9.1 $\quad A_{x}^{2} C_{x}^{x-1}=48$.

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9.1 $\quad A_{x}^{2} C_{x}^{x-1}=48$.
9.1 According to formula (9.1), we find

$$
A_{x}^{2}=\frac{x!}{(x-2)!}=\frac{x(x-1)(x-2)!}{(x-2)!}=x(x-1)
$$

Further, using formulas (9.6) and (9.4), we have

$$
C_{x}^{x-1}=C_{x}^{x-(x-1)}=C_{x}^{1}=\frac{x!}{1!(x-1)!}=x
$$

Therefore, the given equation will take the form

$x^{2}(x-1)=48$, or $x^{2}(x-1)=4^{2} \cdot 3$.

Consequently, $x=4$.

Answer: $x=4$.