Bài giải:

Xét mẫu số:

 $\begin{align} & \frac{1}{2}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{9900} \\& =\frac{1}{1\times 2}+\frac{1}{3\times 4}+\frac{1}{5\times 6}+....+\frac{1}{99\times 100} \\ & =\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100} \\ & =\left( 1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99} \right)-\left( \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100} \right) \\ & =\left( 1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99} \right)+\left( \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100} \right)-\left( \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100} \right)\times 2 \\ & =\left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100} \right)-\left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50} \right) \\ & =\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100} \\\end{align}$

$A=\left( \frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100} \right):\left( \frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100} \right)=1$