由|ab-2|+|a-2|=0得出 ab-2 = 0；a-2 = 0知 a=2,b=1;故1/ab+1/（a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2006)(b+2006) =1/(2*1) + 1/(3*2) + 1/(4*3) +.+1/(2008*2007) =(1-1/2) + (1/2 - 1/3) + (1/3-1/4) +...+(1/2007-1...