Đáp án:
$\begin{array}{l}
c)\sqrt {\dfrac{{2021{x^2} + 3}}{{1 - 5x}}} \\
Dkxd:\dfrac{{2021{x^2} + 3}}{{1 - 5x}} \ge 0\\
\Leftrightarrow 1 - 5x > 0\\
\Leftrightarrow x < \dfrac{1}{5}\\
Vậy\,TXD:D = \left( { - \infty ;\dfrac{1}{5}} \right)\\
2)a)5\sqrt 2 + \dfrac{1}{2}\sqrt {32} + 7\sqrt {72} \\
= 5\sqrt 2 + \dfrac{1}{2}.4\sqrt 2 + 7.6\sqrt 2 \\
= 5\sqrt 2 + 2\sqrt 2 + 42\sqrt 2 \\
= 49\sqrt 2 \\
b)\sqrt {4{x^2} - 12x + 9} \\
= \sqrt {{{\left( {2x - 3} \right)}^2}} \\
= \left| {2x - 3} \right|\\
= 2x - 3\left( {do:2x \ge 3} \right)\\
c)\sqrt {5 - 2\sqrt 6 } + \sqrt {5 + 2\sqrt 6 } \\
= \sqrt {3 - 2.\sqrt 3 .\sqrt 2 + 2} + \sqrt {3 + 2\sqrt 3 .\sqrt 2 + 2} \\
= \sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} + \sqrt {{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}} \\
= \sqrt 3 - \sqrt 2 + \sqrt 3 + \sqrt 2 \\
= 2\sqrt 3 \\
3)Dkxd:x \ge \dfrac{3}{2}\\
\sqrt {4{x^2} + 24x + 16} = 2x - 3\\
\Leftrightarrow 4{x^2} + 24x + 16 = 4{x^2} - 12x + 9\\
\Leftrightarrow 36x = - 7\\
\Leftrightarrow x = - \dfrac{7}{{36}}\left( {ktm} \right)\\
Vậy\,x \in \emptyset
\end{array}$
