Question

rationalize the denominator of

Answer 1

hey there
hope it helps
^_^

Answer 2

Hey buddy !!

Here is yr answer

$( i) \\ = > \frac{ \sqrt{6} }{ \sqrt{3 } - \sqrt{2} } \\ \\ = > \frac{ \sqrt{6} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = > \frac{( \sqrt{6} )( \sqrt{3} + \sqrt{2} ) }{( \sqrt{3} - \sqrt{2})( \sqrt{3} + \sqrt{2} ) } \\ \\ = > \frac{ \sqrt{18} + \sqrt{12} }{3 - 2} \\ \\ = > \sqrt{18} + \sqrt{12} \\ \\ \\ \\ (ii) \\ = > \frac{4}{ \sqrt{7} + \sqrt{3} } \\ \\ = > \frac{4}{ \sqrt{7} + \sqrt{3} } \times \frac{ \sqrt{7} - \sqrt{3} }{ \sqrt{7} - \sqrt{3} } \\ \\ = > \frac{4( \sqrt{7} - \sqrt{3} ) }{( \sqrt{7} + \sqrt{3})( \sqrt{7} - \sqrt{3} )} \\ \\ = > \frac{4 (\sqrt{7} - \sqrt{3}) }{7 - 3} \\ \\ = > \frac{4( \sqrt{7} - \sqrt{3}) }{4} \\ \\ = > \sqrt{7} - \sqrt{3} \\ \\ \\ \\ (iii) \\ = > \frac{1}{3 + \sqrt{2} } \\ \\ = > \frac{1}{3 + \sqrt{2} } \times \frac{3 - \sqrt{2} }{3 - \sqrt{2} } \\ \\ = > \frac{3 - \sqrt{2} }{(3 + \sqrt{2} )(3 - \sqrt{2}) } \\ \\ = > \frac{3 - \sqrt{2} }{9 - 2} \\ \\ = > \frac{ 3 - \sqrt{2} }{7}\\ \\ \\ \\ (iv) \\ = > \frac{1}{2 \sqrt{5} - \sqrt{3} } \\ \\ = > \frac{1}{2 \sqrt{5} - \sqrt{3} } \times \frac{2 \sqrt{5} + \sqrt{3} }{2 \sqrt{5} + \sqrt{3} } \\ \\ = > \frac{2 \sqrt{5} + \sqrt{3} }{(2 \sqrt{5} - \sqrt{3})(2 \sqrt{5} + \sqrt{3} ) } \\ \\ = > \frac{2 \sqrt{5} + \sqrt{3} }{20 - 3} \\ \\ = > \frac{2 \sqrt{5} + \sqrt{3} }{17}$

Hope it hlpz....