giúp em với ạ 5’ nữa em phải nộp ạ

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giúp em với ạ 5’ nữa em phải nộp ạ
giup-em-voi-a-5-nua-em-phai-nop-a

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Phúc Điền 5 years 2021-05-18T07:41:35+00:00 1 Answers 15 views 0

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  1. Amity
    0
    2021-05-18T07:43:14+00:00
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    $\begin{array}{l}
    1)1 – {\sin ^2} + {\cos ^2}x = {\sin ^2}x + {\cos ^2}x – {\sin ^2}x + {\cos ^2}x = 2{\cos ^2}x\\
    2)1 + {\sin ^2}x – {\cos ^2}x = {\sin ^2}x + {\cos ^2}x + {\sin ^2}x – {\cos ^2}x = 2{\sin ^2}x\\
    3)\dfrac{{{{\sin }^2}x – 1}}{{1 – {{\cos }^2}x}} = \dfrac{{{{\sin }^2}x – {{\cos }^2}x – {{\sin }^2}x}}{{{{\sin }^2}x – {{\cos }^2}x + {{\cos }^2}x}} = \frac{{ – {{\cos }^2}x}}{{{{\sin }^2}x}} =  – {\tan ^2}x\\
    4){\sin ^2}x + {\cos ^2}x + {\cot ^2}x = 1 + {\cot ^2}x = \dfrac{1}{{{{\sin }^2}x}}\\
    5)\left( {1 – \cos x} \right){\cot ^2}x\left( {1 + \cos x} \right) = \left( {1 – {{\cos }^2}x} \right){\cot ^2}x = {\sin ^2}x.\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} = {\cos ^2}x\\
    6)1 – \sin \alpha .\cot \alpha .\cos \alpha  = 1 – \sin \alpha .\dfrac{{\cos \alpha }}{{\sin \alpha }}.\cos \alpha  = 1 – {\cos ^2}\alpha  = {\sin ^2}\alpha \\
    7)\dfrac{{\sin x.\sin y}}{{\cos x.\cos y}}.\tan x.\cot y = \dfrac{{\sin x}}{{\cos x}}.\dfrac{{\sin y}}{{\cos y}}.\tan x.\cot y = {\tan ^2}x.\cot y.\tan y = {\tan ^2}x
    \end{array}$

     

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