9.148. $\frac{x^{4}+3 x^{3}+4 x^{2}-8}{x^{2}}<0$.

9.148. $\frac{x^{4}+3 x^{3}+4 x^{2}-8}{x^{2}}<0$. 

The inequality is $\frac{x^{4}+3 x^{3}+4 x^{2}-8}{x^{2}}<0$.
Solution.

The given inequality is equivalent to a system of two inequalities

$$
\begin{aligned}
& \left\{\begin{array}{l}
x^{4}+3 x^{3}+4 x^{2}-8<0, \\
x \neq 0
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
\left(x^{4}-x^{3}\right)+\left(4 x^{3}-4\right)+\left(4 x^{2}-4\right)<0 \\
x \neq 0
\end{array}\right. \\
& \qquad\left\{\begin{array}{l}
x^{3}(x-1)+4(x-1)\left(x^{2}+x+1\right)+4(x-1)(x+1)<0, \\
x \neq 0
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
(x-1)\left(x^{3}+4 x^{2}+8 x+8\right)<0, \\
x \neq 0
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
(x-1)\left(\left(x^{3}+8\right)+\left(4 x^{2}+8 x\right)\right)<0, \\
x \neq 0
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
(x-1)\left((x+2)\left(x^{2}-2 x+4\right)+4 x(x+2)\right)<0, \\
x \neq 0
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ ( x - 1 ) ( x + 2 ) ( x ^ { 2 } + 2 x + 4 ) < 0 , } \\
{ x \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x-1)(x+2)<0 \\
x \neq 0
\end{array}\right.\right.
\end{aligned}
$$

Using the interval method, we get $x \in(-2 ; 0) \cup(0 ; 1)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0671.jpg?height=89&width=452&top_left_y=151&top_left_x=388)

Answer: $x \in(-2 ; 0) \cup(0 ; 1)$.