Question

solve it 2nd question

Answer 1

Hello Mate!

$\frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } + \frac{3 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } - \frac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } \\ \frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} } \\ - (\sqrt{12} - \sqrt{18} ) \\ \sqrt{18} - \sqrt{12} \\ \frac{3 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } \times \frac{ \sqrt{6} - \sqrt{3} }{ \sqrt{6} - \sqrt{3} } \\ \frac{3 \sqrt{2} ( \sqrt{6} - \sqrt{3}) }{3} = \sqrt{12} - \sqrt{6} \\ \frac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } \times \frac{ \sqrt{6} - \sqrt{2} }{ \sqrt{6} - \sqrt{2} } \\ \frac{4 \sqrt{3}( \sqrt{6} - \sqrt{2} )}{4} = \sqrt{18} - \sqrt{6} \\ \sqrt{18} - \sqrt{12} + \sqrt{12} - \sqrt{6} - \sqrt{18} + \sqrt{6} \\ = 0$

Hope it helps☺!