Example 14. Solve the inequality

$$
\frac{|x-3|}{x^{2}-5 x+6} \geq 2
$$
Solution. Inequality (12) is equivalent to the combination of two systems:

![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-137.jpg?height=185&width=891&top_left_y=2052&top_left_x=588)

For the first system of this combination, we get:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x - 3 \geqslant 0 , } \\
{ \frac { x - 3 } { ( x - 2 ) ( x - 3 ) } \geqslant 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x-3>0 \\
\frac{1}{x-2} \geqslant 2
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x - 3 > 0 } \\
{ \frac { 5 - 2 x } { x - 2 } \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x-3>0 \\
\frac{x-5 / 2}{x-2} \leqslant 0
\end{array}\right.\right.
\end{aligned}
$$

The obtained system has no solutions.

Let's solve the second system of the combination. We have:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x - 3 < 0 } \\
{ \frac { - ( x - 3 ) } { ( x - 2 ) ( x - 3 ) } \geqslant 2 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x - 3 < 0 } \\
{ \frac { - 1 } { x - 2 } \geqslant 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x-3<0 \\
\frac{2 x-3}{x-2} \leqslant 0
\end{array} \Leftrightarrow\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x-3<0 \\
\frac{x-3 / 2}{x-2} \leqslant 0
\end{array} \Leftrightarrow 3 / 2 \leqslant x<2\right.
\end{aligned}
$$

Thus, the solutions to inequality (12) are all numbers \( x \) from the interval \( 3 / 2 \leqslant x < 2 \).