Đáp án:
$\begin{array}{l}
y = \sqrt {{{\tan }^2}x} + \dfrac{1}{{\sin 4x}} - 1\\
Dkxd:\left\{ \begin{array}{l}
\cos x\# 0\\
{\tan ^2}x \ge 0\left( {tm} \right)\\
\sin 4x\# 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x\# \dfrac{\pi }{2} + k\pi \\
4x\# k\pi
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\# \dfrac{\pi }{2} + k\pi \\
x\# \dfrac{{k\pi }}{4}
\end{array} \right.\\
\Leftrightarrow x\# \dfrac{{k\pi }}{4}\\
Vậy\,TXD:D = \,R\backslash \left\{ {\dfrac{{k\pi }}{4}/k \in Z} \right\}\\
y = \dfrac{{2x + 1}}{{{{\tan }^2}x + 1}}\\
Dkxd:\left\{ \begin{array}{l}
{\tan ^2}x + 1\# 0\left( {tm} \right)\\
\cos x\# 0
\end{array} \right.\\
\Leftrightarrow x\# \dfrac{\pi }{2} + k\pi \\
Vậy\,TXD:D = R\backslash \left\{ {\dfrac{\pi }{2} + k\pi /k \in Z} \right\}
\end{array}$
