title: 初一数学绝对值问题已知|ab
desc: 已知|ab-2|与|b-1|互为相反数，试求式。
  1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+1/(a+3)(b+3)+……+1/(a+2006)(b+2006)的值。
answer: 解：
  因为|ab-2|与|b-1|互为相反数
  则ab-2=0，b-1=0
  则
  a=2
  b=1
  则
  1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+1/(a+3)(b+3)+……+1/(a+2006)(b+2006)
  =1/2+1/(2*3)+1/(3*4)+.....+1/(2007*2008)
  =1-1/2+1/2-1/3+1/3-1/4+.....+1/2007-1/2008
  =1-1/2008
  =2007/2008 
  
  