Tính giá trị biểu thức ạa, mong có ng trả lời ạ

$\textsf{a)}$

$C = \dfrac{3\sqrt{5} + \sqrt{15}}{\sqrt{5}} - \dfrac{5 + \sqrt{5}}{\sqrt{5} + 1}$

$\quad \ \ = \dfrac{\sqrt{5}(3 + \sqrt{3})}{\sqrt{5}} - \dfrac{\sqrt{5}(\sqrt{5} + 1)}{\sqrt{5} + 1}$

$\quad \ \ = (3 + \sqrt{3}) - \sqrt{5} = 3 + \sqrt{3} - \sqrt{5}$

$\textsf{b)}$

$\quad D = \sqrt{\dfrac{3 - \sqrt{5}}{3 + \sqrt{5}}} + \dfrac{\sqrt{8}}{\sqrt{5} - \sqrt{3}}$

$\quad \texttt{(1) }\ \sqrt{\dfrac{3 - \sqrt{5}}{3 + \sqrt{5}}} = \sqrt{\dfrac{(3 - \sqrt{5})^2}{3^2 - 5}} = \sqrt{\dfrac{(3 - \sqrt{5})^2}{4}} = \dfrac{3 - \sqrt{5}}{2}$

$\quad \texttt{(2) }\ \dfrac{\sqrt{8}}{\sqrt{5} - \sqrt{3}} = \dfrac{2\sqrt{2}(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} = \dfrac{2(\sqrt{10} + \sqrt{6})}{2} = \sqrt{10} + \sqrt{6}$

$\implies D = \dfrac{3 - \sqrt{5}}{2} + \sqrt{10} + \sqrt{6}$