) }^{2} } \\ + \frac{2 \sqrt{3} \times \sqrt{6} - 2 \sqrt{3} \times 2 }{ {( \sqrt{6}) }^{2} - {(2)}^{2} } \\ \\ = \frac{3 \sqrt{2 \times 2 \times 3} + 3 \sqrt{6} }{6 - 3} - \frac{4 \sqrt{3 \times 2 \times 3} + 4 \sqrt{6} }{6 - 2} + \frac{2 \sqrt{3 \times 3 \times 2} - 4 \sqrt{3} }{6 - 4} \\ \\ = \frac{3 \times 2 \sqrt{3} + 3 \sqrt{6} }{3} - \frac{4 \times 3 \sqrt{2} + 4 \sqrt{6} }{4} + \frac{2 \times 3 \sqrt{2} - 4 \sqrt{3} }{2} \\ \\ = \frac{6 \sqrt{2} + 3 \sqrt{6} }{3} - \frac{12 \sqrt{2} + 4 \sqrt{6} }{4} + \frac{6 \sqrt{2} - 4 \sqrt{3} }{2} \\ \\ = 2 \sqrt{2} + \sqrt{6} - 3 \sqrt{2} + \sqrt{6} + 3 \sqrt{2} - 2 \sqrt{3} \\ \\ = 2 \sqrt{2} + 2 \sqrt{6} - 2 \sqrt{3} \\ \\ = 2( \sqrt{2} + \sqrt{6} - \sqrt{3} )$

Hope this helps!!!

@Mahak24

Thanks....
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