2.060. $\left(\frac{3}{2 x-y}-\frac{2}{2 x+y}-\frac{1}{2 x-5 y}\right): \frac{y^{2}}{4 x^{2}-y^{2}}$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq \pm \frac{y}{2}, \\ x \neq \frac{5 y}{2} .\end{array}\right.$

$$
\begin{aligned}
& \left(\frac{3}{2 x-y}-\frac{2}{2 x+y}-\frac{1}{2 x-5 y}\right): \frac{y^{2}}{4 x^{2}-y^{2}}=\left(\frac{3(2 x+y)-2(2 x-y)}{(2 x-y)(2 x+y)}-\frac{1}{2 x-5 y}\right) \\
& \frac{y^{2}}{4 x^{2}-y^{2}}=\left(\frac{6 x+3 y-4 x+2 y}{4 x^{2}-y^{2}}-\frac{1}{2 x-5 y}\right) \cdot \frac{4 x^{2}-y^{2}}{y^{2}}=
\end{aligned}
$$

$$
\begin{aligned}
& =\left(\frac{2 x+5 y}{4 x^{2}-y^{2}}-\frac{1}{2 x-5 y}\right) \cdot \frac{4 x^{2}-y^{2}}{y^{2}}=\frac{(2 x+5 y)(2 x-5 y)-4 x^{2}+y^{2}}{\left(4 x^{2}-y^{2}\right)(2 x-5 y)} \times \\
& \times \frac{4 x^{2}-y^{2}}{y^{2}}=\frac{4 x^{2}-25 y^{2}-4 x^{2}+y^{2}}{\left(4 x^{2}-y^{2}\right)(2 x-5 y)} \cdot \frac{4 x^{2}-y^{2}}{y^{2}}=\frac{-24 y^{2}}{(2 x-5 y) y^{2}}= \\
& =\frac{-24}{2 x-5 y}=\frac{24}{5 y-2 x} .
\end{aligned}
$$

Answer: $\frac{24}{5 y-2 x}$.