Ta có

\(\left( {2x - {x^2}} \right)\left( {2{x^2} - 3x - 2} \right) = 0\) \( \Leftrightarrow \left[ \begin{array}{l}2x - {x^2} = 0\\2{x^2} - 3x - 2 = 0\end{array} \right.\) \( \Leftrightarrow \left[ \begin{array}{l}x\left( {2 - x} \right) = 0\\\left( {x - 2} \right)\left( {2x + 1} \right) = 0\end{array} \right.\) \( \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = 2\\x = - \dfrac{1}{2}\end{array} \right. \Rightarrow A = \left\{ { - \dfrac{1}{2};0;2} \right\}\)

Và \(\left\{ \begin{array}{l}n \in {\mathbb{N}^ * }\\3 < {n^2} < 30\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}n \in {\mathbb{N}^ * }\\\sqrt 3 < n < \sqrt {30} \end{array} \right.\) \( \Rightarrow B = \left\{ {2;3;4;5} \right\}\)

Suy ra \(A \cap B = \left\{ 2 \right\}.\)