8.6 $\quad \frac{x^{2}-3 x+2}{x^{2}+3 x+2} \geq 1$.
8.6 Let's move 1 to the left side of the inequality and perform the transformations:

$$
\frac{x^{2}-3 x+2}{x^{2}+3 x+2} \geq 1 \Rightarrow \frac{x^{2}-3 x+2}{x^{2}+3 x+2}-1 \geq 0 \Rightarrow \frac{6 x}{(x+1)(x+2)} \leq 0
$$

We apply the method of intervals and, using Fig. 8.6, we establish that \( x < -2 \) or \( -1 < x \leq 0 \).

Answer: \( (-\infty, -2) \cup (-1, 0] \).

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-270.jpg?height=100&width=471&top_left_y=902&top_left_x=364)

Fig. 8.6