作业帮若|x-1|+|y-2|=0,求1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+.+1/(x+2009)(y+2009)的值
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若|x-1|+|y-2|=0,求1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+.+1/(x+2009)(y+2009)的值
若|x-1|+|y-2|=0,求1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+.+1/(x+2009)(y+2009)的值
若|x-1|+|y-2|=0,求1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+.+1/(x+2009)(y+2009)的值
由|x-1|+|y-2|=0得x=1,y=2
1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+.+1/(x+2009)(y+2009)
=1/(1*2)+1/(2*3)+1/(3*4) )+.+1/(2010*2011)
=(1-1/2)+(1/2-1/3)+(1/3+1/4) +.+(1/2010-1/2011)
=1-1/2+1/2-1/3+1/3+1/4 +.+1/2010-1/2011
=1-1/2011
=2010/2011
首先得到x=1,y=2;
然后将x,y带入，采用裂相相消法。
即1/(1*2)+1/(2*3)+......+1/(2010*2011)=(1-1/2)+(1/2-2/3)+...+(1/2010-1/2011)=1-1/2011=2010/2011
答案为：2010/2011
x=1,y=2
原式=
(1/1-1/2)+(1/2-1/3)+...+(1/2010-1/2011)
=2010/2011
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