2・112 The equation $2 x^{5}+x^{4}-20 x^{3}-10 x^{2}+2 x+1=0$ has a real root which is
(A) $\sqrt{5}+\sqrt{3}$.
(B) $\sqrt{5}+\sqrt{2}$.
(C) $\sqrt{3}+\sqrt{2}$.
(D) $\sqrt{5}-\sqrt{3}$.
(2nd "Hope Cup" National Mathematics Invitational, 1991)
［Solution］The original equation can be transformed into
$$
\left(2 x^{5}-20 x^{3}+2 x\right)+\left(x^{4}-10 x^{2}+1\right)=0,
$$

which is
$$
(2 x+1)\left(x^{4}-10 x^{2}+1\right)=0 \text {. }
$$

Noting that for $\quad x=\sqrt{3}+\sqrt{2}$, we have $(x-\sqrt{3})^{2}=(\sqrt{2})^{2}$, i.e., $x^{2}-2 \sqrt{3} x+3=2$ or $\left(x^{2}+1\right)^{2}=12 x^{2}$, then
$$
x^{4}-10 x^{2}+1=0 \text {. }
$$

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