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:$${4+ 2\sqrt6} \over {\sqrt{5 + 2\sqrt6}}$$
 A: Start by multiplying the numerator and denominator of $${4+ 2\sqrt6} \over {\sqrt{5 + 2\sqrt6}}$$ by
$\sqrt{5 - 2\sqrt{6}}$. 
Then the denominator evaluates very nicely to 1. Then it's much a matter of simplifying the numerator.
Simpler yet, square the expression, simplify, and then take the square root of the result.
A: Hint: Note that, since $2+3=5$ and $2\cdot 3=6$, we have that $x=5+2\sqrt6=(\sqrt2+\sqrt3)^2$. And that the numerator $=2\sqrt2\cdot (\sqrt2+\sqrt3)$.
A: 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 $(x+1)/\sqrt{x} = 2\sqrt{3}$...
A: Hint $\ \ $ Squaring it, we get $\ \rm\dfrac{40+16\sqrt{6}}{5+2\sqrt{6}}\, =\, 8$
Remark $\ \ $ Generally $\rm\ \ \dfrac{ n\!-\!1 + \sqrt{n^2\!-\!1}}{ \sqrt{ n\ +\ \sqrt{n^2\!-\!1}}}\, =\, \sqrt{2(n\!-\!1)}\ $ for $\rm\: n\ge 1,\:$ with analogous proof.
For another example note $\, \dfrac{\quad\ 3 + \sqrt{11}}{\sqrt{10+ \sqrt{99}}}\, =\, \sqrt{2},\ $ by $\rm\:n = 10\:$ above.
A: $$\frac{x-2}{\sqrt x}=\frac{4+ 2\sqrt6}{\sqrt{5 + 2\sqrt6}}=\frac{4+ 2\sqrt6}{\sqrt{5 + 2\sqrt6}}\cdot\frac{\sqrt{5-2\sqrt 6}}{\sqrt{5-2\sqrt 6}}=\left(4+2\sqrt 6\right)\sqrt{5-2\sqrt 6}\Longrightarrow$$
$$(x-2)^2=x(40+16\sqrt 6)(5-2\sqrt 6)=x(8)\Longrightarrow $$
$$x^2-12x+4=0\ldots...$$
Take it from here
A: The square of $\sqrt{x} - {1 \over \sqrt{x}}$ is $x - 2 + {1 \over x}$. In this case $x = 5 + 2\sqrt{6}$, whose reciprocal is seen to be $5 - 2\sqrt{6}$ by rationalizing the denominator.
So 
$$x - 2 + {1 \over x} = (5 + 2\sqrt{6}) - 2 + (5 - 2\sqrt{6})$$
$$= 8$$
So your answer is $\sqrt{8} = 2\sqrt{2}$. (You take the positive square root since $x > 1$).
