$2 \cdot 57$ The larger root of the equation $(1984 x)^{2}-1983 \cdot 1985 x-1=0$ is $r$, and the smaller root of $1983 x^{2}-1984 x+1=0$ is $s$, then $r-s$ equals
(A) $\frac{1984}{1983}$.
(B) $\frac{1985}{1984}$.
(C) $\frac{1982}{1983}$.
(D) $\frac{1983}{1984}$.
(E) 0 .
(China Beijing Junior High School Mathematics Competition, 1984)
[Solution] Since $1984^{2}-1983 \cdot 1985-1=0$, 1 is a root of the first equation.
Given that $-\frac{1}{1984^{2}}<0$, we know the other root is negative. Therefore, $r=1$.
Observing, we see that 1 is a root of the first equation, and $\frac{1}{1983}$ is the other root, so
$$
s=\frac{1}{1983}, \text { and } \quad r-s=\frac{1982}{1983} .
$$

Thus, the answer is $(C)$.