\(\dfrac{1}{1.2}\) ở đâu đấy bn ?

\(\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)