Đáp án:
\(\left[ \begin{array}{l}
x = - \dfrac{1}{2}\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\sqrt {6{x^2} - 3x + 1} = 10{x^2} - 5x - 3\\
\to \sqrt {3\left( {2{x^2} - x} \right) + 1} = 5\left( {2{x^2} - x} \right) - 3\\
Đặt:\sqrt {3\left( {2{x^2} - x} \right) + 1} = t\left( {t \ge 0} \right)\\
\to 3\left( {2{x^2} - x} \right) + 1 = {t^2}\\
\to 2{x^2} - x = \dfrac{{{t^2} - 1}}{3}\\
Pt \to t = 5.\dfrac{{{t^2} - 1}}{3} - 3\\
\to 5{t^2} - 5 - 9 - 3t = 0\\
\to 5{t^2} - 3t - 14 = 0\\
\to \left[ \begin{array}{l}
t = 2\\
t = - \dfrac{7}{5}\left( l \right)
\end{array} \right.\\
\to 2{x^2} - x = \dfrac{{{2^2} - 1}}{3}\\
\to 2{x^2} - x - 1 = 0\\
\to \left[ \begin{array}{l}
x = - \dfrac{1}{2}\\
x = 1
\end{array} \right.
\end{array}\)