Solve the following equation:

$$
\frac{3}{(x+2)(x-1)}=\frac{1}{x(x-1)^{2}}+\frac{3}{x(x-3)}
$$
A common denominator: $x(x-1)^{2}(x+2)(x+3)$. Bring all fractions to a common denominator, placing all on the left side:

$$
\frac{3 x(x-1)(x-3)-(x+2)(x-3)-3(x-1)^{2}(x+2)}{x(x-1)^{2}(x+1)(x+3)}=0 \ldots
$$

After performing the indicated operations and simplifying:

$$
\frac{13 x^{2}-19 x}{x(x-1)^{2}(x+2)(x+3)}=0 \ldots
$$

The numerator will be zero when $x=0$ and when $x=\frac{19}{13}$.

However, at $x=0$, the denominator also vanishes: if we remove the common factor, $x$, from the numerator and denominator, then the value of the left side will be $\frac{19}{6}$ when $x=0$.

Thus, only $x=\frac{19}{13}$ is a root of the equation.

We can also note that the original equation is satisfied by $x=\infty$.

Tibor Turmezei (Cistercian St. Imre High School, Budapest XI.)

Note. $1^{\circ}$. We have already pointed out several times that if, in solving the equation $f(x)=0$, we multiply both sides of the equation by $g(x)$, then the roots of the equation $g(x) f(x)=0$ are not the same as the roots of the original equation, because among the roots of this equation, in addition to the roots of $f(x)=0$, the roots of $g(x)=0$ also appear. Therefore, if, in the process of rearranging the equation, - to obtain whole expressions in $x$, - we multiply both sides of the equation by $g(x)$, then we must exclude the roots of $g(x)=0$ from among the roots of the equation thus obtained!

This case is present here. If $g(x)$ represents the least common multiple of the denominators of the fractions in the given equation, then $x=0$ is a root of the equation $g(x)=0$, but not a root of the original equation.

In some solutions, the least common multiple of the denominators was not used in rearranging the equation; hence, in addition to $x=0$, $x=1$ also appeared among the roots.

$2^{\circ}$. In solving the equation $a x^{2}+b x=0$ or $a x^{2}=-b x$, we must not say that we simplify by $x$, because this way one root of the quadratic equation (i.e., $x=0$) disappears.