已知(2x-3)/(x+2)(x-3)=A/(x+2)-B/(x-3),求整式AB

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已知(2x-3)/(x+2)(x-3)=A/(x+2)-B/(x-3),求整式AB已知(2x-3)/(x+2)(x-3)=A/(x+2)-B/(x-3),求整式AB已知(2x-3)/(x+2)(x-3

已知(2x-3)/(x+2)(x-3)=A/(x+2)-B/(x-3),求整式AB
已知(2x-3)/(x+2)(x-3)=A/(x+2)-B/(x-3),求整式AB

已知(2x-3)/(x+2)(x-3)=A/(x+2)-B/(x-3),求整式AB
右边 =A(x-3)-B(X+2) /(X+2)(X-3)
此时 分母相同
分子拆开
(2X-3)=A(X-3)-B(X+2)
=(A-B)X-(3A+2B)
所以 A-B=2
3A+2B=3
所以A=-3/5 B=7/5
AB=-21/25