Решите неравенство:
\(\dfrac{x^2}{3}\geq\dfrac{3x-3}{2}.\)
\(x\in (-\infty~; 1{,}5] \cup [3~;+\infty)\)
\(x\in (-\infty~; 3] \cup [1{,}5~;+\infty)\)
\(x\in (1{,}5~;~3)\)
\(x\in [1{,}5~;~3]\)