1/(x+1)(x+2)+1/(x+2)(x+3)+…+1/(x+2001)(x+2002)=2001/3x+6006
题目
1/(x+1)(x+2)+1/(x+2)(x+3)+…+1/(x+2001)(x+2002)=2001/3x+6006
答案
1/(x+1)(x+2)+1/(x+2)(x+3)+…+1/(x+2001)(x+2002)=2001/(3x+6006)
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+…+1/(x+2001)-1/(x+2002)=2001/(3x+6006)
1/(x+1)-1/(x+2002)=2001/(3x+6006)
2001/(x+1)(x+2002)=2001/(3x+6006)
1/(x+1)(x+2002)=1/(3x+6006)
(x+1)(x+2002)=3x+6006
x²+2003x+2002=3x+6006
x²+2000x-4004=0
(x+2002)(x-2)=0
x=-2002,x=2
经检验x=-2002是增根,x=2是原方程的根
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