2. Let $a$ and $b$ be given real numbers such that $a \neq 0, b \neq 0, a \neq b$ and $a \neq -b$. Solve the equation

$$
\frac{1}{x}+\frac{1}{x+a}+\frac{1}{x+b}+\frac{1}{x+a+b}=0
$$
Solution. For the expression $L$ on the left side of the equation, we have:

$$
L=\left(\frac{1}{x}+\frac{1}{x+a+b}\right)+\left(\frac{1}{x+a}+\frac{1}{x+b}\right)=\frac{2 x+a+b}{x(x+a+b)}+\frac{2 x+a+b}{(x+a)(x+b)}=\frac{(2 x+a+b)\left[2 x^{2}+2(a+b) x+a b\right]}{x(x+a)(x+b)(x+a+b)}
$$

According to this, the given equation is equivalent to the equation

$$
(2 x+a+b)\left[2 x^{2}+2(a+b) x+a b\right]=0
$$

whose solutions are:

$$
x_{1}=-\frac{a+b}{2}, x_{2 / 3}=\frac{-(a+b) \pm \sqrt{a^{2}+b^{2}}}{2}
$$

From the assumptions about $a$ and $b$ given in the problem, it follows that the roots $x_{1}, x_{2}, x_{3}$ are different from zero, from $-a$, from $-b$, and from $-(a+b)$.