11) $$\frac{2m-3n}{m^2n} + \frac{7m-2m}{mn^2}$$
Приведем дроби к общему знаменателю $$m^2n^2$$.
$$\frac{2m-3n}{m^2n} + \frac{7m-2m}{mn^2} = \frac{n(2m-3n)}{m^2n^2} + \frac{m(7m-2m)}{m^2n^2} = \frac{2mn - 3n^2 + 7m^2 - 2m^2}{m^2n^2} = \frac{5m^2 + 2mn - 3n^2}{m^2n^2}$$
Ответ: $$\frac{5m^2 + 2mn - 3n^2}{m^2n^2}$$