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

        √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 .