Đáp án:
$\begin{array}{l}
9)\sqrt {\dfrac{{\sqrt 6 - 4}}{{m + 2}}} \\
Dkxd:\dfrac{{\sqrt 6 - 4}}{{m + 2}} \ge 0\\
\Leftrightarrow m + 2 < 0\left( {do:\sqrt 6 - 4 < 0} \right)\\
\Leftrightarrow m < - 2\\
Vậy\,m < - 2\\
10)\dfrac{{16x - 1}}{{\sqrt {x - 7} }}\\
Dkxd:x - 7 > 0\\
\Leftrightarrow x > 7\\
Vậy\,x > 7\\
13)\sqrt {\sqrt 5 - \sqrt 3 x} \\
Dkxd:\sqrt 5 - \sqrt 3 x \ge 0\\
\Leftrightarrow \sqrt 3 x \le \sqrt 5 \\
\Leftrightarrow x \le \dfrac{{\sqrt 5 }}{{\sqrt 3 }}\\
\Leftrightarrow x \le \dfrac{{\sqrt {15} }}{3}\\
Vậy\,x \le \dfrac{{\sqrt {15} }}{3}\\
14)\sqrt {\sqrt 6 x - 4x} \\
Dkxd:\sqrt 6 x - 4x \ge 0\\
\Leftrightarrow \left( {\sqrt 6 - 4} \right).x \ge 0\\
\Leftrightarrow x \le 0\\
Vậy\,x \le 0\\
15)\sqrt {\left( {\sqrt x - 7} \right)\left( {\sqrt x + 7} \right)} \\
Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
\left( {\sqrt x - 7} \right)\left( {\sqrt x + 7} \right) \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\sqrt x \ge 7
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ge 49
\end{array} \right.\\
Vậy\,x \ge 49\\
16)\sqrt {{{\left( {x - 6} \right)}^6}} \\
DKxd:{\left( {x - 6} \right)^6} \ge 0\left( {tm} \right)\\
Vậy\,x \in R\\
19)\sqrt {\dfrac{{ - 2\sqrt 6 + \sqrt {23} }}{{ - x + 5}}} \\
Dkxd: - x + 5 < 0\left( {do: - 2\sqrt 6 + \sqrt {23} < 0} \right)\\
\Leftrightarrow x > 5\\
Vậy\,x > 5\\
20)\sqrt {2011 - m} \\
Dkxd:2011 - m \ge 0\\
\Leftrightarrow m \le 2011\\
Vậy\,m \le 2011\\
21)\\
\sqrt {\dfrac{{2\sqrt {15} - \sqrt {59} }}{{x - 7}}} \\
Dkxd:x - 7 > 0\\
\Leftrightarrow x > 7\\
Vậy\,x > 7\\
22)\sqrt {4{z^2} + 4z + 1} \\
Dkxd:4{z^2} + 4z + 1 \ge 0\\
\Leftrightarrow {\left( {2z + 1} \right)^2} \ge 0\left( {tm} \right)\\
Vậy\,z \in R
\end{array}$
