Môn Toán Lớp 11 Giúp em với anh chị

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Môn Toán Lớp 11 Giúp em với anh chị Giúp em bài này với ạ em cần gấp, đừng copy nguồn trên mạng nha. Em xin cảm ơn thầy cô và các bạn nhiều.
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Ella 9 phút 2022-09-19T10:28:24+00:00 2 Answers 0 views 0

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    0
    2022-09-19T10:30:14+00:00

    Đáp án + Giải thích các bước giải:

    `15.`

    `sin^4“x/2-sin^2“x/2(sinx+3)+sinx+2=0`

    `<=>sin^4“x/2-sin^2“x/2(sinx+2)-sin^2“x/2“+sinx+2=0`

    `<=>sin^2“x/2(sin^2“x/2-sinx-2)-(sin^2“x/2-sinx-2)=0`

    `<=>(sin^2“x/2-sinx-2)(sin^2“x/2-1)=0`

    `<=>((1-cosx)/2-sinx-2)((1-cosx)/2-1)=0`

    `<=>(-sinx-1/2cosx-3/2)(-1/2cosx-1/2)=0`

    `<=>`\(\left[ \begin{array}{l}-\sin x-\dfrac{1}{2}\cos x-\dfrac{3}{2}(VN)\\-\dfrac{1}{2}\cos x-\dfrac{1}{2}=0\end{array} \right.\)

    `<=>cosx=-1`

    `<=>x=pi+k2pi(kinZZ)`

    `16.`

    `3cos4x+(cos2x-sinx)^2=7`

    `<=>3(2cos^2 2x-1)+cos^2 2x-2cos2xsinx+sin^2x=7`

    `<=>7cos^2 2x-2(1-2sin^2x)sinx+sin^2x-10=0`

    `<=>7(1-2sin^2x)^2-2sinx+4sin^3x+sin^2x-10=0`

    `<=>7-28sin^2 x+28sin^4x-2sinx+4sin^3x+sin^2x-10=0`

    `<=>28sin^4x+4sin^3x-27sin^2x-2sinx-3=0`

    `<=>(sinx-1)(28sin^3x+32sin^2x+5sinx+3)=0`

    `<=>`\(\left[ \begin{array}{l}\sin x-1=0\\28\sin^3x+32\sin^2x+5\sin x+3=0(VN)\end{array} \right.\)

    `<=>sinx=1`

    `<=>x=pi/2+k2pi(kinZZ)`

  1. minhuyen
    0
    2022-09-19T10:30:17+00:00

    Câu 15:

    $\begin{array}{l} {\sin ^4}\left( {\dfrac{x}{2}} \right) – {\sin ^2}\dfrac{x}{2}\left( {\sin x + 3} \right) + \sin x + 2 = 0\\  \Leftrightarrow {\sin ^4}\left( {\dfrac{x}{2}} \right) – {\sin ^2}\dfrac{x}{2}\left( {\sin x + 3} \right) + \left( {\sin x + 3} \right) – 1 = 0\\  \Leftrightarrow \left( {\sin x + 3} \right)\left( {1 – {{\sin }^2}\dfrac{x}{2}} \right) + {\sin ^4}\left( {\dfrac{x}{2}} \right) – 1 = 0\\  \Leftrightarrow \left( {\sin x + 3} \right)\left( {1 – {{\sin }^2}\dfrac{x}{2}} \right) + \left[ {{{\sin }^2}\left( {\dfrac{x}{2}} \right) – 1} \right]\left[ {{{\sin }^2}\left( {\dfrac{x}{2}} \right) + 1} \right] = 0\\  \Leftrightarrow \left( {\sin x + 3} \right){\cos ^2}\dfrac{x}{2} – {\cos ^2}\dfrac{x}{2}\left[ {{{\sin }^2}\left( {\dfrac{x}{2}} \right) + 1} \right] = 0\\  \Leftrightarrow {\cos ^2}\dfrac{x}{2}\left[ {\sin x + 3 – {{\sin }^2}\left( {\dfrac{x}{2}} \right) – 1} \right] = 0\\  \Leftrightarrow {\cos ^2}\dfrac{x}{2}\left[ {\sin x + 2 – {{\sin }^2}\dfrac{x}{2}} \right] = 0\\  \Leftrightarrow \left[ \begin{array}{l} {\cos ^2}\dfrac{x}{2} = 0\\ \sin x + 2 – {\sin ^2}\dfrac{x}{2} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \cos \dfrac{x}{2} = 0\\ \sin x + 2 – \dfrac{{1 – \cos x}}{2} = 0 \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} \dfrac{x}{2} = \dfrac{\pi }{2} + k\pi \\ 2\sin x + 4 – \left( {1 – \cos x} \right) = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \pi  + k2\pi \\ 2\sin x + \cos x + 3 = 0 \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = \pi  + k2\pi \\ 2\sin x + \cos x =  – 3\left( 1 \right) \end{array} \right.\\ \left( 1 \right): \Rightarrow a = 2,b = 1 \Rightarrow {a^2} + {b^2} = 5 < {c^2} = {\left( { – 3} \right)^2} = 9\\  \Rightarrow \left( 1 \right)VN\\  \Rightarrow x = \pi  + k2\pi\,(k\in \mathbb{Z}) \end{array}$

    Câu 16:

    $\begin{array}{l} 3\cos 4x + {\left( {\cos 2x – \sin x} \right)^2} = 7\\  \Leftrightarrow \left[ \begin{array}{l} \cos 4x \le 1,\cos 2x \le 1,\sin x \ge  – 1\\ \cos 4x \le 1,\cos 2x \ge  – 1,\sin x \le 1 \end{array} \right.\\  \Rightarrow \left[ \begin{array}{l} 3\cos 4x \le 3,{\left( {\cos 2x – \sin x} \right)^2} \le 4\\ 3\cos 4x \le 3,{\left( {\cos 2x – \sin x} \right)^2} \le 4 \end{array} \right. \end{array}$

    mà theo đề $3\cos 4x+(\cos 2x-\sin x)^2=7$ nên dấu bằng xảy ra.

    Vậy dấu bằng xảy ra khi và chỉ khi:

    $\begin{array}{l} \left\{ \begin{array}{l} \cos 4x = 1\\ \cos 2x – \sin x = 2 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} \cos 4x = 1\\ \left[ \begin{array}{l} \left\{ \begin{array}{l} \cos 2x =  – 1\\ \sin x = 1 \end{array} \right.\\ \left\{ \begin{array}{l} \cos 2x = 1\\ \sin x =  – 1 \end{array} \right. \end{array} \right. \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} 4x = k2\pi \\ \left[ \begin{array}{l} \left\{ \begin{array}{l} 2x = \pi  + k2\pi \\ x = \dfrac{\pi }{2} + k2\pi  \end{array} \right.\\ \left\{ \begin{array}{l} 2x = k2\pi \\ x =  – \dfrac{\pi }{2} + k2\pi  \end{array} \right.(L) \end{array} \right. \end{array} \right.\\  \Rightarrow \left\{ \begin{array}{l} \cos 4x = 1\\ \cos 2x =  – 1\\ \sin x = 1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 4x = k2\pi \\ 2x = \pi  + k2\pi \\ x = \dfrac{\pi }{2} + k2\pi  \end{array} \right.\\  \Rightarrow x = \dfrac{\pi }{2} + k\pi  \end{array}$

     Thử lại ta được $x=\dfrac{\pi} 2+k2\pi$ thỏa mãn yêu cầu bài toán.

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