Đáp án:
$a) \dfrac{1}{2} \le x \le \dfrac{5}{2}\\ b) \left[\begin{array}{l} x\ge \dfrac{-1}{2} \\x \le \dfrac{-7}{2}\end{array} \right.$
Giải thích các bước giải:
$a)(2x-1)(2x-5) \le 0 \\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} 2x-1 \ge 0\\ 2x-5 \le 0\end{array} \right.\\ \left\{\begin{array}{l} 2x-1 \le 0\\ 2x-5 \ge 0\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} 2x\ge 1\\ 2x \le 5\end{array} \right.\\ \left\{\begin{array}{l} 2x \le 1\\ 2x \ge 5\end{array} \right.\end{array} \right.\\\Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x\ge \dfrac{1}{2}\\ x \le \dfrac{5}{2}\end{array} \right.\\ \left\{\begin{array}{l} x \le \dfrac{1}{2}\\ x \ge \dfrac{5}{2}\end{array} \right.(\text{Vô lí})\end{array} \right.\\ \Leftrightarrow \dfrac{1}{2} \le x \le \dfrac{5}{2}\\ b)(2x+1)(2x+7) \ge 0 \\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} 2x+1 \ge 0\\ 2x+7 \ge 0\end{array} \right.\\ \left\{\begin{array}{l} 2x+1 \le 0\\ 2x+7 \le 0\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} 2x\ge -1\\ 2x \ge -7\end{array} \right.\\ \left\{\begin{array}{l} 2x \le -1\\ 2x \le -7\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x\ge \dfrac{-1}{2}\\ x \ge \dfrac{-7}{2}\end{array} \right.\\ \left\{\begin{array}{l} x \le \dfrac{-1}{2}\\ x \le \dfrac{-7}{2}\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x\ge \dfrac{-1}{2} \\x \le \dfrac{-7}{2}\end{array} \right.$
