Solve the following equation:

$$
\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x} .
$$
Let's eliminate the fractions:

$$
\begin{gathered}
a b x=(a+b+x)(a x+b x+a b) \\
x^{2}(a+b)+x(a+b)^{2}+a b(a+b)=0 \\
x^{2}+x(a+b)+a b=0
\end{gathered}
$$

or

$$
(x+a)(x+b)=0
$$

from which

$$
x_{1}=-a \quad \text { and } \quad x_{2}=-b \text {. }
$$

(Adolf Viola, Budapest.)

Number of solutions: 51.