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.
 A: 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,
$$A= \sqrt2 (\sqrt5+1)^2+2\sqrt{(5-\sqrt5)(\sqrt5 + 1)^2}$$
$$=\sqrt2 (6+2\sqrt5)+2\sqrt{(5-\sqrt5)(6+2\sqrt5 )}$$
$$=6\sqrt2 + 2\sqrt{10}+4\sqrt{5+\sqrt5}$$
A: Let use
$$\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}} \frac{\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}{\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}} \frac{\sqrt{4-{\frac{5+\sqrt{5}}{2}}}}{\sqrt{4-{\frac{5+\sqrt{5}}{2}}}}$$
A: Multiply top and bottom by the denominator to get $$\frac{8}{ \sqrt{2-\sqrt{\frac{5+\sqrt 5}{2}} } }=\frac{8 {\sqrt{2-\sqrt{\frac{5+\sqrt 5}{2}}}} } {2-\sqrt{\frac{5+\sqrt 5}{2}} }.$$ Then multiply top and bottom by $$2+\sqrt{\frac{5+\sqrt 5}{2}},$$ the conjugate of the denominator, to get $$\frac{8 {\sqrt{2-\sqrt{\frac{5+\sqrt 5}{2}}}} \left(2+\sqrt{\frac{5+\sqrt 5}{2}}\right)}{4-\frac{5+\sqrt 5}{2}}.$$ This simplifies to give $$\frac{16 {\sqrt{2-\sqrt{\frac{5+\sqrt 5}{2}}}} \left(2+\sqrt{\frac{5+\sqrt 5}{2}}\right)}{3-\sqrt 5}.$$ Finally multiply by $3+\sqrt 5$ again to get $$\frac{16 {\sqrt{2-\sqrt{\frac{5+\sqrt 5}{2}}}} \left(2+\sqrt{\frac{5+\sqrt 5}{2}}\right)(3+\sqrt 5)}{9-5},$$ which simplifies to give $$4{\sqrt{2-\sqrt{\frac{5+\sqrt 5}{2}}}} \left(2+\sqrt{\frac{5+\sqrt 5}{2}}\right)(3+\sqrt 5).$$
By the way what you did was not rationalise the expression (you can't, since it's not rational). You've only rationalised the denominator.
A: Call the fraction by $r$.
$$r:=\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}=\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}\frac{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$$
$$=\frac{8\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}{2-\sqrt{\frac{5+\sqrt{5}}{2}}}=\frac{8\left(\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}\right)}{2-\sqrt{\frac{5+\sqrt{5}}{2}}}\frac{\left(2+\sqrt{\frac{5+\sqrt{5}}{2}}\right)}{2+\sqrt{\frac{5+\sqrt{5}}{2}}}$$
Call $u:=8\left(\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}\right)\left(2+\sqrt{\frac{5+\sqrt{5}}{2}}\right)$. Then, 
$$r=\frac{u}{4-\frac{5+\sqrt{5}}{2}}=\left(\frac{1}{2}\right)\frac{u}{8-5-\sqrt{5}}=\left(\frac{1}{2}\right)\frac{u}{3-\sqrt{5}}=\left(\frac{1}{2}\right)\frac{u}{3-\sqrt{5}}\frac{3+\sqrt{5}}{3+\sqrt{5}}.$$
Can you procced from here?
