Đáp án:
$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
{x^2} - 9 \ge 0\\
x + 3 \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x^2} \ge 9\\
x \ge - 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 3\\
x \le - 3
\end{array} \right.\\
x \ge - 3
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = - 3\\
x \ge 3
\end{array} \right.\\
\sqrt {{x^2} - 9} + 2\sqrt {x + 3} = 0\\
\Rightarrow \sqrt {\left( {x - 3} \right)\left( {x + 3} \right)} + 2\sqrt {x + 3} = 0\\
\Rightarrow \sqrt {x + 3} .\sqrt {x - 3} + 2\sqrt {x + 3} = 0\\
\Rightarrow \sqrt {x + 3} .\left( {\sqrt {x - 3} + 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x + 3 = 0\\
\sqrt {x - 3} = - 2\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow x = - 3\left( {tmdk} \right)\\
\text{Vậy}\,x = - 3
\end{array}$
