$1 \cdot 86$ Simplify $\frac{\sqrt{6}+4 \sqrt{3}+3 \sqrt{2}}{(\sqrt{6}+\sqrt{3})(\sqrt{3}+\sqrt{2})}$, the result is
(A) $\frac{2}{3}$.
(B) $\frac{2 \sqrt{3}}{3}$.
(C) $\sqrt{6}+\sqrt{2}$.
(D) $\sqrt{6}-\sqrt{2}$.
(China Beijing Junior High School Mathematics Competition, 1986)
[Solution]
$$
\begin{aligned}
\text { Original expression } & =\frac{(\sqrt{6}+\sqrt{3})+3(\sqrt{3}+\sqrt{2})}{(\sqrt{6}+\sqrt{3})(\sqrt{3}+\sqrt{2})} \\
& =\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{3}{\sqrt{6}+\sqrt{3}} \\
& =\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{3(\sqrt{6}-\sqrt{3})}{3} \\
& =\sqrt{6}-\sqrt{2} .
\end{aligned}
$$

Therefore, the correct choice is $(D)$.