If $$x=\frac{2\sqrt{15}}{\sqrt{3}+\sqrt{5}}$$, then what is the value of $$\frac{x+\sqrt{5}}{x-\sqrt{5}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}$$

 $$x=\frac{2\sqrt{3}\sqrt{5}}{\sqrt{3}+\sqrt{5}}$$

 $$\frac{x}{\sqrt{5}}=\frac{2\sqrt{3}}{\sqrt{3}+\sqrt{5}}$$

By C-D rule,

 $$\frac{x+\sqrt{5}}{x-\sqrt{5}}=\frac{3\sqrt{3}+\sqrt{5}}{\sqrt{3}-\sqrt{5}}$$-------(1)

and 

 $$\frac{x}{\sqrt{3}}=\frac{2\sqrt{5}}{\sqrt{3}+\sqrt{5}}$$

 $$\frac{x+\sqrt{3}}{x-\sqrt{3}}=\frac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$-------(2)

add (1) & (2)

$$\frac{x+\sqrt{5}}{x-\sqrt{5}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}$$

 = $$\frac{3\sqrt{3}+\sqrt{5}}{\sqrt{3}-\sqrt{5}}+\frac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$

 =  $$\frac{2\sqrt{3}-2\sqrt{5}}{\sqrt{3}-\sqrt{5}}$$

 = $$2$$

so the answer is option D.