2) \(\frac{3y}{4y-4}+\frac{2y}{5-5y}\)

Разложим знаменатели: \(4y-4=4(y-1)\), а \(5-5y=-5(y-1)\). Ограничение: \(y\ne1\).

\[\frac{3y}{4(y-1)}+\frac{2y}{-5(y-1)}=\frac{15y-8y}{20(y-1)}=\frac{7y}{20(y-1)}.\]

Ответ: \(\frac{7y}{20(y-1)}\), \(y\ne1\).