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I've tried to see the root as

$\sqrt{\frac{5+\sqrt{5}}{2}} = \sqrt{a}+\sqrt{b},$

but this method doesn't give me something good.

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We can write it so: $$\sqrt{\frac{5+\sqrt{5}}{2}}=\sqrt{\frac{5+\sqrt{5}}{2}}+\sqrt0.$$

Let there be rationals $a$ and $b$ for which $\sqrt{\frac{5+\sqrt{5}}{2}}=\sqrt{a}+\sqrt{b}.$

Thus, $$\frac{5+\sqrt5}{2}=a+b+2\sqrt{ab},$$ which gives $$a+b=\frac{5}{2}$$ and $$ab=\frac{5}{16},$$ which says that $a$ and $b$ are roots of the equation $$x^2-\frac{5}{2}x+\frac{5}{16}=0,$$ which is a contradiction because it is easy to see that this equation has no rational roots.

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It can't give you anything good.

Say you have $\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}$, all numbers rational and $b$ not a rational square. Then

$a+\sqrt{b}=(\sqrt{x}+\sqrt{y})^2=(x+y)+2\sqrt{xy}$

The uniqueness of quadratic surds forces $a=x+y,b=4xy$. Then

$a^2-b=(x+y)^2-4xy=(x-y)^2,\therefore\text{ a rational square.}$

But here, $a=5/2$ and $b=5/4$, giving $a^2-b=5$. So we're having a bad day.

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