Đáp án:
\[\left[ \begin{array}{l}
x = \frac{{ - 1 + 2\sqrt {10} }}{3}\\
x = \frac{{ - 1 - 2\sqrt {10} }}{3}
\end{array} \right.\]
Giải thích các bước giải:
ĐK: \({x^2} - x - 1 \ge 0 \Leftrightarrow \left[ \begin{array}{l}
x \ge \frac{{1 + \sqrt 5 }}{2}\\
x \le \frac{{1 - \sqrt 5 }}{2}
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
6x - 8 = \left( {3x - 1} \right)\sqrt {{x^2} - x - 1} \\
\Leftrightarrow {\left( {6x - 8} \right)^2} = {\left( {3x - 1} \right)^2}\left( {{x^2} - x - 1} \right)\\
\Leftrightarrow 36{x^2} - 96x + 64 = \left( {9{x^2} - 6x + 1} \right)\left( {{x^2} - x - 1} \right)\\
\Leftrightarrow 36{x^2} - 96x + 64 = 9{x^4} - 9{x^3} - 9{x^2} - 6{x^3} + 6{x^2} + 6x + {x^2} - x - 1\\
\Leftrightarrow 36{x^2} - 96x + 64 = 9{x^4} - 15{x^3} - 2{x^2} + 5x - 1\\
\Leftrightarrow 9{x^4} - 15{x^3} - 38{x^2} + 101x - 65 = 0\\
\Leftrightarrow \left( {9{x^4} + 6{x^3} - 39{x^2}} \right) - \left( {21{x^3} + 14{x^2} - 91x} \right) + \left( {15{x^2} + 10x - 65} \right) = 0\\
\Leftrightarrow 3{x^2}\left( {3{x^2} + 2x - 13} \right) - 7x\left( {3{x^2} + 2x - 13} \right) + 5\left( {3{x^2} + 2x - 13} \right) = 0\\
\Leftrightarrow \left( {3{x^2} + 2x - 13} \right)\left( {3{x^2} - 7x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
3{x^2} + 2x - 13 = 0\\
3{x^2} - 7x + 5 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 1 + 2\sqrt {10} }}{3}\\
x = \frac{{ - 1 - 2\sqrt {10} }}{3}
\end{array} \right.\left( {t/m} \right)
\end{array}\)
