Bài giải:

Xét mẫu số:   $\frac{1}{1\times 2}+\frac{1}{3\times 4}+....+\frac{1}{99\times 100}$

       = $\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$

       = $(1+\frac{1}{3}+...+\frac{1}{99})-(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100})$

       = $(1+\frac{1}{3}+...+\frac{1}{99})+\left( \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100} \right)-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100})\times 2$

       = $(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{99}+\frac{1}{100})-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50})$

       =$\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}$ (Đơn giản số trừ)

Vậy: $\frac{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}{\frac{1}{1\times 2}+\frac{1}{3\times 4}+....+\frac{1}{99\times 100}}$ 

= $\frac{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}$          

 = 1