Đáp án:
$\begin{array}{l}
f)y = \tan \left( {4x + \dfrac{\pi }{6}} \right)\\
Dkxd:\cos \left( {4x + \dfrac{\pi }{6}} \right) \ne 0\\
\Leftrightarrow 4x + \dfrac{\pi }{6} \ne \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)\\
\Leftrightarrow 4x \ne \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)\\
\Leftrightarrow x \ne \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{4}\left( {k \in Z} \right)\\
Vậy\,TXD:D = R\backslash \left\{ {\dfrac{\pi }{{12}} + \dfrac{{k\pi }}{4}/k \in Z} \right\}\\
g)y = \cot \left( {\dfrac{x}{2} - \dfrac{\pi }{3}} \right)\\
Dkxd:\sin \left( {\dfrac{x}{2} - \dfrac{\pi }{3}} \right) \ne 0\\
\Leftrightarrow \dfrac{x}{2} - \dfrac{\pi }{3} \ne k\pi \\
\Leftrightarrow x \ne \dfrac{{2\pi }}{3} + k2\pi \\
Vậy\,TXD:D = R\backslash \left\{ {\dfrac{{2\pi }}{3} + k2\pi /k \in Z} \right\}\\
h)y = \dfrac{x}{{\sin 2x - 1}}\\
Dkxd:\sin 2x \ne 1\\
\Leftrightarrow 2x \ne \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow x \ne \dfrac{\pi }{4} + k\pi \\
Vậy\,TXD:D = R\backslash \left\{ {\dfrac{\pi }{4} + k\pi /k \in Z} \right\}
\end{array}$
