Question

Помогите номер два пожалуйста

Answer 1

1)(3b+7)/3b -(b²-5)/b²=(3b²+7b-3b²+15)/3b²=(7b+15)/3b²
2)1/(4p+q)-1/(4p-q)=(4p-q-4p-q)/(16p²-q²)=-2q/(16p²-q²)=2q/(q²-16p²)
3)(5-4y)/y(y-6)+4/(y-6)=(5-4y+4y)/y(y-6)=5/y(y-6)