Question

2√3-3√2/3√2+2√3=a-b√6

Answer 1

Heya !!!

$\frac{2 \sqrt{3} - 3 \sqrt{2} }{3 \sqrt{2} + 2 \sqrt{3} } = a - b \sqrt{6}$

L.H.S.

On rationalizing the denominator we get,

$= \frac{2 \sqrt{3} - 3 \sqrt{2} }{3 \sqrt{2} + 2 \sqrt{3} } \times \frac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \\ \\ = \frac{2 \sqrt{3}(3 \sqrt{2} - 2 \sqrt{3} ) - 3 \sqrt{2} (3 \sqrt{2} - 2 \sqrt{3}) }{ {(3 \sqrt{2} )}^{2} - {(2 \sqrt{3}) }^{2} } \\ \\ = \frac{6 \sqrt{6} - 12 - 18 + 6 \sqrt{6} }{18 - 12} \\ \\ = \frac{12 \sqrt{2} - 30}{6} \\ \\ = \frac{6(2 \sqrt{6} - 5)}{6} \\ \\ = 2 \sqrt{6} - 5$

On comparing both the sides we get,

$2 \sqrt{6} - 5 = a - b \sqrt{6} \\ \\ - 5 + 2 \sqrt{6} = a - b \sqrt{6} \\ \\ a = - 5 \: : \:b = - 2$

Hope this helps ☺

Answer 2

$hope \: this \: is \: helpful \: to \: u \:$