3. $$rac{b+1}{abc} + \frac{1-a}{a^2c} =$$

$$\frac{b+1}{abc} + \frac{1-a}{a^2c} = \frac{a(b+1)}{a^2bc} + \frac{b(1-a)}{a^2bc} = \frac{ab+a+b-ab}{a^2bc} = \frac{a+b}{a^2bc}$$ $$\textbf{Ответ: } \frac{a+b}{a^2bc}$$