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

Toán Lớp 11: tìm tập xác định của hàm số sau (căn 2+sinx)-(1/(tan^2x-1))

Toán Lớp 11: tìm tập xác định của hàm số sau (căn 2+sinx)-(1/(tan^2x-1))

Comments ( 2 )

  1. ~rai~
    \(\sqrt{2+\sin x}-\dfrac{1}{tan^2x-1}\\ĐKXĐ:\begin{cases}2+\sin x\ge 0\\\cos x\ne 0\\\tan^2x\ne 1.\end{cases}\quad(1)\\\text{Ta có:}-1\le \sin x\le 1\quad\forall x\\\Leftrightarrow 1\le 2+\sin x\le 3\quad\forall x\\\Rightarrow 2+\sin x>0\quad\forall x\quad nên \\(1)\Leftrightarrow \begin{cases}\cos x\ne 0\\\tan^2x\ne 1\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\\dfrac{\sin^2x}{\cos^2x}\ne 1\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\\sin^2x\ne \cos^2x\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\\cos^2x-\sin^2x\ne 0\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\\cos2x\ne 0\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\2x\ne\dfrac{\pi}{2}+k\pi\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\x\ne\dfrac{\pi}{4}+k\dfrac{\pi}{2}.\end{cases}\quad(k\in\mathbb{Z})\\D=\mathbb{R}\backslash\left\{\dfrac{\pi}{2}+k\pi;\dfrac{\pi}{4}+k\dfrac{\pi}{2}\Big|k\in\mathbb{Z}\right\}.\)

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