$7 \cdot 100$ If the function $f$ is defined as
$$
f(x)=\frac{c x}{2 x+3}, \quad x \neq-\frac{3}{2} .
$$

and for all real numbers $x$ except $-\frac{3}{2}$, it satisfies $f[f(x)]=x$, then $c$ is
(A) -3 .
(B) $-\frac{3}{2}$.
(C) $\frac{3}{2}$.
(D) 3 .
(E) No unique solution according to the given conditions.
(31st American High School Mathematics Examination, 1980)
[Solution] For all real numbers $x \neq -\frac{3}{2}$, according to the problem, we have
$$
x=f[f(x)]=\frac{c\left(\frac{c x}{2 x+3}\right)}{2\left(\frac{c x}{2 x+3}\right)+3}=\frac{c^{2} x}{2 c x+6 x+9} \text {, }
$$

From this, we can derive $(2 c+6) x+\left(9-c^{2}\right)=0$.
Therefore, by the arbitrariness of $x$, we have $2 c+6=0, 9-c^{2}=0$.
Solving these equations, we get
$$
c=-3 \text {. }
$$

Thus, the answer is $(A)$.