Tính đạo mà giúp mình với mình cảm ơn

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Tính đạo mà giúp mình với mình cảm ơn
tinh-dao-ma-giup-minh-voi-minh-cam-on

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Jezebel 5 years 2021-05-21T02:32:25+00:00 2 Answers 15 views 0

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  1. linhdan
    0
    2021-05-21T02:34:05+00:00
    Reply

    $y=\dfrac{\sqrt[]{1-2x}}{x}$

    $y’=\dfrac{(\sqrt[]{1-2x})’.x-(\sqrt[]{1-2x}).x’}{x^2}$

    $=\dfrac{\dfrac{1}{2\sqrt[]{1-2x}}.(1-2x)’.x-(\sqrt[]{1-2x})}{x^2}$

    $=(\dfrac{-2x}{2\sqrt[]{1-2x}}-\sqrt[]{1-2x}):x^2$

    $=\dfrac{\dfrac{-x}{\sqrt[]{1-2x}}-\sqrt[]{1-2x}}{x^2}$

    $=\dfrac{-x-(1-2x)}{\sqrt[]{1-2x}}.$ $\dfrac{1}{x^2}$

    $=\dfrac{x-1}{x^2(\sqrt[]{1-2x})}$

    BẠN THAM KHẢO.

  2. Adela
    0
    2021-05-21T02:34:15+00:00
    Reply

    $y’=\dfrac{(\sqrt{1-2x})’x-\sqrt{1-2x} }{x^2}$

    $=\dfrac{ \dfrac{-2x}{2\sqrt{1-2x}}-\sqrt{1-2x} }{x^2}$

    $=\dfrac{ \dfrac{-x}{\sqrt{1-2x}}-\sqrt{1-2x} }{x^2}$

    $=\dfrac{-x-(1-2x) }{x^2\sqrt{1-2x}}$

    $=\dfrac{x-1}{x^2\sqrt{1-2x}}$

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