Question

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

拜托！

Answer 1

解：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

Answer 2

x=1,y=2
原式=
(1/1-1/2)+(1/2-1/3)+...+(1/2010-1/2011)
=2010/2011

Answer 3

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