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

ОДЗ: \(y\ne1\). Так как \(4y-4=4(y-1)\), а \(5-5y=-5(y-1)\), имеем:

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

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