Question

rationalizing the denominator
$12 + 4 \sqrt{3} by3 \sqrt{2 } - \sqrt{6}$
ok​

Answer 1

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

$\frac{36 \sqrt{2} + 12 \sqrt{6} + 12 \sqrt{6} + 12 \sqrt{2}}{18 - 6}$

$= \frac{48 \sqrt{2} + 24 \sqrt{6} }{12} = \frac{12(4 \sqrt{2} + 2 \sqrt{6})}{12}$

$= 4 \sqrt{2} + 2 \sqrt{6}$