$2 \cdot 33$ Given the quadratic equation
$$
(2 \sqrt{3}+\sqrt{2}) x^{2}+2(\sqrt{3}+\sqrt{2}) x+(\sqrt{2}-2 \sqrt{3})=0 \text {, }
$$

then
(A) The sum of the roots is $-\frac{4+\sqrt{6}}{5}$, and the product is $\frac{7-2 \sqrt{6}}{5}$.
(B) The sum of the roots is $\frac{4+\sqrt{6}}{5}$, and the product is $\frac{2 \sqrt{6}-7}{5}$.
(C) The sum of the roots is $-2(\sqrt{3}+\sqrt{2})$, and the product is $2 \sqrt{3}-\sqrt{2}$.
(D) None of the above answers is correct.
(China Fujian Province Junior High School Mathematics Competition, 1983)
[Solution] Let the two roots of the equation be $x_{1}, x_{2}$, then by Vieta's formulas and the given conditions, we have
$$
\begin{aligned}
x_{1}+x_{2} & =-\frac{2(\sqrt{3}+\sqrt{2})}{2 \sqrt{3}+\sqrt{2}}=-1-\frac{\sqrt{2}}{2 \sqrt{3}+\sqrt{2}}=-\frac{4+\sqrt{6}}{5}, \\
x_{1} x_{2} & =\frac{\sqrt{2}-2 \sqrt{3}}{2 \sqrt{3}+\sqrt{2}}=-\frac{2 \sqrt{3}-\sqrt{2}}{2 \sqrt{3}+\sqrt{2}}=-1+\frac{2 \sqrt{2}}{2 \sqrt{3}+\sqrt{2}} \\
& =\frac{2 \sqrt{6}-7}{5} .
\end{aligned}
$$

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