若|x-1|+|y-3|=0,求1/xy+1/(x+2)(y+2)+1/(x+4)(y+4)+1/(x+6)(y+6).

1个回答

  • 绝对值项恒非负,两绝对值项之和=0,两绝对值项分别=0

    x-1=0 x=1

    y-3=0 y=3

    y=x+2

    1/(xy)+1/[(x+2)(y+2)]+1/[(x+4)(y+4)]+...+1/[(x+2010)(y+2010)]

    =1/[x(x+2)]+1/[(x+2)(x+4)]+1/[(x+4)(x+6)]+...+1/[(x+2010)(x+2012)]

    =(1/2)[1/x -1/(x+2)+1/(x+2)-1/(x+4)+1/(x+4)-1/(x+6)+...+1/(x+2010)-1/(x+2012)]

    =(1/2)[1/x -1/(x+2012)]

    =(1/2)[1/1 -1/(1+2012)]

    =(1/2)(1-1/2013)

    =1006/2013

    老师扣你分的原因是你算错了.1/[n(n+2)]=(1/2)[1/n -1/(n+2)]而不是1/[n(n+2)]=1/n -1/(n+2)