因为|(ab-2)|与|(b-1)|互为相反数,所以得
|(ab-2)|+|(b-1)| = 0
所以 abs(ab-2)=0 abs(b-1)=0
即ab-2=0 b-1=0
解得a=2 b=1
所以 1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2005)(b+2005)
=1/(1*2)+1/(2*3)+.+1/(2006*2007)
=1-1/2+1/2-1/3+.+1/2006-1/2007
=1-1/2007
=2006/2007
因为|(ab-2)|与|(b-1)|互为相反数,所以得
|(ab-2)|+|(b-1)| = 0
所以 abs(ab-2)=0 abs(b-1)=0
即ab-2=0 b-1=0
解得a=2 b=1
所以 1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2005)(b+2005)
=1/(1*2)+1/(2*3)+.+1/(2006*2007)
=1-1/2+1/2-1/3+.+1/2006-1/2007
=1-1/2007
=2006/2007