9.144. $\frac{\left(x-\frac{1}{2}\right)(3-x)}{\log _{2}|x-1|}>0$.
Solution.

Domain of definition: $|x-1|>0, x \neq 1$.

The roots of the equations $\left(x-\frac{1}{2}\right)(3-x)=0$ and $\log _{2}|x-1|=0$ are the numbers $x_{1}=\frac{1}{2}, x_{2}=3, x_{3}=2, x_{4}=\dot{0}$, which are not solutions to the given inequality, so we mark them on the number line with hollow circles:

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0668.jpg?height=133&width=460&top_left_y=1223&top_left_x=448)

(including $x=1$). These points divide the number line into six intervals. Substituting a value of $x$ from each interval into the given inequality, we determine the sign of the inequality in the considered interval. Using the diagram, we find the answer: $x \in\left(0 ; \frac{1}{2}\right) \cup(2 ; 3)$.

$$
\text { Answer: } x \in\left(0 ; \frac{1}{2}\right) \cup(2 ; 3)
$$