作业帮已知：(2x-3)/(x-1)(x+2)=(A/x-1)+(B/x+2)求A,B的值（完整的）
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已知：(2x-3)/(x-1)(x+2)=(A/x-1)+(B/x+2)求A,B的值（完整的）
已知：(2x-3)/(x-1)(x+2)=(A/x-1)+(B/x+2)求A,B的值（完整的）
已知：(2x-3)/(x-1)(x+2)=(A/x-1)+(B/x+2)求A,B的值（完整的）
(A/x-1)+(B/x+2)
=[A(x+2)+B(x-1)]/(x-1)(x+2)
=[(A+B)x+(2A-B]/(x-1)(x+2)
=(2x-3)/(x-1)(x+2)
所以(A+B)x+(2A-B)=2x-3
则A+B=2
2A-B=-3
相加
3A=-1
A=-1/3
B=2-A=7/3
(2x-3)/[(x-1)(x-2)]=A/x-1+B/x+2A/(x-1)+B/(x+2)
=[A(x+2)+B(x-1)]/(x-1)(x+2)
则A(X+2)+B(X-1)=2X-3
令x=1则3A=-1
A=-1/3
令x=-2
则-3B=-7
B=7/3
或者
=(Ax+2A+Bx-B)/(x-1)(x+2)
全部展开
(2x-3)/[(x-1)(x-2)]=A/x-1+B/x+2A/(x-1)+B/(x+2)
=[A(x+2)+B(x-1)]/(x-1)(x+2)
则A(X+2)+B(X-1)=2X-3
令x=1则3A=-1
A=-1/3
令x=-2
则-3B=-7
B=7/3
或者
=(Ax+2A+Bx-B)/(x-1)(x+2)
=[(A+B)x+(2A-B)/(x-1)(x+2)
=(2x-3)/[(x-1)(x-2)]=[(A+B)x+(2A-B)/(x-1)(x+2)
A+B=2
A-B=-3
A=-1/3
B=7/3
收起
(2x-3)/(x-1)(x+2)=(A/x-1)+(B/x+2)
两边同乘以x-1,再令x=1,得
A=(2-3)/(1+2)=-1/3
两边同乘以x+2,再令x=-2,得
B=(-4-3)/(-2-1)=7/3
左右都乘以（x-1）（x+2） 得 2x-3=A(x+2)+B(x-1)=(A+B)x+2A-B=2x-3 所以A+B=2 2A-B=-3 A=-1/3 B=7/3 希望能采纳
等号右边通分：(A/x-1)+(B/x+2)=（Ax+2A+Bx-B）/(x-1)(x+2)=[(A+B)x+2A-B]/(x-1)(x+2)
左边等于右边：(2x-3)/(x-1)(x+2）=[(A+B)x+2A-B]/(x-1)(x+2)
即：分子相等：(2x-3)=(A+B)x+2A-B
即：A+B=2
2A-B=-3
解方程就可以了