如果lab-2l+(b-1)的2次幂=0，试求：1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...1/(a+99)(b+99)的值

lab-2l+(b-1)^2=0
lab-2l>=0且(b-1)^2>=0
所以ab-2=0且b-1=0
即a=2，b=1

又有1/n*(n+1)=1/n-1/(n+1)

所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...1/(a+99)(b+99)
=1/2+1/2*3+1/3*4+……+1/100*101
=1-1/2+1/2-1/3+1/3-1/4+……+1/100-1/101
=1-1/101
=100/101

|ab-2|+(b-1)^2=0,即
ab-2=0
b-1=0
a=2、b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...1/(a+99)(b+99)
=1/(1*2)+1/(2*3)+1/(3*4)+......+1/(100*101)
=（1-1/2)+(1/2-1/3)+(1/3-1/4)+.........+(1/100-1/101)
=1-1/101
=100/101