Question

1/(x+1)(x+2)+1/(x+2)(x+3)+…+1/(x+2001)(x+2002)=2001/3x+6006

Answer 1

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