|ab-2|>=0,(b-1)²>=0,两个式子互为相反数.所以都等于0,b=1,a=2
首先已知1/(a*(a+1))=1/a-1/(a+1),例如1/(2*3)=1/2-1/3
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)
=1/(1*2)+1/（2*3）+1/(3*4)+.+1/(2006*2007)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.+(1/2005-1/2006)+(1/2006-1/2007)
=1-1/2007
=2006/2007