Trục căn thức ở mẫu

a) \(\dfrac{4 + 3\sqrt{5}}{\sqrt{5}} = \dfrac{(4 + 3\sqrt{5}) \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} = \dfrac{4\sqrt{5} + 15}{5}\) b) \(\dfrac{1}{\sqrt{5} - 2} = \dfrac{1 \cdot (\sqrt{5} + 2)}{(\sqrt{5} - 2)(\sqrt{5} + 2)} = \dfrac{\sqrt{5} + 2}{5 - 4} = \sqrt{5} + 2\) c) \(\dfrac{3 + \sqrt{3}}{1 - \sqrt{3}} = \dfrac{(3 + \sqrt{3})(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})}\) Tính tử: \((3 + \sqrt{3})(1 + \sqrt{3}) = 3 + 3\sqrt{3} + \sqrt{3} + 3 = 6 + 4\sqrt{3}\) Tính mẫu: \((1 - \sqrt{3})(1 + \sqrt{3}) = 1 - 3 = -2\) Vậy \(\dfrac{3 + \sqrt{3}}{1 - \sqrt{3}} = \dfrac{6 + 4\sqrt{3}}{-2} = -3 - 2\sqrt{3}\) d) \(\dfrac{\sqrt{2}}{\sqrt{3} + \sqrt{2}} = \dfrac{\sqrt{2}(\sqrt{3} - \sqrt{2})}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} = \dfrac{\sqrt{6} - 2}{3 - 2} = \sqrt{6} - 2\)