a) Ta có
$\dfrac{10}{3\sqrt{5}} = \dfrac{2.5}{3\sqrt{5}} = \dfrac{2\sqrt{5}}{3}$
b) Ta có
$\dfrac{\sqrt{3} - \sqrt{6}}{1 - \sqrt{2}} - \dfrac{2 + \sqrt{8}}{1 + \sqrt{2}} = \dfrac{\sqrt{3} (1 - \sqrt{2})}{1 - \sqrt{2}} - \dfrac{2(1 + \sqrt{2})}{1 + \sqrt{2}} = \sqrt{3} - 2$
c) Ta có
$\left( 2 + \dfrac{3 + \sqrt{3}}{\sqrt{3} + 1} \right) \left( 2 - \dfrac{3 - \sqrt{3}}{\sqrt{3} - 1} \right) = \left( 2 + \dfrac{\sqrt{3}( 1 + \sqrt{3})}{\sqrt{3} + 1} \right)\left( 2 - \dfrac{\sqrt{3}(\sqrt{3}-1)}{\sqrt{3} - 1} \right)$
$= (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$
