$-\sqrt{3-2 \sqrt{2}}$, then $N=(\quad)$.
(A) 1
(B) $2 \sqrt{2}-1$
(C) $\frac{\sqrt{5}}{2}$
(D) $\sqrt{\frac{5}{2}}$
\begin{array}{l}-1 . \ A. \\ \because\left(\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\right)^{2} \\ =\frac{2 \sqrt{5}+2}{\sqrt{5}+1}=2 . \\ \therefore N=\sqrt{2}-\sqrt{3-2 \sqrt{2}} \\ =\sqrt{2}-(\sqrt{2}-1)=1 . \\\end{array}