9.162. $0.3^{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots}<\sqrt[3]{0.3^{3 x^{2}+5 x}}<1$.
Solution.
Let's write the inequality as $0.3^{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots}\frac{3 x^{2}+5 x}{3}>0 \Leftrightarrow\left\{\begin{array}{l}
1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots>\frac{3 x^{2}+5 x}{3} \\
\frac{3 x^{2}+5 x}{3}>0
\end{array}\right.
$

The sum of the terms of an infinite decreasing geometric series, where

$b_{1}=1, q=-\frac{1}{2}, S=\frac{b_{1}}{1-q}$, therefore, $1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots=\frac{1}{1-\left(-\frac{1}{2}\right)}=\frac{2}{3}$.
Then

$$
\left\{\begin{array} { l } 
{ \frac { 2 } { 3 } > \frac { 3 x ^ { 2 } + 5 x } { 3 } , } \\
{ \frac { 3 x ^ { 2 } + 5 x } { 3 } > 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 3 x ^ { 2 } + 5 x - 2 < 0 , } \\
{ 3 x ^ { 2 } + 5 x > 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
-2 < x < \frac{1}{3} \\
x < -\frac{5}{3} \text{ or } x > 0
\end{array}\right.}
\end{array}\right.\right.\right.
$$

Using the number line, we find $x \in\left(-2 ;-\frac{5}{3}\right) \cup\left(0 ; \frac{1}{3}\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0678.jpg?height=173&width=504&top_left_y=1615&top_left_x=402)

Answer: $x \in\left(-2 ;-\frac{5}{3}\right) \cup\left(0 ; \frac{1}{3}\right)$.