$1.82 \frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$ is equal to
(A) $\sqrt{2}+\sqrt{3}-\sqrt{5}$.
(B) $4-\sqrt{2}-\sqrt{3}$.
(C) $\sqrt{2}+\sqrt{3}+\sqrt{6}-\sqrt{5}$.
(D) $\frac{1}{2}(\sqrt{2}+\sqrt{5}-\sqrt{3})$.
(E) $\frac{1}{3}(\sqrt{3}+\sqrt{5}-\sqrt{2})$.
(35th American High School Mathematics Examination, 1984)
[Solution] Notice that $(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{3}-\sqrt{5})$
$$
\begin{array}{l}
=(\sqrt{2}+\sqrt{3})^{2}-5 \\
=(5+2 \sqrt{6})-5 \\
=2 \sqrt{6}
\end{array}
$$

Then
$$
\begin{aligned}
\text { Original expression } & =\frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}} \cdot \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{2}+\sqrt{3}-\sqrt{5}} \\
& =\sqrt{2}+\sqrt{3}-\sqrt{5} .
\end{aligned}
$$

Therefore, the answer is $(A)$.