已知x为任何有理数时,都有等式:(2-4x)/x(x+1)(x+2)=A/(x+2)+B/x+C/(x+1)成立,求A,B,C.

题目

已知x为任何有理数时,都有等式:(2-4x)/x(x+1)(x+2)=A/(x+2)+B/x+C/(x+1)成立,求A,B,C.

答案

A/(x+2)+B/x+C/(x+1)
=[Ax(x+1)+B(x+1)(x+2)+Cx(x+2)]/x(x+1)(x+2)
=[(A+B+C)x^2+(A+3B+2C)x+(2B+2C)]/x(x+1)(x+2)
=(-4x+2)/x(x+1)(x+2)=
所以
A+B+C=0 (1)
A+3B+2C=-4 (2)
2B+2C=2 (3)
(2)-(1)
2B+C=-4 (4)
(3)-(4)
C=6
代入 (3)
2B+12=2,B=-5
A=0-(B+C)=-1
A=-1,B=-5,C=6