If

then $x$ equals:

Let's simplify the numerator of the first term, $N=\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}$.
Squaring $N$:

$$N^2=(\sqrt{5}+2)+(\sqrt{5}-2)+2\sqrt{(\sqrt{5}+2)(\sqrt{5}-2)}$$

$$N^2=2\sqrt{5}+2\sqrt{(\sqrt{5}{)}^2-2^2}$$

$$N^2=2\sqrt{5}+2\sqrt{5-4}=2\sqrt{5}+2=2(\sqrt{5}+1)$$

Taking the square root on both sides:

$$N=\sqrt{2}\sqrt{\sqrt{5}+1}$$

Substitute $N$ back into the first term of the expression for $x$:

$$\text{First Term}=\frac{\sqrt{2}\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}}=\sqrt{2}$$

Now simplify the second term:

$$\sqrt{3-2\sqrt{2}}=\sqrt{2+1-2\sqrt{2}}=\sqrt{(\sqrt{2}{)}^2+(1{)}^2-2(\sqrt{2})(1)}=\sqrt{(\sqrt{2}-1{)}^2}=\sqrt{2}-1$$

Combining both parts:

$$x=\sqrt{2}-(\sqrt{2}-1)=1$$