若有理数a,b满足/ab-2/+(1-b）的平方=0,试求：1/ab+1/(a+1）(b+1)+1/（a+2)(b+2)+...(1/(a+99)(b+99)的值

由于/ab-2/+(1-b）的平方=0,所以b=1,a=2
所以
1/ab+1/(a+1）(b+1)+1/（a+2)(b+2)+...(1/(a+99)(b+99)
=1/2+1/(3*2)+1/(4*3)+...+1/(101*100)
=(1-1/2)+(1/2-1/3)+...+(1/100-1/101)
=1-1/101
=100/101