\[\begin{array}{l}
Bai\,\,6:\\
1)\,\,\,y = \frac{1}{{\sqrt {\sin x + 1} }}\\
DK:\,\,\,\sin x + 1 > 0 \Leftrightarrow \sin x > - 1\\
\Leftrightarrow \sin x \ne - 1\,\,\,\left( {do\,\,\, - 1 \le \sin x \le 1\,\,\,\forall x} \right)\\
\Leftrightarrow x \ne - \frac{\pi }{2} + k2\pi .\\
TXD:\,\,\,D = R\backslash \left\{ { - \frac{\pi }{2} + k2\pi ,\,\,k \in Z} \right\}.\\
6)\,\,\,y = \frac{{\tan x}}{{\sqrt {2 - \cos x} }}\\
DK:\,\,\,\cos x \ne 0\\
\Leftrightarrow x \ne \frac{\pi }{2} + k\pi \\
TXD:\,\,\,D = R\backslash \left\{ {\frac{\pi }{2} + k\pi \,\,\,\left( {k \in Z} \right)} \right\}.
\end{array}\]
\[\begin{array}{l}
Bai\,\,7:\,\,\\
d)\,\,y = \frac{{\sin x}}{{\cos \sqrt {{x^2} - 2x} }}\\
DK:\,\,\,\left\{ \begin{array}{l}
{x^2} - 2x \ge 0\\
\cos \sqrt {{x^2} - 2x} \ne 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.\\
\sqrt {{x^2} - 2x} \ne \frac{\pi }{2} + k\pi
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.\\
\left[ \begin{array}{l}
k \le - 1\\
\left\{ \begin{array}{l}
k \ge 0\\
{x^2} - 2x \ne {\left( {\frac{\pi }{2} + k\pi } \right)^2}
\end{array} \right.
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.\\
\left[ \begin{array}{l}
k \le - 1\\
\left\{ \begin{array}{l}
k \ge 0\\
{x^2} - 2x \ne {\left( {\frac{\pi }{2} + k\pi } \right)^2}
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.\\
\left[ \begin{array}{l}
k \le - 1\\
\left\{ \begin{array}{l}
k \ge 0\\
{x^2} - 2x - a \ne 0\,\,\,\left( {a = {{\left( {\frac{\pi }{2} + k\pi } \right)}^2}} \right)
\end{array} \right.
\end{array} \right.
\end{array} \right.
\end{array}\]
Những câu khác em làm tương tự nhé!
