Question

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

过程 过程

Answer 1

解：
(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

追问

[A(x²+x-2)+B(x²+2x)+C(x²-x)]/[x(x-1)(x+2)]
这一步怎么的得来的

追答

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)]