Question

2√5 -√3 / √5 +√3 + √6 + √2 / 2√6 -√2

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{2 \sqrt{5} - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } + \frac{ \sqrt{6} + \sqrt{2} }{2 \sqrt{6} - \sqrt{2} }$

On rationalizing the denominators we get :

$= \frac{2 \sqrt{5} - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \times \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} - \sqrt{3} } + \frac{ \sqrt{6} + \sqrt{2} }{2 \sqrt{6} - \sqrt{2} } \times \frac{2 \sqrt{6} - \sqrt{2} }{2 \sqrt{6} - \sqrt{2} }$

Using the identity :

$(x + y)(x - y) = {x}^{2} - {y}^{2}$
$= \frac{2 \sqrt{5}( \sqrt{5} - \sqrt{3} ) - \sqrt{3} ( \sqrt{5} - \sqrt{3} )}{ {( \sqrt{5} )}^{2} - {( \sqrt{3}) }^{2} } + \frac{ \sqrt{6}(2 \sqrt{6} + \sqrt{2}) + \sqrt{2} (2 \sqrt{6} + \sqrt{2} )}{ {(2 \sqrt{6}) }^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{2 \times 5 - 2 \sqrt{15} - \sqrt{15} + 3}{5 - 3} + \frac{2 \times 6 + \sqrt{12} + 2 \sqrt{12} + 2}{24 - 2} \\ \\ = \frac{10 + 3 - 3 \sqrt{15} }{2} + \frac{12 + 2 + 3 \sqrt{12} }{22} \\ \\ = \frac{13 - 3 \sqrt{15} }{2} + \frac{14 + 3 \sqrt{2 \times 2 \times 3} }{22} \\ \\ = \frac{13 - 3 \sqrt{15} }{2} + \frac{14 + 3 \times 2 \sqrt{3} }{22} \\ \\ = \frac{13 - 3 \sqrt{15} }{2} + \frac{14 + 6 \sqrt{3} }{22}$

Taking LCM of 22 and 2 we get 22

$= \frac{13 \times 11 - 3 \sqrt{15} \times 11 + 14 \times 1 + 6 \sqrt{3} \times 1}{22} \\ \\ = \frac{143 - 33 \sqrt{15} + 14 + 6 \sqrt{3} }{22} \\ \\ = \frac{157 - 33 \sqrt{15} + 6 \sqrt{3} }{22}$

Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
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