a. $\dfrac{2 - \sqrt{3}}{3\sqrt{6}} = \dfrac{\sqrt{6}(2 - \sqrt{3})}{3\sqrt{6}.\sqrt{6}}$
$= \dfrac{2\sqrt{6} - \sqrt{18}}{3.6} = \dfrac{2\sqrt{6} - 3\sqrt{2}}{18}$
b. $\dfrac{1}{\sqrt{2} + \sqrt{3}} = \dfrac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} = \dfrac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}$
c. $\dfrac{1}{2\sqrt{2} - 3\sqrt{3}} = \dfrac{2\sqrt{2} + 3\sqrt{3}}{(2\sqrt{2} - 3\sqrt{3})(2\sqrt{2} + 3\sqrt{3})}$
$= \dfrac{2\sqrt{2} - 3\sqrt{3}}{(2\sqrt{2})^2 - (3\sqrt{3})^2} = \dfrac{2\sqrt{2} - 3\sqrt{3}}{8 - 27}$
$= \dfrac{3\sqrt{3} - 2\sqrt{2}}{19}$
d. $\dfrac{1 + \sqrt{a}}{\sqrt{a}} = \dfrac{\sqrt{a}(1 + \sqrt{a})}{\sqrt{a}.\sqrt{a}} = \dfrac{\sqrt{a} + a}{a}$
(Với a > 0)
