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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: titm txđ hàm số y=tanx/sin^2x-cos^2x y=tan2x/sinx.căncos^2x+1

Home/ Toán Lớp 11: titm txđ hàm số y=tanx/sin^2x-cos^2x y=tan2x/sinx.căncos^2x+1

Toán Lớp 11: titm txđ hàm số y=tanx/sin^2x-cos^2x y=tan2x/sinx.căncos^2x+1

Tháng Một 2, 2023 Toán học 2 Comments 0 views

Toán Lớp 11: titm txđ hàm số
y=tanx/sin^2x-cos^2x
y=tan2x/sinx.căncos^2x+1

Comments ( 2 )

  1. maingocquynhnhu2k1
    maingocquynhnhu2k1
    2 Tháng Một, 2023 at 12:14 sáng
    Reply
    a) Điều kiện xác định:
    $\begin{array}{l} \left\{ \begin{array}{l} \cos x \ne 0\\ {\sin ^2}x – {\cos ^2}x \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \cos x \ne 0\\ \cos 2x \ne 0 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} x \ne \dfrac{\pi }{2} + k\pi \\ 2x \ne \dfrac{\pi }{2} + k\pi  \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ne \dfrac{\pi }{2} + k\pi \\ x \ne \dfrac{\pi }{4} + \dfrac{{k\pi }}{2} \end{array} \right.\\ D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{2} + k\pi ;\dfrac{\pi }{4} + \dfrac{{k\pi }}{2}|k \in Z} \right\} \end{array}$
    b) Điều kiện xác định:
    $\begin{array}{l} \left\{ \begin{array}{l} \cos 2x \ne 0\\ \sin x \ne 0\\ \sqrt {{{\cos }^2}x + 1}  > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \cos 2x \ne 0\\ \sin x \ne 0\\ {\cos ^2}x + 1 > 0\left( {do\,{{\cos }^2}x + 1 \ge 1} \right) \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} 2x \ne \dfrac{\pi }{2} + k\pi \\ x \ne k\pi  \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ne \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\ x \ne k\pi  \end{array} \right.\\ D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{4} + \dfrac{{k\pi }}{2};k\pi |k \in Z} \right\} \end{array}$

  2. lanhuongthu
    lanhuongthu
    2 Tháng Một, 2023 at 12:14 sáng
    Reply
    ~rai~
    \(a)y=\dfrac{\tan x}{\sin^2x-\cos^2x}\\ĐKXĐ:\begin{cases}\cos x\ne 0\\\sin^2x-\cos ^2x\ne 0\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\\cos^2x-\sin^2x\ne 0\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\\cos2x\ne 0\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\2x\ne\dfrac{\pi}{2}+k\pi\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\x\ne\dfrac{\pi}{4}+k\dfrac{\pi}{2}\end{cases}\quad(k\in\mathbb{Z})\\TXĐ:D=\mathbb{R}\backslash\left\{\dfrac{\pi}{2}+k\pi;\dfrac{\pi}{4}+k\dfrac{\pi}{2}\Big|k\in\mathbb{Z}\right\}.\\b)y=\dfrac{\tan2x}{\sin x\sqrt{\cos^2x+1}}\\ĐKXĐ:\begin{cases}\cos2x\ne 0\\\sin x\sqrt{\cos^2x+1}\ne 0\end{cases}\\\Leftrightarrow \begin{cases}2x\ne\dfrac{\pi}{2}+k\pi\\\sin x\ne 0\\\cos^2x+1>0\quad\text{(luôn đúng vì }\cos^2x+1\ge 1\quad\forall x\in\mathbb{R})\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{4}+k\dfrac{\pi}{2}\\x\ne k\pi.\end{cases}(k\in\mathbb{Z})\\TXĐ:D=\mathbb{R}\backslash\left\{\dfrac{\pi}{4}+k\dfrac{\pi}{2};k\pi\Big|k\in\mathbb{Z}\right\}.\)

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222-9+11+12:2*14+14 = ? ( )

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