Question

√ 3+√2/√3-√2+√3-√2/√3+√2

Answer 1

Hi friend!!

$\frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$


=(√3+√2)²+(√3-√2)²/(3-2)

=(3+2+2√6+3+2-2√6)/1

=10

I hope this will help you ;)

Answer 2

Heya there !!!
Here is the answer you were looking for:

Identities used :

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ (x + y)(x - y) = {x}^{2} - {y}^{2}$

$\frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } + \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ = \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{2}) }^{2} + 2( \sqrt{3})( \sqrt{2} ) }{ {( \sqrt{3} )}^{2} - {( \sqrt{2}) }^{2} } + \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{2} )}^{2} - 2( \sqrt{3} )( \sqrt{2}) }{ {( \sqrt{3}) }^{2} - {( \sqrt{2}) }^{2} } \\ \\ = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} + \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} \\ \\ = 5 + 2 \sqrt{6} + 5 - 2 \sqrt{6} \\ \\ = 5 + 5 \\ \\ = 10$

Hope this helps!!!

If you have any doubt regarding to my answer, please ask in the comment section or inbox me ^_^

@Mahak24

Thanks...
☺☺