$9.2 \quad C_{x+1}^{x-2}+2 C_{x-1}^{3}=7(x-1)$.
9.2 Let's use formula (9.4):

$$
\frac{(x+1)!}{(x-2)!3!}+\frac{2(x-1)!}{(x-4)!3!}=7(x-1)
$$

from which, after simplifying the first fraction by $(x-2)!$ and the second fraction by $(x-4)!$, we get

$$
(x+1) x(x-1)+2(x-1)(x-2)(x-3)=42(x-1)
$$

Further, considering that $x \neq 1$, we arrive at the quadratic equation

$$
x^{2}+x+2\left(x^{2}-5 x+6\right)=42, \text{ or } x^{2}-3 x-10=0
$$

which has roots $x_{1}=5, x_{2}=-2$ (not suitable by the context of the problem).

Answer: $x=5$.