Question

solve the question in the attachment and give step by step explanation... ​

Answer 1

Here is your answer.

Check out the attachment

Thanks

Answer 2

Hlw...

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

Again..

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