9.131. $\log _{|x-1|} 0.5>0.5$.
## Solution.

From the condition, we have two systems of inequalities:

1) $\left\{\begin{array}{l}01, \\ 0.5>\sqrt{|x-1|}\end{array} \Leftrightarrow\left\{\begin{array}{l}|x-1|>1, \\ |x-1|1.25, \\
x<0.75
\end{array}\right.}
\end{array}\right.
$

Using the interval method, we get $x \in(0 ; 0.75) \cup(1.25 ; 2)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0660.jpg?height=129&width=503&top_left_y=141&top_left_x=435)

Answer: $x \in(0 ; 0.75) \cup(1.25 ; 2)$.