Đáp án:
$\begin{array}{l}
a)\sqrt {{x^2} + 11} \\
Dkxd:{x^2} + 11 \ge 0\\
\Leftrightarrow {x^2} \ge - 11\left( {tmdk} \right)\\
Vậy\,x \in R\\
b)\sqrt {{x^2} + 5x + 6} \\
Dkxd:{x^2} + 5x + 6 \ge 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {x + 3} \right) \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge - 2\\
x \le - 3
\end{array} \right.\\
Vậy\,x \le - 3\,hoặc\,x \ge - 2\\
c)\sqrt {\dfrac{{10 - 3x}}{{3{x^2} + 1}}} \\
Dkxd:\dfrac{{10 - 3x}}{{3{x^2} + 1}} \ge 0\\
\Leftrightarrow 10 - 3x \ge 0\\
\Leftrightarrow x \le \dfrac{{10}}{3}\\
Vậy\,x \le \dfrac{{10}}{3}\\
d)\sqrt {\dfrac{{4x + 2}}{{{x^2} + 4x + 5}}} \\
DKxd:\dfrac{{4x + 2}}{{{x^2} + 4x + 5}} \ge 0\\
\Leftrightarrow \dfrac{{4x + 2}}{{{{\left( {x + 2} \right)}^2} + 1}} \ge 0\\
\Leftrightarrow 4x + 2 \ge 0\\
\Leftrightarrow x \ge - \dfrac{1}{2}\\
Vậy\,x \ge - \dfrac{1}{2}
\end{array}$
