3. The irrational root of the fractional equation $\frac{1}{x-5}+\frac{2}{x-4}+\frac{3}{x-3}+\frac{4}{x-2}$ $=4$ is
3. $\frac{7 \pm \sqrt{5}}{2}$.

Transform the original equation to
$$
\begin{array}{l}
\left(1-\frac{1}{x-5}\right)+\left(1-\frac{2}{x-4}\right)+\left(1-\frac{3}{x-3}\right)+\left(1-\frac{4}{x-2}\right)=0 \\
\Leftrightarrow \frac{x-6}{x-5}+\frac{x-6}{x-4}+\frac{x-6}{x-3}+\frac{x-6}{x-2}=0 \\
\Leftrightarrow(x-6)\left[\left(\frac{1}{x-5}+\frac{1}{x-2}\right)+\left(\frac{1}{x-4}+\frac{1}{x-3}\right)\right]=0 \\
\Leftrightarrow(x-6)\left(\frac{2 x-7}{x^{2}-7 x+10}+\frac{2 x-7}{x^{2}-7 x+12}\right)=0 \\
\Leftrightarrow(x-6)(2 x-7)\left(\frac{1}{x^{2}-7 x+10}+\frac{1}{x^{2}-7 x+12}\right)=0 \\
\Leftrightarrow(x-6)(2 x-7)\left(x^{2}-7 x+11\right)=0 . \\
\text { Therefore, } x_{1}=6, x_{2}=\frac{7}{2}, x_{3}=\frac{7+\sqrt{5}}{2}, x_{4}=\frac{7-\sqrt{5}}{2} .
\end{array}
$$

Upon verification, they are all roots of the original equation.
Thus, the irrational roots are $x=\frac{7 \pm \sqrt{5}}{2}$.