bạn sai ở đoạn cuối
A=\(\frac{3\cdot3}{15\cdot18}+\frac{3\cdot3}{18\cdot21}+...+\frac{3\cdot3}{96\cdot99}=3\cdot\left(\frac{3}{15\cdot18}+\frac{3}{18\cdot21}+...+\frac{3}{96\cdot99}\right)=3\cdot\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{96}-\frac{1}{99}\right)=3\cdot\left(\frac{1}{15}-\frac{1}{99}\right)=3\cdot\frac{28}{495}=\frac{28}{165}\)