已知2x+3/x(x-1)(x+2)=A/x+B/x-1+C/x+2,A,B,C为常数,求A,B,C的值

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/17 02:14:20

/>(2x+3)/[x(x-1)(x+2)]=A/x+B/(x-1)+C/(x+2)
=[A(x²+x-2)+B(x²+2x)+C(x²-x)]/[x(x-1)(x+2)]
=[(A+B+C)x²+(A+2B-C)x-2A]/[x(x-1)(x+2)]
于是
A+B+C=0
A+2B-C=2
-2A=3
联立解得
A=-3/2
B=5/3
C=-1/6
再问： [A(x²+x-2)+B(x²+2x)+C(x²-x)]/[x(x-1)(x+2)] 这一步怎么的得来的
再答：不好意思，昨晚有事出去了 A/x+B/(x-1)+C/(x+2) 最小公分母是x(x-1)(x+2)，通分得 A(x-1)(x+2)/[x(x-1)(x+2)]+Bx(x-2)/[x(x-1)(x+2)]+Cx(x-1)/[x(x-1)(x+2)] =[A(x-1)(x+2)+Bx(x+2)+Cx(x-1)]/[x(x-1)(x+2)] =[A(x²+x-2)+B(x²+2x)+C(x²-x)]/[x(x-1)(x+2)]

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