原式=[1/x-1/(x+1)]+[1/(x+1)-1/(x+2)]+[1/(x+2)-1/(x+3)]
=1/x-1/(x+3)
=2/(x^2+3x)
=2/(-1)
=-2
化简1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3),其中x^2+3x+1=0
化简1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3),其中x^2+3x+1=0
数学人气：124 ℃时间：2020-05-16 03:19:16
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