1952 AMC 12 第 35 题

First treat the denominator as $(\sqrt2+\sqrt3)-\sqrt5$ and multiply by its conjugate

After the first rationalization, a denominator containing $\sqrt6$ remains; use another conjugate

Multiply first by the conjugate $\sqrt2+\sqrt3+\sqrt5.$ This gives $$\begin{aligned} &\frac{\sqrt2(\sqrt2+\sqrt3+\sqrt5)} {(\sqrt2+\sqrt3)^2-5}\\ &\qquad=\frac{2+\sqrt6+\sqrt{10}}{2\sqrt6}. \end{aligned}$$ Multiplying numerator and denominator by $\sqrt6$ yields $$\begin{gathered} \frac{2\sqrt6+6+\sqrt{60}}{12} \\ =\frac{3+\sqrt6+\sqrt{15}}6. \end{gathered}$$

Thus, the correct answer is A.