如果有理数ab满足丨ab-2丨+(1-b)²=0,试求1/ab+1/(a+2)(b+2)+L+1/(a＋200

∵丨ab-2丨+(1-b)²=0.
∴ab-2=0; 1-b=0.
解得:b=1,a=2.
◆估计题中抄写有误,原题可能是求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2007)(b+2007)的值
所以：原式=1/(1*2)+1/(2*3)+1/(3*4)+…+1/(2008*2009)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+…+(1/2008-1/2009)
=1-1/2009 --------------------上式去括号,合并同类项后只剩下首项和末项了!
=2008/2009