Đáp án:

 đặt 2x+3=a

       x-1=b

=> $\sqrt[3]{a}$ -$\sqrt[3]{b}$ +a-b=0h các bước giải:

 <=>$\sqrt[3]{a}$ -$\sqrt[3]{b}$  +( $\sqrt[3]{a}$ -$\sqrt[3]{b})(a+$\sqrt[]{a}$ $\sqrt[]{b}$ +b)

<=> ($\sqrt[3]{a}$ -$\sqrt[3]{b}$)(a+$\sqrt[]{a}$ *$\sqrt[]{b}$ +b+1)=0

<=> $\sqrt[3]{a}$ -$\sqrt[3]{b}$=0

<=> $\sqrt[3]{a}$ =$\sqrt[3]{b}$

<=> a=b

<=> 2x+3=x-1

<=> x=-4

vậy x = -4

$\begin{array}{l} \sqrt[3]{{2x + 3}} - \sqrt[3]{{x - 1}} + x + 4 = 0\\  \Leftrightarrow \dfrac{{2x + 3 - x + 1}}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x - 1} \right)}} + \sqrt[3]{{{{\left( {x - 1} \right)}^2}}}}} + x + 4 = 0\\  \Leftrightarrow \dfrac{{x + 4}}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x - 1} \right)}} + \sqrt[3]{{{{\left( {x - 1} \right)}^2}}}}} + x + 4 = 0\\  \Leftrightarrow \left( {x + 4} \right)\left[ {\dfrac{1}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x - 1} \right)}} + \sqrt[3]{{{{\left( {x - 1} \right)}^2}}}}} + 1} \right] = 0\\  \Leftrightarrow \left[ \begin{array}{l} x + 4 = 0 \Leftrightarrow x =  - 4\\ \dfrac{1}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x - 1} \right)}} + \sqrt[3]{{{{\left( {x - 1} \right)}^2}}}}} + 1 = 0\left( * \right) \end{array} \right.\\ \left( * \right) \Rightarrow \sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x - 1} \right)}} + \sqrt[3]{{{{\left( {x - 1} \right)}^2}}} = {a^2} + ab + {b^2}\left( {a = \sqrt[3]{{2x + 3}},b = \sqrt[3]{{x - 1}}} \right)\\  \Rightarrow {a^2} + ab + {b^2} > 0\forall a,b \in R\\  \Rightarrow \dfrac{1}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x - 1} \right)}} + \sqrt[3]{{{{\left( {x - 1} \right)}^2}}}}} + 1 > 1\\  \Rightarrow \left( * \right)VN\\  \Rightarrow S = \left\{ { - 4} \right\} \end{array}$