Đáp án + Giải thích các bước giải:
$\dfrac{6}{\sqrt{2}-\sqrt{3}+3}$
$=\dfrac{6[(\sqrt{2}-\sqrt{3})-3]}{[(\sqrt{2}-\sqrt{3})+3][(\sqrt{2}-\sqrt{3})-3]}$
$=\dfrac{6[(\sqrt{2}-\sqrt{3})-3]}{(\sqrt{2}-\sqrt{3})^{2}-9}$
$=\dfrac{6[(\sqrt{2}-\sqrt{3})-3]}{5-2\sqrt{6}-9}$
$=\dfrac{6[(\sqrt{2}-\sqrt{3})-3]}{-4-2\sqrt{6}}$
$=\dfrac{-3(\sqrt{2}-\sqrt{3}-3)}{2+\sqrt{6}}$
$=\dfrac{(-3\sqrt{2}+3\sqrt{3}+9)(2-\sqrt{6})}{(2-\sqrt{6})(2+\sqrt{6})}$
$=\dfrac{-6\sqrt{2}+6\sqrt{3}+18+6\sqrt{3}-9\sqrt{2}-9\sqrt{6}}{4-6}$
$=\dfrac{-15\sqrt{2}+12\sqrt{3}+18-9\sqrt{6}}{-2}$
$=\dfrac{15\sqrt{2}-12\sqrt{3}-18+9\sqrt{6}}{2}$
