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
问题描述：
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
答
(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=0A+2B-C=2-2A=3联立解得A=-3/2B=5/3C=-1/6[A(x²+x-2)+B(x²+2x)+C(x²-x)]/[x(x-1)(x+2)]这一步怎么的得来的不好意思，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)]