<s>[INST] 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.
 [/INST] 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,
$$A= \sqrt2 (\sqrt5+1)^2+2\sqrt{(5-\sqrt5)(\sqrt5 + 1)^2}$$
$$=\sqrt2 (6+2\sqrt5)+2\sqrt{(5-\sqrt5)(6+2\sqrt5 )}$$
$$=6\sqrt2 + 2\sqrt{10}+4\sqrt{5+\sqrt5}$$
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