Tính: $\frac{1}{2√1+1√2}$ + $\frac{1}{3√2+2√3}$ +….+ $\frac{1}{100√99+99√100}$

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Tính: $\frac{1}{2√1+1√2}$ + $\frac{1}{3√2+2√3}$ +….+ $\frac{1}{100√99+99√100}$

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Athena 4 tháng 2021-08-02T11:59:24+00:00 1 Answers 10 views 0

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    2021-08-02T12:00:57+00:00

    Đáp án:

    $\dfrac{9}{10}$

    Giải thích các bước giải:

    $\dfrac{1}{2\sqrt1 + 1\sqrt2} + \dfrac{1}{3\sqrt2 + 2\sqrt3} + \cdots + \dfrac{1}{100\sqrt{99} + 99\sqrt{100}}$

    $= \dfrac{1}{\sqrt1.\sqrt2.(\sqrt2 + \sqrt1)} + \dfrac{1}{\sqrt2.\sqrt3.(\sqrt3 + \sqrt2)} + \cdots+ \dfrac{1}{\sqrt{99}.\sqrt{100}.(\sqrt{100} + \sqrt{99})}$

    $= \dfrac{\sqrt2 – \sqrt1}{\sqrt1.\sqrt2.(\sqrt2 + \sqrt1)(\sqrt2 – \sqrt1)} + \dfrac{\sqrt3 – \sqrt2}{\sqrt2.\sqrt3.(\sqrt3 + \sqrt2)(\sqrt3 – \sqrt2)} + \cdots+ \dfrac{\sqrt{100} – \sqrt{99}}{\sqrt{99}.\sqrt{100}.(\sqrt{100} + \sqrt{99})(\sqrt{100} – \sqrt{99})}$

    $= \dfrac{\sqrt2 – \sqrt1}{\sqrt1.\sqrt2} + \dfrac{\sqrt3 – \sqrt2}{\sqrt2.\sqrt3} + \cdots+ \dfrac{\sqrt{100} – \sqrt{99}}{\sqrt{99}.\sqrt{100}}$

    $= \dfrac{1}{\sqrt1} – \dfrac{1}{\sqrt2} + \dfrac{1}{\sqrt2} – \dfrac{1}{\sqrt3} + \cdots + \dfrac{1}{\sqrt{99}} – \dfrac{1}{\sqrt{100}}$

    $= 1 – \dfrac{1}{10}$

    $= \dfrac{9}{10}$

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