9.171. $0.6^{\lg ^{2}(-x)+3} \leq\left(\frac{5}{3}\right)^{2 \lg x^{2}}$.
## Solution.

Domain of definition: $x<0$.

Since $\lg x^{2 k}=2 k \lg |x|$, taking into account the domain of definition, we can rewrite the given inequality as

$\left(\frac{3}{5}\right)^{\lg ^{2}(-x)+3} \leq\left(\frac{3}{5}\right)^{-4 \lg (-x)} \Leftrightarrow \lg ^{2}(-x)+3 \geq-4 \lg (-x), \lg ^{2}(-x)+4 \lg (-x)+3 \geq 0$.

Solving it as a quadratic inequality in terms of $\lg (-x)$, we get

$\left[\begin{array}{l}\lg (-x) \geq-1, \\ \lg (-x) \leq-3\end{array} \Leftrightarrow\left[\begin{array}{l}-x \geq 0.1, \\ -x \leq 0.001\end{array} \Leftrightarrow\left[\begin{array}{l}x \leq-0.1, \\ x \geq-0.001\end{array}\right.\right.\right.$

Answer: $x \in(-\infty ;-0.1] \cup[-0.001 ; 0)$.