1. Solve the equation

$$
\frac{3}{x}+\frac{1}{x-1}+\frac{4}{x-2}+\frac{4}{x-3}+\frac{1}{x-4}+\frac{3}{x-5}=0
$$
Solution. We have

$$
\begin{aligned}
& \frac{3}{x}+\frac{1}{x-1}+\frac{4}{x-2}+\frac{4}{x-3}+\frac{1}{x-4}+\frac{3}{x-5}=0 \\
& \frac{3}{x}+\frac{3}{x-5}+\frac{1}{x-1}+\frac{1}{x-4}+\frac{4}{x-2}+\frac{4}{x-3}=0 \\
& \frac{3(2 x-5)}{x^{2}-5 x}+\frac{2 x-5}{x^{2}-5 x+4}+\frac{4(2 x-5)}{x^{2}-5 x+6}=0
\end{aligned}
$$

From here, it is clear that one solution to the equation is $x_{1}=\frac{5}{2}$. If for $x \neq \frac{5}{2}$ we divide the equation by $2 x-5$ and introduce the substitution $x^{2}-5 x=t$, we obtain the equation

$$
\frac{3}{t}+\frac{1}{t+4}+\frac{4}{t+6}=0
$$

which, after multiplying by $t(t+4)(t+6)$, reduces to a quadratic equation with solutions $t_{1}=-\frac{9}{2}$ and $t_{2}=-2$. Substituting these values back into the substitution, we get two quadratic equations whose solutions are $x_{2 / 3}=\frac{5 \pm \sqrt{7}}{2}$ and $x_{4 / 5}=\frac{5 \pm \sqrt{17}}{2}$.

Since all obtained solutions satisfy the original equation, we conclude that the equation has five solutions.