Question

simplify (3√2/√6-√3)-(4√3/√6-√2)+(2√3/√6+√2)

Answer 1

Heya there !!!
Here is the answer you were looking for:
$\frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } - \frac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } + \frac{2 \sqrt{3} }{ \sqrt{6} + \sqrt{2} }$

On rationalizing the denominator we get,

$\frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } \times \frac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} + \sqrt{3} } - \frac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } \times \frac{ \sqrt{6} + \sqrt{2} }{ \sqrt{6} + \sqrt{2} } + \frac{2 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } \times \frac{ \sqrt{6} - \sqrt{2} }{ \sqrt{6} - \sqrt{2} } \\ \\ = \frac{3 \sqrt{2} ( \sqrt{6} + \sqrt{3} )}{ {( \sqrt{6} )}^{2} - {( \sqrt{3} )}^{2} } - \frac{4 \sqrt{3}( \sqrt{6} + \sqrt{2} ) }{ {( \sqrt{6} )}^{2} - {( \sqrt{4} )}^{2} } + \frac{2 \sqrt{3} ( \sqrt{6} - \sqrt{2} )}{ {( \sqrt{6}) }^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{3 \sqrt{12} + 3 \sqrt{6} }{6 - 3} - \frac{4 \sqrt{18} - 4 \sqrt{6} }{6 - 2} + \frac{3 \sqrt{18} - 2 \sqrt{6} }{6 - 2} \\ \\ = \frac{3 \sqrt{2 \times 2 \times 3} + 3 \sqrt{6} }{3} - \frac{4 \sqrt{3 \times 3 \times 2} - 4 \sqrt{6} }{4} + \frac{3 \sqrt{3 \times 3 \times 2} - 2 \sqrt{6} }{4} \\ \\ = \frac{6 \sqrt{3} + 3 \sqrt{6} }{3} - \frac{12 \sqrt{2} - 4 \sqrt{6} }{4} + \frac{9 \sqrt{2} - 2 \sqrt{6} }{4} \\ \\ = 2 \sqrt{3} + \sqrt{6} - 3 \sqrt{2} + \sqrt{6} + \frac{9 \sqrt{2} - 2 \sqrt{6} }{4} \\ \\ = \frac{2 \sqrt{3} \times 4 + 2 \sqrt{6} \times 4 - 3 \sqrt{2} \times 4 + 9 \sqrt{2} - 2 \sqrt{6} }{4} \\ \\ = \frac{8 \sqrt{3} + 8 \sqrt{6} - 12 \sqrt{2} + 9 \sqrt{2} - 2 \sqrt{6} }{4} \\ \\ = \frac{ - 3 \sqrt{2} + 6 \sqrt{6} + 8 \sqrt{3} }{4}$

Hope this helps!!!

@Mahak24

Thanks...
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