1.83 When rationalizing the denominator, the expression $\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$ is equivalent to
(A) $\frac{3+\sqrt{6}+\sqrt{15}}{6}$.
(B) $\frac{\sqrt{6}-2+\sqrt{10}}{6}$.
(C) $\frac{2+\sqrt{6}+\sqrt{10}}{10}$.
(D) $\frac{2+\sqrt{6}-\sqrt{10}}{6}$.
(E) None of the above.
(3rd American High School Mathematics Examination, 1952)
$$
\begin{aligned}
& \frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}} \\
= & \frac{\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{5})}{(\sqrt{2}+\sqrt{3}-\sqrt{5})(\sqrt{2}+\sqrt{3}+\sqrt{5})} \\
= & \frac{2+\sqrt{6}+\sqrt{10}}{(\sqrt{2}+\sqrt{3})^{2}-5} \\
= & \frac{2+\sqrt{6}+\sqrt{10}}{2 \sqrt{6}} \\
= & \frac{\sqrt{6}(2+\sqrt{6}+\sqrt{10})}{2 \sqrt{6} \cdot \sqrt{6}} \\
= & \frac{\sqrt{6}+3+\sqrt{15}}{6} .
\end{aligned}
$$

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