9.050. $\frac{x^{2}-3 x+2}{x^{2}+3 x+2} \geq 1$.

Solve the inequality:

9.050. $\frac{x^{2}-3 x+2}{x^{2}+3 x+2} \geq 1$.
Solution.

Domain of definition: $x \neq -2, x \neq -1$.

We have

$\frac{x^{2}-3 x+2}{x^{2}+3 x+2}-1 \geq 0 \Leftrightarrow \frac{-6 x}{x^{2}+3 x+2} \geq 0, \frac{x}{(x+2)(x+1)} \leq 0,\left\{\begin{array}{l}x(x+2)(x+1) \leq 0, \\ x \neq -2, \\ x \neq -1 .\end{array}\right.$

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

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-535.jpg?height=150&width=612&top_left_y=1097&top_left_x=320)

Answer: $\quad x \in(-\infty ;-2) \cup(-1 ; 0]$

9.05i. $\left(\left(\frac{3}{7}\right)^{\frac{1}{x^{2}}}\right)^{x^{2}-2 x} \geq 1$.

Solution.

Domain of definition: $\quad x \neq 0$.

From the condition

$\left(\frac{3}{7}\right)^{\frac{x^{2}-2 x}{x^{2}}} \geq\left(\frac{3}{7}\right)^{0} \Leftrightarrow \frac{x^{2}-2 x}{x^{2}} \leq 0, \frac{x-2}{x} \leq 0,\left\{\begin{array}{l}(x-2) x \leq 0, \\ x \neq 0 .\end{array}\right.$

Answer: $\quad x \in(0 ; 2]$