Rationalize nested radical expression $\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$ I have a college task to rationalize this fraction.
$$\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$$
I do not know how to simplify this fraction. 
Please, explain how to remove the radical from the denominator. Thanks, for your help.

As shown below, the expression can be rationalized and simplified to,
$$\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}} =6\sqrt2 + 2\sqrt{10}+4\sqrt{5+\sqrt5}$$
First, apply the denesting formula $\sqrt{a-\sqrt c}=\sqrt{\frac{a+\sqrt{a^2-c}}{2}}-\sqrt{\frac{a-\sqrt{a^2-c}}{2}}$ to the denominator,
$$\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}
=\frac{\sqrt{3+\sqrt5}}{2} -\frac{\sqrt{5-\sqrt5}}{2} $$
The expression then becomes,
$$A=\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}} 
= \frac{16}{\sqrt{3+\sqrt5}-\sqrt{5-\sqrt5}}
$$
Next, apply the conjugate $\sqrt{3+\sqrt5}+\sqrt{5-\sqrt5}$ to the denominator,
$$A= \frac{8(\sqrt{3+\sqrt5}+\sqrt{5-\sqrt5})}{\sqrt5 -1}
$$
Apply the conjugate $\sqrt5 +1$ to the denominator again to obtain,
$$A= 2\left(\sqrt{3+\sqrt5}+\sqrt{5-\sqrt5}\right)(\sqrt5 + 1)
$$
Recognize $\sqrt{3+\sqrt5} = \frac{\sqrt5+1}{\sqrt2}$ to simplify,
$$A= \sqrt2 (\sqrt5+1)^2+2\sqrt{(5-\sqrt5)(\sqrt5 + 1)^2}$$
$$=\sqrt2 (6+2\sqrt5)+2\sqrt{(5-\sqrt5)(6+2\sqrt5 )}$$
$$=6\sqrt2 + 2\sqrt{10}+4\sqrt{5+\sqrt5}$$
