$33 \cdot 46$ Let $a$ be the fractional part of $\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}$, and $b$ be the fractional part of $\sqrt{6+3 \sqrt{3}}-$ $\sqrt{6-3 \sqrt{3}}$, then the value of $\frac{2}{b}-\frac{1}{a}$ is
(A) $\sqrt{6}-\sqrt{2}+1$.
(B) $\sqrt{6}+\sqrt{2}-1$.
(C) $\sqrt{6}-\sqrt{2}-1$.
(D) $\sqrt{6}+\sqrt{2}+1$.
(Chinese Zhejiang Province Junior High School Mathematics Competition, 1991)
[Solution] From $(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}})^{2}=2$, we have $\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{2}$, thus $a=\sqrt{2}-1$. Similarly, from $(\sqrt{6+3 \sqrt{3}}-\sqrt{6-3 \sqrt{3}})^{2}=6$, we have $b=\sqrt{6}-2$. Therefore, $\frac{2}{b}-\frac{1}{a}=\frac{2}{\sqrt{6}-2}-\frac{1}{\sqrt{2}-1}=\sqrt{6}-\sqrt{2}+1$. Hence, the answer is $(A)$.