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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Trúc Chi 5 years 2021-05-16T13:54:40+00:00 1 Answers 33 views 0

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  1. diemthu
    0
    2021-05-16T13:56:27+00:00
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    $\quad \dfrac{1 -2\sin^2x}{\sin^2x\cos^2x}$

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

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

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

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

    $= \cot^2x + 1 – (\tan^2x +1)$

    $=\cot^2x – \tan^2x$

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