有理数A,B满足|AB-2|+(1-B)^2=0,求1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+...+(A+2007)(

因为|AB-2|+(1-B)^2=0;
而 |AB-2|>=0 且 (1-B)^2>=0;
则要满足条件,只能,
|AB-2|=0 且 (1-B)^2=0;
即 AB=2 且 B=1 →A=2;B=1
所以原式可化为
1/2+1/(2*3)+1/(3*4)+1/(4*5)+...+1/(2008*2009)
=1/2+(3-2)/(2*3)+(4-3)/(3*4)+(5-4)/(4*5)+...+(2009-2008)/(2008*2009)
=1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/2008-1/2009
=1/2+1/2-1/2009
=1-1/2009
=2008/2009