Đáp án:
$\begin{array}{l}
1)y = \tan x + \dfrac{{3\cos x}}{{1 - \sin x}}\\
Dkxd:\left\{ \begin{array}{l}
\cos x \ne 0\\
1 - \sin x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{2} + k\pi \\
x \ne \dfrac{\pi }{2} + k2\pi
\end{array} \right. \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \\
Vậy\,TXD:D = R\backslash \left\{ {\dfrac{\pi }{2} + k\pi ;k \in Z} \right\}\\
2)y = \sqrt {\dfrac{{5 + \sin x}}{{2 - 2\cos x}}} \\
Dkxd:\dfrac{{5 + \sin x}}{{2 - 2\cos x}} \ge 0\\
\Leftrightarrow 2 - 2\cos x > 0\left( {do:\sin x + 5 > 0} \right)\\
\Leftrightarrow \cos x < 1\\
\Leftrightarrow \cos x \ne 1\\
\Leftrightarrow x \ne k2\pi \\
Vậy\,TXD:D = R\backslash \left\{ {k2\pi ;k \in Z} \right\}\\
3)y = \sqrt {\dfrac{{3 - \sin x}}{{4 + \cos x}}} \\
Dkxd:\dfrac{{3 - \sin x}}{{4 + \cos x}} \ge 0\left( {tm} \right)\\
Do:\left\{ \begin{array}{l}
3 - \sin x > 0\\
4 + \cos x > 0
\end{array} \right.\\
Vậy\,TXD:D = R
\end{array}$
