<s>[INST] How to simplify $\frac{4 + 2\sqrt6}{\sqrt{5 + 2\sqrt{6}}}$? I was tackling through an olympiad practice book when I saw one of these problems:

If  $x = 5 + 2\sqrt6$, evaluate $\Large{x  \ - \ 1 \over\sqrt{x}}$?

The answer written is $2\sqrt2$, but I can't figure my way out through the manipulations. I just know that I have the following:$${4+ 2\sqrt6} \over {\sqrt{5 + 2\sqrt6}}$$
 [/INST] When you know the answer, it is often easier...
Indeed, begin with $(x-1)/\sqrt{x} = 2\sqrt{2}$
It is equivalent to $x-1 = 2\sqrt{2x}$
Equivalent to $(x-1)^2 = 8x$
Equivalent to $x^2-2x+1 = 8x$
Equivalent to $x^2-10x+1 = 0$
Equivalent to $(x-a)(x-b) = 0$ where $a = 5+2\sqrt{6}$ and $b = 5-2\sqrt{6}$
Equivalent to {$x=a$ or $x=b$}
Of course, since these are equivalences, this suffices to prove the fact. However, if you want to write it in a more straightforward way, it suffices to go back from the end to the beginning:
Let $x = 5+2\sqrt{6}$. Then $(x-(5+2\sqrt{6})(x-(5-2\sqrt{6}))=0$. But this product can be expanded as $x^2-10x+1$. So, we have $x^2+1 = 10x$. At this point, you can generate various identities. For instance, you could conclude $x^2+2x+1 = 12x$. Therefore $(x+1)^2=12x$ and $x+1 = 2\sqrt{3x}$ and so $(x+1)/\sqrt{x} = 2\sqrt{3}$...
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