$9.261 \log _{1 / 2} \frac{\left|x^{2}-2 x\right|+4}{|x+2|+x^{2}} \leq 0$
Solution. The inequality is equivalent to the following:

$$
\begin{aligned}
& \frac{\left|x^{2}-2 x\right|+4}{|x+2|+x^{2}} \geq 1 \Leftrightarrow\left|x^{2}-2 x\right| \geq x^{2}+|x+2|-4 \Leftrightarrow \\
& \Leftrightarrow\left[\begin{array} { l } 
{ x ^ { 2 } - 2 x \geq x ^ { 2 } + | x + 2 | - 4 , } \\
{ x ^ { 2 } - 2 x \leq - x ^ { 2 } - | x + 2 | + 4 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
|x+2| \leq 4-2 x, \\
|x+2| \leq-2 x^{2}+2 x+4
\end{array} \Leftrightarrow\right.\right.
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-206.jpg?height=247&width=1066&top_left_y=1489&top_left_x=110)
$\Leftrightarrow\left[\begin{array}{l}x \leq \frac{2}{3}, \\ \frac{1-\sqrt{17}}{4} \leq x \leq \frac{1+\sqrt{17}}{4}\end{array} \Leftrightarrow x \leq \frac{1+\sqrt{17}}{4}\right.$.

Answer: $\quad x \in\left(-\infty ; \frac{1+\sqrt{17}}{4}\right]$.