a) \(\left\{ \begin{array}{l}{x^2} \ge 4x\\{(2x - 1)^2} < 9\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 4x \ge 0\\ - 3 < 2x - 1 < 3\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x \in ( - \infty ;0] \cup {\rm{[}}4; + \infty )\\ - 1 < x < 2\end{array} \right.\) \( \Leftrightarrow - 1 < x \le 0\)
b) \(\left\{ \begin{array}{l}2x - 3 < (x + 1)(x - 2)\\{x^2} - x \le 6\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 3x + 1 > 0\\{x^2} - x - 6 \le 0\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x \in ( - \infty ;\dfrac{{3 - \sqrt 5 }}{2}) \cup (\dfrac{{3 + \sqrt 5 }}{2};3]\\ - 2 \le x \le 3\end{array} \right.\)
\( \Leftrightarrow x \in {\rm{[ - 2;}}\dfrac{{3 - \sqrt 5 }}{2}) \cup (\dfrac{{3 + \sqrt 5 }}{2};3{\rm{]}}\)
