Đáp án:
$\begin{array}{l}
a)\sqrt {4 - {x^2}} \\
Dkxd:4 - {x^2} \ge 0\\
\Leftrightarrow {x^2} \le 4\\
\Leftrightarrow - 2 \le x \le 2\\
Vậy\,TXD:D = \left[ { - 2;2} \right]\\
b)\sqrt {{x^2} - 16} \\
Dkxd:{x^2} - 16 \ge 0\\
\Leftrightarrow {x^2} \ge 16\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 4\\
x \le - 4
\end{array} \right.\\
Vậy\,TXD:D = R\backslash \left( { - 4;4} \right)\\
c)\sqrt {{x^2} - 3} \\
Dkxd:{x^2} - 3 \ge 0\\
\Leftrightarrow {x^2} \ge 3\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge \sqrt 3 \\
x \le - \sqrt 3
\end{array} \right.\\
Vậy\,TXD:D = R\backslash \left( { - \sqrt 3 ;\sqrt 3 } \right)
\end{array}$
