Giúp mình câu 2a với Thanks ❤️❤️

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Giúp mình câu 2a với
Thanks ❤️❤️
giup-minh-cau-2a-voi-thanks-️-️

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Ngọc Diệp 5 years 2021-05-17T02:17:55+00:00 1 Answers 19 views 0

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  1. vankhanh
    0
    2021-05-17T02:19:51+00:00
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    $\quad \dfrac{1 -2\cos^2x}{\sin^2x\cos^2x}$

    $=\dfrac{\sin^2x + \cos^2x – 2\cos^2x}{\sin^2x\cos^2x}$

    $=\dfrac{\sin^2x – \cos^2x}{\sin^2x\cos^2x}$

    $=\dfrac{\sin^2x}{\sin^2x.\cos^2x} – \dfrac{\cos^2x}{\sin^2x.\cos^2x}$

    $=\dfrac{1}{\cos^2x} – \dfrac{1}{\sin^2x}$

    $= \tan^2x + 1 – (\cot^2x +1)$

    $=\tan^2x – \cot^2x$

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )