问题描述：
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/(x+2010)(y+2010)的值!我做了,就是最后两步错了,不知道怎么错了,我把X=1,Y=3带入最后得到1/1*3+1/3*5+1/5*7+...+1/2011*2013=1-1/2013=2012/2013,老师给我扣了三分,
问题解答：
我来补答
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)
再问： 1/[n(n+2)]=(1/2)[1/n -1/(n+2)]，为什么要乘1/2
