Câu 3:
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2020}-\dfrac{1}{2021}=\dfrac{2020}{2021}\)
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Câu 3:
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2020}-\dfrac{1}{2021}=\dfrac{2020}{2021}\)