Đáp án:
$C.\ D =\Bbb R\backslash\left\{k\dfrac{\pi}{4},\ k\in\Bbb Z\right\}$
Giải thích các bước giải:
$y = \sqrt{2 +\sin x} - \dfrac{1}{\tan^2\left(x -\dfrac{\pi}{4}\right) -1} + \tan\left(x +\dfrac{\pi}{4}\right)$
Hàm số xác định khi và chỉ khi
$\quad \begin{cases}\tan\left(x -\dfrac{\pi}{4}\right)\ne \pm 1\\\cos\left(x -\dfrac{\pi}{4}\right)\ne 0\\\cos\left(x +\dfrac{\pi}{4}\right) \ne 0\end{cases}$
$\Leftrightarrow \begin{cases}x - \dfrac{\pi}{4} \ne -\dfrac{\pi}{4} + k\pi\\x -\dfrac{\pi}{4} \ne \dfrac{\pi}{4} + k\pi\\x -\dfrac{\pi}{4}\ne \dfrac{\pi}{2} + k\pi\\x +\dfrac{\pi}{4} \ne \dfrac{\pi}{2} + k\pi\end{cases}$
$\Leftrightarrow \begin{cases}x \ne k\pi\\x \ne \dfrac{\pi}{2} + k\pi\\x \ne \dfrac{3\pi}{4} + k\pi\\x \ne \dfrac{\pi}{4} + k\pi\end{cases}$
$\Leftrightarrow x \ne k\dfrac{\pi}{4}\quad (k\in\Bbb Z)$
Vậy $D =\Bbb R\backslash\left\{k\dfrac{\pi}{4},\ k\in\Bbb Z\right\}$
