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,

$= \frac{2 \sqrt{3} - 3 \sqrt{2} }{3 \sqrt{2} + 2 \sqrt{3} } \times \frac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} }$

$\bf \: using \: the \: identity \\ \\ (a + b)(a - b) = {a}^{2} - {b}^{2}$

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

$\bf \: on \: comparing \: both \: the \: sides \: we \: get \\ \\ 2 \sqrt{6} - 5 = a + b \sqrt{6} \\ \\ a = - 5 \: : \: b = 2$

Hope this helps ☺

Answer 2

$\underline{\bf HELLO \: \: DEAR,}$

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

$\bf \frac{2\sqrt{3} - 3\sqrt{2}}{3\sqrt{2} + 2\sqrt{3}} * \frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} - 2\sqrt{3}}$

$\to \bf \frac{(6\sqrt{6} -12- 18+ 6\sqrt{6})}{(3\sqrt{2})^{2} - (2\sqrt{3})^{2}}$

$\to \bf \frac{- 12 - 18 + 12\sqrt{6}}{18 - 12}$

$\to \bf \frac{-30 +12 \sqrt6}{6}$

$\to \bf \frac{-30}{6} + \frac{12\sqrt{6}}{6}$

$\bf on \: \: comparing \: \: with \: \: BOTH \: \: SIDE$

$\bf we \: \: get$

$\bf (-5) + 2\sqrt{6} = a + b\sqrt{6}$

$\bf a = (-5) \: \: , \: \: b = (2)$

$\underline{\bf I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}$