Example 12. Solve the equation

$$
\left|\frac{x}{x-1}\right|+|x|=\frac{x^{2}}{|x-1|}
$$
Solution. Since

$$
\begin{aligned}
& \frac{x^{2}}{x-1}=\frac{x+x^{2}-x}{x-1}=\frac{x+x(x-1)}{x-1}=\frac{x}{x-1}+x \\
& \frac{x^{2}}{|x-1|}=\frac{\left|x^{2}\right|}{|x-1|}=\left|\frac{x^{2}}{x-1}\right|
\end{aligned}
$$

the equation (18) can be rewritten as

$$
\left|\frac{x}{x-1}\right|+|x|=\left|\frac{x}{x-1}+x\right|
$$

Using statement 2, we obtain the equivalent inequality

$$
\frac{x}{x-1} x \geqslant 0
$$

i.e.

$$
\frac{x^{2}}{x-1} \geqslant 0
$$

The solutions to this inequality, and thus to the original equation, are the number 0 and all numbers from the interval $(1 ;+\infty)$.