若│ab-2│+(b-1）二次方=0

│ab-2│+(b-1）二次方=0,则
ab-2=0,b-1=0
所以b=1,a*1-2=0,a=2
所以
ab分之1 + （a+1)×(b+1)分之1 +……………………+(a+2006)×（b+2006)分之1
=2*1分之1 + （2+1)×(1+1)分之1 +……………………+(2+2006)×（1+2006)分之1
=1/(1*2)+1/(2*3)+1/(3*4)+.+1/(2006*2007)+1/(2007*2008)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.+(1/2006-1/2007)+(1/2007-1/2008)
=1-1/2008
=2007/2008

│ab-2│+(b-1）二次方=0
|ab-2|=0 ,(b-1)二次方=0
ab-2=0,b-1=0
ab=2,b=1,a=2
ab分之1 + （a+1)×(b+1)分之1+...+(a+2006)×（b+2006)分之1
=1/1*2+1/2*3+1/3*4+1/4*5+...+1/2007*2008
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/2007-1/2008
=1-1/2008
=2007/2008