Đáp án:
$\begin{array}{l}
1)y = \dfrac{{2\sin x + 1}}{{{{\cos }^2}x - 4}}\\
Dkxd:{\cos ^2}x - 4 \ne 0\\
\Leftrightarrow {\cos ^2}x \ne 4\\
\Leftrightarrow \cos x \ne 2;\cos x \ne - 2\left( {tm} \right)\\
Do: - 1 \le \cos x \le 1\\
Vậy\,TXD:D = R\\
2)y = \tan 3x - \dfrac{{\sin x - 1}}{{\cos x}}\\
Dkxd:\left\{ \begin{array}{l}
\cos 3x \ne 0\\
\cos x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3x \ne \dfrac{\pi }{2} + k\pi \\
x \ne \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
\Leftrightarrow x \ne \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}\\
Vậy\,TXD:D = R\backslash \left\{ {\dfrac{\pi }{6} + \dfrac{{k\pi }}{3}} \right\}\\
3)y = \sin \sqrt {4 - {x^2}} \\
Dkxd:4 - {x^2} \ge 0\\
\Leftrightarrow {x^2} \le 4\\
\Leftrightarrow - 2 \le x \le 2\\
Vậy\,TXD:D = \left[ { - 2;2} \right]\\
4)y = \cos \sqrt {x + 1} + \dfrac{1}{{{x^2} - 4x}}\\
DKxd:\left\{ \begin{array}{l}
x + 1 \ge 0\\
{x^2} - 4x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 1\\
x \ne 0\\
x \ne 4
\end{array} \right.\\
Vậy\,TXD:D = \left[ { - 1; + \infty } \right)\backslash \left\{ {0;4} \right\}\\
11)y = \dfrac{{\sin x}}{{{{\sin }^2}x - {{\cos }^2}x}}\\
Dkxd:{\sin ^2}x - {\cos ^2}x \ne 0\\
\Leftrightarrow 1 - {\cos ^2}x - {\cos ^2}x \ne 0\\
\Leftrightarrow {\cos ^2}x \ne \dfrac{1}{2}\\
\Leftrightarrow x \ne \pm \dfrac{\pi }{3} + k2\pi ;x \ne \pm \dfrac{{2\pi }}{3} + k2\pi \\
Vậy\,TXD:D = R\backslash \left\{ { \pm \dfrac{\pi }{3} + k2\pi ; \pm \dfrac{{2\pi }}{3} + k2\pi } \right\}
\end{array}$
