B) $$\frac{3y}{4y-4} + \frac{2y}{5-5y} = \frac{3y}{4(y-1)} - \frac{2y}{5(y-1)} = \frac{15y - 8y}{20(y-1)} = \frac{7y}{20(y-1)}$$
Ответ: $$\frac{7y}{20(y-1)}$$