$8.8 \quad \frac{1}{x^{2}-5 x+6} \leq \frac{1}{2}$.
8.8 Transform the given inequality into an equivalent one:

$$
\frac{1}{x^{2}-5 x+6}-\frac{1}{2} \leq 0 ; \frac{x^{2}-5 x+4}{2\left(x^{2}-5 x+6\right)} \geq 0
$$

The roots \( x_{1}=1, x_{2}=4 \) of the equation \( x^{2}-5 x+4=0 \) are solutions to the inequality (on Fig. 8.7, we mark them with filled circles); the roots \( x_{3}=2, x_{4}=3 \) of the equation \( x^{2}-5 x+6=0 \) are not solutions to the inequality (on Fig. 8.7, we mark them with open circles). Using the sign curve, we obtain the solution to the original inequality.

Answer: $(-\infty, 1] \cup(2,3) \cup[4, \infty)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-271.jpg?height=103&width=586&top_left_y=485&top_left_x=307)

Fig. 8.7