Question

(1/√3+√2)-(2/√5-√3)-(3/√2-√5)
Simplify by rationalising denominator.

Answer 1

Hello Mate!

1 / ( √3 + √2 ) = 1 / ( √3 + √2 ) × ( √3 - √2 ) / ( √3 - √2 )
= √3 - √2

2 / ( √5 - √3 ) × ( √5 + √3 ) / ( √5 + √3 )

= √5 + √3

3 / ( √2 - √5 ) × ( √2 + √5 ) / ( √2 + √5 )
= - √2 - √5

√3 - √2 - √5 - √3 + √2 + √5
= 0

Hope it helps☺!

Answer 2

Hey Friend,
Here is the answer you were looking for:
$\frac{1}{ \sqrt{3} + \sqrt{2} } - \frac{2}{ \sqrt{5} - \sqrt{3} } - \frac{3}{ \sqrt{2} - \sqrt{5} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{1}{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } - \frac{2}{ \sqrt{5} - \sqrt{3} } \times \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} } - \frac{3}{ \sqrt{2} - \sqrt{5} } \times \frac{ \sqrt{2} + \sqrt{5} }{ \sqrt{2} + \sqrt{5} } \\ \\ using \: the \: denominator \: we \: get \\ \\ = \frac{ \sqrt{3} - \sqrt{2} }{ {( \sqrt{3} )}^{2} - {( \sqrt{2}) }^{2} } - \frac{2( \sqrt{5} ) + 2( \sqrt{3} )}{ {( \sqrt{5}) }^{2} - {( \sqrt{3} )}^{2} } - \frac{3( \sqrt{2}) + 3( \sqrt{5}) }{ {( \sqrt{2}) }^{2} - {( \sqrt{5} )}^{2} } \\ \\ = ( \frac{ \sqrt{3} - \sqrt{2} }{3 - 2}) - (\frac{2 \sqrt{5} + 2 \sqrt{3} }{5 - 3} ) - ( \frac{3 \sqrt{2} + 3 \sqrt{5} }{2 - 5}) \\ \\ = \sqrt{3} - \sqrt{2} - ( \frac{2( \sqrt{5} + \sqrt{3}) }{2} ) - ( \frac{3( \sqrt{2} + \sqrt{5}) }{ - 3} ) \\ \\ = \sqrt{3} - \sqrt{2} - ( \sqrt{5} + \sqrt{3} ) - ( - \sqrt{2} - \sqrt{5} ) \\ \\ = \sqrt{3} - \sqrt{2} - \sqrt{5} - \sqrt{3} + \sqrt{2} + \sqrt{5} \\ \\ = 0$

Hope this helps!!!

@Mahak24

Thanks...
☺☺