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Tính A=1/99.97-1/97.95-...-1/5.3-1/3.1

\(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-....-\dfrac{1}{3.1}\)

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  • \(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

    \(=\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)

    Đặt \(B=\dfrac{1}{97.95}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\)

    \(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{95.97}\)

    \(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\)

    \(2B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{95}-\dfrac{1}{97}\)

    \(2B=1-\dfrac{1}{97}\)

    \(2B=\dfrac{96}{97}\)

    \(B=\dfrac{96}{97}:2=\dfrac{48}{97}\)

    \(\Rightarrow A=\dfrac{1}{99.97}-B=\dfrac{1}{9603}-\dfrac{48}{97}=\dfrac{-4751}{9603}\)

      bởi Nguyen Linh 19/12/2018
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