9.142. $\left|\frac{x^{2}-5 x+4}{x^{2}-4}\right| \leq 1$.

Solve the inequality $\left|\frac{x^{2}-5 x+4}{x^{2}-4}\right| \leq 1$.
## Solution.

Domain of definition: $x \neq \pm 2$.

The given inequality is equivalent to the system of two inequalities

$$
\left\{\begin{array} { l } 
{ \frac { x ^ { 2 } - 5 x + 4 } { x ^ { 2 } - 4 } \leq 1 , } \\
{ \frac { x ^ { 2 } - 5 x + 4 } { x ^ { 2 } - 4 } \geq - 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{x^{2}-5 x+4}{x^{2}-4}-1 \leq 0, \\
\frac{x^{2}-5 x+4}{x^{2}-4}+1 \geq 0
\end{array} \Leftrightarrow\right.\right.
$$

$$
\begin{aligned}
& \Leftrightarrow\left\{\begin{array} { l } 
{ \frac { x ^ { 2 } - 5 x + 4 - x ^ { 2 } + 4 } { x ^ { 2 } - 4 } \leq 0 , } \\
{ \frac { x ^ { 2 } - 5 x + 4 + x ^ { 2 } - 4 } { x ^ { 2 } - 4 } \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{-5 x+8}{x^{2}-4} \leq 0, \\
\frac{2 x^{2}-5 x}{x^{2}-4} \geq 0
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
\left(x-\frac{8}{5}\right)(x-2)(x+2) \geq 0 \\
x\left(x-\frac{5}{2}\right)(x-2)(x+2) \geq 0,
\end{array} \Leftrightarrow\right.
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0667.jpg?height=228&width=460&top_left_y=673&top_left_x=462)

For the system of inequalities, graphically we obtain $x \in\left[0 ; \frac{8}{5}\right] \cup\left[\frac{5}{2} ; \infty\right)$.

Answer: $x \in\left[0 ; \frac{8}{5}\right] \cup\left[\frac{5}{2} ; \infty\right)$.