Solve the following equation:

$$
(x-5)\left|\frac{x-2}{x-5}\right|+\left|\frac{1}{x-2}\right|=2
$$
As $x=5$ is clearly not a solution to the equation. If $x>5$, then

$$
\frac{x-5}{|x-5|}=1
$$

therefore, in this case, our equation becomes:

$$
|x-2|+\frac{1}{|x-2|}=2
$$

Based on the relationship between the arithmetic mean and the geometric mean, we know that the sum of a positive number and its reciprocal is only 2 if the number is 1, i.e., $|x-2|=1$.

If $x>5$, then $|x-2|>3$, so we do not get a solution in this case either.

If $x<5$, then $\frac{x-5}{|x-5|}=-1$, so in this case

$$
-|x-2|+\frac{1}{|x-2|}=2
$$

from which we get the quadratic equation

$$
|x-2|^{2}+2|x-2|-1=0
$$

in terms of $|x-2|$. The only positive root of this equation is $|x-2|=\sqrt{2}-1$. Therefore, $x=1+\sqrt{2}$, or $x=3-\sqrt{2}$. Both numbers are less than 5, so they are also solutions to the original equation.

Note. With a bit more calculation, we can obtain the result if we divide the domain of the equation into the following three parts: $x<2, 2<x<5, 5<x$.