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x-1的绝对值+(xy-2)平方=0,求1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+.+1/(x+20

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x-1的绝对值+(xy-2)平方=0,求1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+.+1/(x+2008)(y+2008)=
x-1的绝对值+(xy-2)平方=0
x-1的绝对值和(xy-2)平方均为非负数,现在和为0,则均为0
有:x-1=0
xy-2=0
x=1
y=2
1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+.+1/(x+2008)(y+2008)=
=1/1*2+1/2*3+..+1/2009*2010
=1-1/2+1/2-1/3+...+1/2009-1/2010
=1-1/2010
=2009/2010
补充:1/1*2=1-1/2
1/2*3=1/2-1/3
...
1/2009*2010=1/2009-1/2010