9.082. $\frac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}>1$.
## Solution.

Domain of definition: $x \neq-3, x \neq-2, x \neq-1$.

We have

$$
\begin{aligned}
& \frac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}-1>0 \Leftrightarrow \frac{-12 x^{2}-6}{(x+1)(x+2)(x+3)}>0 \Leftrightarrow \\
& \Leftrightarrow \frac{2 x^{2}+1}{(x+1)(x+2)(x+3)}<0 \text{. Since } 2 x^{2}+1>0 \text{ for } x \in \mathbb{R}, \text{ the obtained inequality is equivalent to the inequality } (x+1)(x+2)(x+3)<0. \text{ Using the interval method, we find } x \in(-\infty ;-3) \cup(-2 ;-1).
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-546.jpg?height=148&width=600&top_left_y=1626&top_left_x=318)

Answer: $\quad x \in(-\infty ;-3) \cup(-2 ;-1)$