Solve the following equation:

$$
\frac{x+2}{x-2}-\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x-2}{x+2}=2\left(\frac{1}{x-2}-\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x+2}\right)
$$
First, let's rearrange - after combining - the fractions with the same denominator, and then those whose denominators differ only in sign:

$$
\frac{x}{x-2}+\frac{x-4}{x+2}=1+\frac{x-3}{x+1}, \quad \frac{x^{2}-2 x+4}{x^{2}-4}=\frac{x-1}{x+1}
$$

Subtracting one from both sides and simplifying again, we get

$$
\frac{x-4}{x^{2}-4}=\frac{1}{x+1}
$$

Since the difference between the two sides here is $3 x /\left(x^{2}-4\right)(x+1)$, the only solution to the equation is $x=0$.

Remark. If we start the solution by eliminating the fractions, the resulting equation will also have the value $x=1$ as a solution. In our solution, we not only avoided this by keeping the fractions throughout, but also because, by proceeding mechanically, we would now, as usual, need significantly more numerical work.