Đáp án đúng: B
Phương pháp giải:
Bất phương trình \(\sqrt {f\left( x \right)} \le g\left( x \right) \Leftrightarrow \left\{ \begin{array}{l}g\left( x \right) \ge 0\\f\left( x \right) \ge 0\\f\left( x \right) \le {g^2}\left( x \right)\end{array} \right.\)
Giải chi tiết:\(\sqrt {3{x^2} - 2x - 5} \le x + 1\)\( \Leftrightarrow \left\{ \begin{array}{l}3{x^2} - 2x - 5 \ge 0\\x + 1 \ge 0\\3{x^2} - 2x - 5 \le {\left( {x + 1} \right)^2}\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}3{x^2} - 2x - 5 \ge 0\\x + 1 \ge 0\\3{x^2} - 2x - 5 \le {x^2} + 2x + 1\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}\left( {x + 1} \right)\left( {3x - 5} \right) \ge 0\,\,\,\,\\x \ge - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\2{x^2} - 4x - 6 \le 0\,\,\,\,\,\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x \ge \frac{5}{3}\\x \le - 1\end{array} \right.\\x \ge - 1\\\left( {x - 3} \right)\left( {x + 1} \right) \le 0\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x \ge \frac{5}{3}\\x = - 1\end{array} \right.\\ - 1 \le x \le 3\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - 1\\\frac{5}{3} \le x \le 3\end{array} \right..\)
Vậy bất phương trình có tập nghiệm là : \(S = \left\{ { - 1} \right\} \cup \left[ {\frac{5}{3};3} \right].\)
Chọn B.
