Question

one by root 3 + root 2 minus 2 by root 5 minus root 3 minus 3 by root 2 minus root 5 simplified

Answer 1

Hey dear!

Here is yr answer.......

$= > \frac{1}{ \sqrt{3} + \sqrt{2} } - \frac{2}{ \sqrt{5} - \sqrt{3} } - \frac{3}{ \sqrt{2} - \sqrt{5} } \\ \\ \\ on \: rationalising........ \\ \\ \frac{1}{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } - \frac{2}{ \sqrt{5} - \sqrt{3} } \times \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} } - \frac{3}{ \sqrt{2} - \sqrt{5} } \times \frac{ \sqrt{2} + \sqrt{5} }{ \sqrt{2} + \sqrt{5} } \\ \\ = > \frac{ \sqrt{3} - \sqrt{2} }{3 - 2} - \frac{2 \sqrt{5} + 2 \sqrt{3} }{5 - 3} - \frac{3 \sqrt{2} + 3 \sqrt{5} }{2 - 5} \\ \\ = > \sqrt{3} - \sqrt{2} - \frac{2( \sqrt{5} + \sqrt{3}) }{2} - \frac{3( \sqrt{2} + \sqrt{5} )}{ - 3} \\ \\ = > \sqrt{3} + \sqrt{2} - \sqrt{5} - \sqrt{3} + \sqrt{2} + \sqrt{5} \\ \\ = > 2\sqrt{2}$

Hope it hlpz....