If $x = 3 + 2\sqrt{2}$, then the value of $\sqrt{x} -\frac{1}{\sqrt{x}}$ is:

2

When a fraction is in the form of $x = a + b$ and the difference between the square of a and square of b is equal to 1 then we can say, \(\frac{1}{x}\) = a - b

If $x = 3 + 2\sqrt{2}$,

then the value of $\sqrt{x} -\frac{1}{\sqrt{x}}$

If $x = 3 + 2\sqrt{2}$

then, \(\frac{1}{x}\) = 3 - $2\sqrt{2}$

So, x + \(\frac{1}{x}\) = 3 + $2\sqrt{2}$ + 3 - $2\sqrt{2}$ = 6

and, $\sqrt{x} -\frac{1}{\sqrt{x}}$ = \(\sqrt {6 - 2}\) = 2