Question

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.

Answer 1

(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