$\begin{cases} mx-1<0\\2x-1 >= \frac{x+1}{2} +3 \end{cases}$ tìm m để bpt sau vô nghiệm

Đáp án: $\,m \ge \dfrac{1}{3}$

Giải thích các bước giải:

$\begin{array}{l} \left\{ \begin{array}{l} mx - 1 < 0\\ 2x - 1 \ge \dfrac{{x + 1}}{2} + 3 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} mx < 1\\ 2x - \dfrac{1}{2}x \ge \dfrac{1}{2} + 3 + 1 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} mx < 1\\ \dfrac{3}{2}x \ge \dfrac{9}{2} \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} mx < 1\\ x \ge 3 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x < \dfrac{1}{m}\left( {m > 0} \right)\\ x > \dfrac{1}{m}\left( {m < 0} \right) \end{array} \right.\\ x \ge 3 \end{array} \right.\\ Khi:x \in \emptyset \\  \Leftrightarrow \dfrac{1}{m} \le 3\left( {m > 0} \right)\left( {de:\left\{ \begin{array}{l} x < 3\\ x \ge 3 \end{array} \right.} \right)\\  \Leftrightarrow \dfrac{1}{m} - 3 \le 0\left( {m > 0} \right)\\  \Leftrightarrow \dfrac{{1 - 3m}}{m} \le 0\left( {m > 0} \right)\\  \Leftrightarrow 1 - 3m \le 0\\  \Leftrightarrow 3m \ge 1\\  \Leftrightarrow m \ge \dfrac{1}{3}\left( {tm} \right)\\ Vậy\,m \ge \dfrac{1}{3} \end{array}$