$2 \cdot 58$ If the larger root of the equation $(1992 x)^{2}-1991 \cdot 1993 x-1=0$ is $m$, and the smaller root of the equation $x^{2}+1991 x-1992=0$ is $n$, then $m-n$ equals
(A) 1993.
(B) 1992.
(C) $\frac{1992}{1993}$.
(D) $\frac{1991}{1992}$.
(China Sichuan Junior High School Mathematics Competition, 1992)
[Solution] For the first equation, the larger root is
$$
m=\frac{1991 \cdot 1993+\sqrt{1991^{2} \cdot 1993^{2}+4 \cdot 1992^{2}}}{2 \cdot 1992^{2}}.
$$

For the second equation, the smaller root is
$$
n=\frac{-1991-\sqrt{1991^{2}+4 \cdot 1992}}{2},
$$

then
$$
\begin{array}{l}
m-n \\
=\frac{(1992-1)(1992+1)+\sqrt{(1992-1)^{2}(1992+1)^{2}+4 \cdot 1992^{2}}}{2 \cdot 1992^{2}} \\
+\frac{1991+\sqrt{(1992-1)^{2}}+4 \cdot 1992}{2} \\
=\frac{\left(1992^{2}-1\right)+\left(1992^{2}+1\right)}{2 \cdot 1992^{2}}+\frac{1991+(1992+1)}{2} \\
=1+1992 \\
=1993. \\
\text { Therefore, the answer is }(A). \\
\end{array}
$$