$\sqrt {3{x^2} - 5x - 1} = x - 1$ .

$S = \left\{ 1 \right\}.$

$S = \left\{ 2 \right\}.$

$S = \left\{ 3 \right\}.$

$S = \left\{ 4 \right\}.$

Đáp án

- Hướng dẫn giải

Phương pháp giải:

$\sqrt {f\left( x \right)} = g\left( x \right) \Leftrightarrow \left\{ \begin{array}{l}f\left( x \right) \ge 0\\g\left( x \right) \ge 0\\f\left( x \right) = {g^2}\left( x \right)\end{array} \right.$

Giải chi tiết:

ĐKXĐ: $3{x^2} - 5x - 1 \ge 0 \Leftrightarrow \left[ \begin{array}{l}x \ge \frac{{5 + \sqrt {37} }}{6}\\x \le \frac{{5 - \sqrt {37} }}{6}\end{array} \right.$

$\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}x - 1 \ge 0\\3{x^2} - 5x - 1 = {(x - 1)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 1\\3{x^2} - 5x - 1 = {x^2} - 2x + 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ge 1\\2{x^2} - 3x - 2 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 1\\\left[ \begin{array}{l}x = 2\\x = - \frac{1}{2}\end{array} \right.\end{array} \right. \Leftrightarrow x = 2\,\,\left( {tm} \right)\end{array}$