Trả lời bởi giáo viên
Đáp án đúng: c
\(\begin{array}{l}\,\,\,\,\,\,\,\sqrt {3{x^2} - 8x + 1} = 5 - x\\ \Leftrightarrow \left\{ \begin{array}{l}3{x^2} - 8x + 1 = 25 - 10x + {x^2}\\5 - x \ge 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}2{x^2} + 2x - 24 = 0\\x \le 5\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = 3\\x = - 4\end{array} \right.\\x \le 5\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 3\\x = - 4\end{array} \right.\end{array}\).
Vậy \(S = \left\{ {3; - 4} \right\}\).
Hướng dẫn giải:
\(\sqrt {f\left( x \right)} = g\left( x \right) \Leftrightarrow {\left\{ \begin{array}{l}f\left( x \right) = \left( {g\left( x \right)} \right)\\g\left( x \right) \ge 0\end{array} \right.^2}\)