9.150. $\frac{\log _{0.3}|x-2|}{x^{2}-4 x}<0$.
## Solution.

Domain of definition: $|x-2|>0, x \neq 2, x \neq 0, x \neq 4$.

The roots of the equations $\log _{0.3}|x-2|=0$ and $x^{2}-4 x=0$ are the numbers $x_{1}=1$, $x_{2}=3, x_{3}=0, x_{4}=4, x_{5}=2$ - does not fit the domain of definition.

These points divide the number line into 6 intervals. We select a value of $x$ from each interval and, substituting it into the given inequality, determine the sign of the inequality on the considered interval. Using the interval method, we get that $x \in(-\infty ; 0) \cup(1 ; 2) \cup(2 ; 3) \cup(4 ; \infty)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0672.jpg?height=87&width=495&top_left_y=336&top_left_x=431)

Answer: $x \in(-\infty ; 0) \cup(1 ; 2) \cup(2 ; 3) \cup(4 ; \infty)$ :