Đáp án:

$S = \left\{ { - 1;2} \right\}$

Giải thích các bước giải:

$\begin{array}{l}
2x + 3 = {x^2} + \left( {x - 1} \right)\sqrt[3]{{3{x^3} + 3}}\\
 \Leftrightarrow {x^2} - 2x - 3 + \left( {x - 1} \right)\sqrt[3]{{3{x^3} + 3}} = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {x - 3} \right) + \left( {x - 1} \right)\sqrt[3]{{3\left( {{x^3} + 1} \right)}} = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {x - 3} \right) + \left( {x - 1} \right)\sqrt[3]{{3\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {x - 3} \right) + \left( {x - 1} \right)\sqrt[3]{{x + 1}}.\sqrt[3]{{3\left( {{x^2} - x + 1} \right)}} = 0\\
 \Leftrightarrow \sqrt[3]{{x + 1}}\left( {\left( {x - 3} \right){{\left( {\sqrt[3]{{x + 1}}} \right)}^2} + \left( {x - 1} \right)\sqrt[3]{{3\left( {{x^2} - x + 1} \right)}}} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt[3]{{x + 1}} = 0\\
\left( {x - 3} \right){\left( {\sqrt[3]{{x + 1}}} \right)^2} + \left( {x - 1} \right)\sqrt[3]{{3\left( {{x^2} - x + 1} \right)}} = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
\left( {x - 3} \right)\sqrt[3]{{{{\left( {x + 1} \right)}^2}}} = \left( {1 - x} \right)\sqrt[3]{{3\left( {{x^2} - x + 1} \right)}}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
{\left( {x - 3} \right)^3}{\left( {x + 1} \right)^2} = 3{\left( {1 - x} \right)^3}\left( {{x^2} - x + 1} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
{x^5} - 7{x^4} + 10{x^3} + 18{x^2} - 27x - 27 = 3\left( { - {x^5} + 4{x^4} - 7{x^3} + 7{x^2} - 4x + 1} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
4{x^5} - 19{x^4} + 31{x^3} - 3{x^2} - 15x - 30 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
\left( {x - 2} \right)\left( {4{x^4} - 11{x^3} + 9{x^2} + 15x + 15} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
x = 2\\
4{x^4} - 11{x^3} + 9{x^2} + 15x + 15 = 0\left( 1 \right)
\end{array} \right.
\end{array}$

Lại có:

$\begin{array}{l}
\left( 1 \right) \Leftrightarrow 16{x^4} - 44{x^3} + 36{x^2} + 60x + 60 = 0\\
 \Leftrightarrow \left( {16{x^4} - 8{x^2} + 1} \right) - 44{x^3} + 44{x^2} + 60x + 59 = 0\\
 \Leftrightarrow {\left( {4{x^2} - 1} \right)^2} - 11x\left( {4{x^2} - 1} \right) + 44{x^2} + 49x + 59 = 0\\
 \Leftrightarrow {\left( {4{x^2} - 1} \right)^2} - 2.\left( {4{x^2} - 1} \right).\dfrac{{11x}}{2} + \dfrac{{121{x^2}}}{4} + \dfrac{{55}}{4}{x^2} + 49x + 59 = 0\\
 \Leftrightarrow {\left( {4{x^2} - 1 - \dfrac{{11x}}{2}} \right)^2} + \dfrac{{55}}{4}\left( {{x^2} + \dfrac{{196}}{{55}}x + {{\left( {\dfrac{{98}}{{55}}} \right)}^2}} \right) + \dfrac{{844}}{{55}} = 0\\
 \Leftrightarrow {\left( {4{x^2} - \dfrac{{15}}{2}x - 1} \right)^2} + \dfrac{{55}}{4}{\left( {x + \dfrac{{98}}{{55}}} \right)^2} + \dfrac{{844}}{{55}} = 0\left( {vn} \right)
\end{array}$

Vậy phương trình có tập nghiệm là: $S = \left\{ { - 1;2} \right\}$