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计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+.+1/(x+2003)(x+200

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计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+.+1/(x+2003)(x+2004)
一道分式加法.在线等~~~~~
计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+.+1/(x+2003)(x+200
[1/(x+1)-1/(x+2)]+[1/(x+2)-1/(x+3)]+[1/(x+3)-1/(x+4)]+...[1/(x+2003)-1/(x+2004)]
把相邻的抵消
剩下1/(x+1)-1/(x+2004)=2003/(x+1)(x+2004)