9.173. $\left(\frac{3}{5}\right)^{13 x^{2}} \leq\left(\frac{3}{5}\right)^{x^{4}+36}<\left(\frac{3}{5}\right)^{12 x^{2}}$.
Solution.

The given inequality is equivalent to the inequality $13 x^{2} \geq x^{4}+36>12 x^{2} \Leftrightarrow$ $\Leftrightarrow\left\{\begin{array}{l}x^{4}+36>12 x^{2}, \\ x^{4}+36 \leq 13 x^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{4}-12 x^{2}+36>0, \\ x^{4}-13 x^{2}+36 \leq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}\left(x^{2}-6\right)^{2}>0, \\ 4 \leq x^{2} \leq 9\end{array} \Leftrightarrow\right.\right.\right.$ $\Leftrightarrow\left\{\begin{array}{l}x^{2}-6 \neq 0, \\ x^{2} \leq 9, \\ x^{2} \geq 4\end{array} \Leftrightarrow\left\{\begin{array}{l}x \neq \pm \sqrt{6}, \\ -3 \leq x \leq 3, \\ {\left[\begin{array}{l}x \geq 2, \\ x \leq-2 .\end{array}\right.}\end{array}\right.\right.$

Using the interval method, we get $x \in[-3 ;-\sqrt{6}) \cup(-\sqrt{6} ;-2] \cup[2 ; \sqrt{6}) \cup(\sqrt{6} ; 3]$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0686.jpg?height=134&width=504&top_left_y=809&top_left_x=402)

Answer: $x \in[-3 ;-\sqrt{6}) \cup(-\sqrt{6} ;-2] \cup[2 ; \sqrt{6}) \cup(\sqrt{6} ; 3]$.