Đặt $\sqrt{x^3+3}=t(t\ge 0)$
Phương trình trở thành:
$\begin{array}{l} {x^3} + 6{x^2} - 2x + 3 = \left( {5x - 1} \right)\sqrt {{x^3} + 3} \left( {DK:x \ge \sqrt[3]{{ - 3}}} \right)\\ \left( {{x^3} + 3} \right) + 6{x^2} - 2x - \left( {5x - 1} \right)\sqrt {{x^3} + 3} = 0\\ \Leftrightarrow {t^2} - \left( {5x - 1} \right)t + 6{x^2} - 2x = 0\left( {t = \sqrt {{x^3} + 3} ,t \ge 0} \right)\\ \Rightarrow \Delta = {\left( {5x - 1} \right)^2} - 4\left( {6{x^2} - 2x} \right)\\ = 25{x^2} - 10x + 1 - 24{x^2} + 8x = {x^2} - 2x + 1 = {\left( {x - 1} \right)^2}\\ \Rightarrow \left[ \begin{array}{l} t = \dfrac{{5x - 1 - \left( {x - 1} \right)}}{2} = 2x\\ t = \dfrac{{5x - 1 + \left( {x - 1} \right)}}{2} = 3x - 1 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} \sqrt {{x^3} + 3} = 2x\left( {x \ge 0} \right)\\ \sqrt {{x^3} + 3} = 3x - 1\left( {x \ge \dfrac{1}{3}} \right) \end{array} \right. \Rightarrow \left[ \begin{array}{l} {x^3} + 3 = 4{x^2}\\ {x^3} + 3 = 9{x^2} - 6x + 1 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} {x^3} - 4{x^2} + 3 = 0\\ {x^3} - 9{x^2} + 6x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left( {x - 1} \right)\left( {{x^2} - 3x - 3} \right) = 0\\ \left( {x - 1} \right)\left( {{x^2} - 8x - 2} \right) = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = \dfrac{{3 \pm \sqrt {21} }}{2}\\ x = 4 \pm 3\sqrt 2 \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = 1\\ x = \dfrac{{3 + \sqrt {21} }}{2}\\ x = 4 + 3\sqrt 2 \end{array} \right.\\ \Rightarrow S = \left\{ {1;\dfrac{{3 + \sqrt {21} }}{2};4 + 3\sqrt 2 } \right\} \end{array}$
