9.048. $\frac{4-x}{x-5}>\frac{1}{1-x}$.
## Solution.

Domain of definition: $\quad x \neq 1, x \neq 5$.

From the condition

$$
\begin{aligned}
& \frac{4-x}{x-5}-\frac{1}{1-x}>0, \frac{(4-x)(1-x)-x+5}{(x-5)(1-x)}>0, \frac{x^{2}-6 x+9}{(x-5)(x-1)}<0 \\
& \frac{(x-3)^{2}}{(x-5)(x-1)}<0,(x-3)^{2}(x-5)(x-1)<0
\end{aligned}
$$

Using the interval method, we find $x \in(1 ; 3) \cup(3 ; 5)$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-534.jpg?height=158&width=600&top_left_y=1425&top_left_x=394)

Answer: $\quad x \in(1 ; 3) \cup(3 ; 5)$.