Question

已知x^2-2/x^3-3x^2+2x=A/x+B/x-1+C/x-2,试确定整数A,B,C的值。要具体过程，谢谢啊

Answer 1

x³-3x²+2x=x(x²-3x+2)=x(x-1)(x-2)
(x²-2)/(x³-3x²+2x)=A/x+B/(x-1)+C/(x-2)
等式两边同乘x(x-1)(x-2)得：
x²-2=A(x-1)(x-2)+Bx(x-2)+Cx(x-1)
x²-2=A(x²-3x+2)+Bx²-2Bx+Cx²-Cx
x²-2=(A+B+C)x²-(3A+2B+C)x+2A
则：A+B+C=1
3A+2B+C=0
2A=-2
解得：A=-1，B=1，C=1

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Answer 2

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+B+C)x²-(2B+C-3A)x+2A]/(x³-3x²+2x)
=(x²-2)/(x³-3x²+2x)
A+B+C=1
2B+C-3A=0
2A=-2
解得
A=-1 B=-5 C=7