Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
Bpt \to \left[ \begin{array}{l}
\left( {x - 4} \right)\sqrt {3x + 1} \ge 0\\
\left( {x - 4} \right)\sqrt {3x + 1} \le 0
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 4 \ge 0\\
\sqrt {3x + 1} \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 4 \le 0\\
\sqrt {3x + 1} \le 0
\end{array} \right.\\
\left\{ \begin{array}{l}
\left( {x - 4} \right) \ge 0\\
\sqrt {3x + 1} \le 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 4 \le 0\\
\sqrt {3x + 1} \ge 0
\end{array} \right.
\end{array} \right.\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 4\\
x \ge \frac{{ - 1}}{3}
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 4\\
x = \frac{{ - 1}}{3}
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge 4\\
x = \frac{{ - 1}}{3}
\end{array} \right.(l)\\
\left\{ \begin{array}{l}
x \le 4\\
x \ge \frac{{ - 1}}{3}
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x \ge 4\\
x = \frac{{ - 1}}{3}\\
\frac{{ - 1}}{3} \le x \le 4
\end{array} \right.
\end{array}\)
