Đáp án:
$\begin{array}{l}
a)Dkxd:{x^2} - 3 \ge 0\\
\Rightarrow {x^2} \ge 3\\
\sqrt {{x^2} - 3} = 1\\
\Rightarrow {x^2} - 3 = {1^2}\\
\Rightarrow {x^2} = 4\left( {tm} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = - 2
\end{array} \right.\\
Vậy\,x = 2;x = - 2\\
b)Dkxd:\left\{ \begin{array}{l}
3{x^2} - 4x - 4 \ge 0\\
2x + 5 \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {3x + 2} \right)\left( {x - 2} \right) \ge 0\\
x \ge - \frac{5}{2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le - \frac{2}{3}
\end{array} \right.\\
x \ge - \frac{5}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x \ge 2\\
- \frac{5}{2} \le x \le - \frac{2}{3}
\end{array} \right.\\
\sqrt {3{x^2} - 4x - 4} = \sqrt {2x + 5} \\
\Rightarrow 3{x^2} - 4x - 4 = 2x + 5\\
\Rightarrow 3{x^2} - 6x - 9 = 0\\
\Rightarrow {x^2} - 2x - 3 = 0\\
\Rightarrow \left( {x - 3} \right)\left( {x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 3\left( {ktm} \right)\\
x = - 1\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = - 1
\end{array}$
