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

Đáp án: $\,m \ge \dfrac{1}{3}$

Giải thích các bước giải:

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