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=0x-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=0xy-2=0x=1y=21/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+...+1/20...