\[\begin{array}{l}
\sqrt {3{x^2} - 4x - 4} = \sqrt {2x + 5} \\
dk:\left\{ \begin{array}{l}
3{x^2} - 4x - 4 \ge 0\\
2{\rm{x}} + 5 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le \frac{{ - 2}}{3}
\end{array} \right.\\
x \ge \frac{{ - 5}}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 2\\
\frac{{ - 5}}{2} \le x \le \frac{{ - 2}}{3}
\end{array} \right.\\
\Leftrightarrow 3{{\rm{x}}^2} - 4{\rm{x}} - 4 = 2{\rm{x}} + 5\\
\Leftrightarrow 3{{\rm{x}}^2} - 6{\rm{x}} - 9 = 0\\
\Leftrightarrow 3{{\rm{x}}^2} - 9{\rm{x}} + 3{\rm{x}} - 9 = 0\\
\Leftrightarrow 3{\rm{x}}\left( {x - 3} \right) + 3\left( {x - 3} \right) = 0\\
\Leftrightarrow 3\left( {x - 3} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\left( {tm} \right)\\
vay:S = \{ 3; - 1\}
\end{array}\]

        √3x^2 - 4x - 4  = √2x + 5

<=> √2x+5 = √2x^2 + x^2 - 4x - 4 

<=> √2x+5 = √2x(x-2) + (x - 2)(x+2)

<=> √2x+5 = √(x-2)(3x+2)

<=> √(2x+5)^2 = √[(x-2)(3x+2)]^2

<=> |2x+5| = |x-2|.|3x+2|

 đến đây mình tịt rồi . bạn nghĩ đc cách nào khác thì tự làm nha .