已知(x+2)(x-3)分之2x-3=x+2分之A减x-3分之B,求整式A,B.

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已知(x+2)(x-3)分之2x-3=x+2分之A减x-3分之B,求整式A,B.

(2x-3)/(x+2)(x-3)=A/(x+2)-B/(x-3)
(2x-3)/(x+2)(x-3)=[A(x-3)-B(x+2)]/(x+2)(x-3)
(2x-3)/(x+2)(x-3)=[(A-B)x-2B-3A]/(x+2)(x-3)
(2x-3)=[(A-B)x-2B-3A]
A-B=2
3A+2B=3
A=7/5
B=-3/5