Question

若(x^2-x-12)分之(6x+5)=（x-4)分之A+(x+3)分之B,则A+B=?

Answer 1

(6x-5)/(x^2-x-12)=A/(x-4)+B/(x+3)

(6x-5)/(x^2-x-12)=[A(x+3)+B(x-4)]/(x-4)(x+3)

(6x-5)/(x^2-x-12)=[(A+B)x+(3A-4B)]/(x^2-x-12)

所以

A+B=6

3A-4B=5

所以

A+B=6