已知|ab-2|+（b+1）^2=0求1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)...+1/(a-2009)(b-2009)的值.

因为绝对值和平方都大于等于零，所以ab=2,b=-1,a=-2
因此下面的式子等于
1/(1*2)+1/(2*3)+1/(3*4)+...+1/(2010*2011)=1/1-1/2+1/2-1/3+...+1/2010-1/2011=1-1/2011=2010/2011

∵|ab-2|+（b+1）^2=0
∴|ab-2|=0，（b+1）^2=0
∴ab-2=0，b+1=0
∴ab=2,b=-1
∴a=-2
∴原式=1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)...+1/(a-2009)(b-2009)
=1/(-2)(-1)+1/(-3)(-2)+1/(-4)(-3)+...+1/(-2011)(-2010)
=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
=1-1/2011
=2010/2011
加油！祝你学习进步

已知：|ab-2|+(b+1)^2=0
而：|ab-2|≥0；(b+1)^2≥0
所以：|ab-2|=0；(b+1)^2=0
得：ab=2，或者b=-1
即：a=-2，b=-1
代入所求，有：
1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)...+1/(a-2009)(b-2009)
=1/2+1/(-3)(-2)+1/(-4)(-3)+……+1/(-2011)(-2010)
=1/2+(1/2-1/3)+(1/3-1/4)+……+(1/2010-1/2011)
=1/2+1/2-1/2011
=1-1/2011
=2010/2011

a=-2!b=-1

已知|ab-2|+（b+1）^2=0
所以ab-2=0,b+1=0
所以b=-1,a=-2
所以
1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)...+1/(a-2009)(b-2009)
=1/(-2)(-1)+1/(-3)(-2)+1/(-4)(-3)+...+1/(-2011)(-2010)
=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
=1-1/2011
=2010/2011