\(A=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dots+\dfrac{1}{9900}\)
\(=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dots+\dfrac{1}{99\cdot100}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dots+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{3}-\dfrac{1}{100}=\dfrac{97}{300}\)
Vậy \(A=\dfrac{97}{300}\).
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