9.167. $\frac{1}{x+1}-\frac{2}{x^{2}-x+1} \leq \frac{1-2 x}{x^{3}+1}$.
## Solution.

Domain of definition: $x \neq-1$.

From the condition we have $\frac{x^{2}-x+1-2 x-2}{x^{3}+1} \leq \frac{1-2 x}{x^{3}+1}$,

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

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