2) a) $$\frac{y^2 + 3y}{4} \cdot \frac{y}{2y+6} = \frac{y(y + 3)}{4} \cdot \frac{y}{2(y+3)} = \frac{y^2(y+3)}{8(y+3)} = \frac{y^2}{8}$$
Ответ: $$\frac{y^2}{8}$$