已知x^2+2/x(x+1)(x+2)=A/x+B/x+1+C/x+2,求A.B.C的值

Question

已知x^2+2/x(x+1)(x+2)=A/x+B/x+1+C/x+2,求A.B.C的值

Answer 1

A/x+B/(x+1)+C/(x+2)
=[A(x+1)(x+2)+Bx(x+2)+Cx(x+1)]/x(x+1)(x+2)
=[A(x^2+3x+2)+B(x^2+2x)+C(x^2+x)]/x(x+1)(x+2)
=[(A+B+C)x^2+(3A+2B+C)x+2A]/x(x+1)(x+2)
因为(x^2+2)/x(x+1)(x+2)=A/x+B/(x+1)+C/(x+2)
所以对应系数相等可得：
A+B+C=1,3A+2B+C=0,2A=2
解得:A=1
则有:B+C=0 (1)
2B+C=-3 (2)
(2)-(1)得：B=-3
C=-B=3
综上：A=1,B=-3,C=3