题目内容：

已知根号a-1+(ab-2)平方=0.求1/ab+1/(a+2)(b+2)+...+1/(a+2011)(b+2011)的值

优质解答

a-1=0ab-2=0∴a=1b=21/ab +1/(a+1)(b+1) +1/(a+2)(b+2)+...+1/(a+2011)(b+2011)=1/1×2+1/2×3+1/3×4+……+1/2012×2013=1-1/2+1/2-1/3+1/3-1/4+1/4-1/2012+1/2012-1/2013=2012/2013