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

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

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

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