Упростим выражение $$\frac{4}{y} + \frac{2}{y-5} + \frac{2y}{25-y^2} - \frac{10}{y^2-25}$$.
$$\frac{4}{y} + \frac{2}{y-5} - \frac{2y}{y^2-25} + \frac{10}{y^2-25} = \frac{4}{y} + \frac{2}{y-5} + \frac{10-2y}{y^2-25} = \frac{4}{y} + \frac{2}{y-5} + \frac{2(5-y)}{(y-5)(y+5)} = \frac{4}{y} + \frac{2}{y-5} - \frac{2}{y+5} = \frac{4(y-5)(y+5) + 2y(y+5) - 2y(y-5)}{y(y-5)(y+5)} = \frac{4(y^2-25) + 2y^2+10y - 2y^2+10y}{y(y-5)(y+5)} = \frac{4y^2 - 100 + 20y}{y(y-5)(y+5)} = \frac{4y^2 + 20y - 100}{y(y-5)(y+5)} = \frac{4(y^2+5y-25)}{y(y-5)(y+5)}$$
Ответ: $$\frac{4(y^2+5y-25)}{y(y-5)(y+5)}$$