Đáp án:
\[\left[ \begin{array}{l}
x = 2 \pm \sqrt 2 \\
x = 8 \pm 3\sqrt 6
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ: \(x \ge \frac{1}{2}\)
Ta có:
\(\begin{array}{l}
{x^2} - 6x + 2 = 2\left( {2 - x} \right)\sqrt {2x - 1} \\
\Leftrightarrow {\left( {{x^2} - 6x + 2} \right)^2} = 4{\left( {2 - x} \right)^2}\left( {2x - 1} \right)\\
\Leftrightarrow {x^4} + 36{x^2} + 4 - 12{x^3} + 4{x^2} - 24x = 4\left( {{x^2} - 4x + 4} \right)\left( {2x - 1} \right)\\
\Leftrightarrow {x^4} - 12{x^3} + 40{x^2} - 24x + 4 = 4\left( {2{x^3} - {x^2} - 8{x^2} + 4x + 8x - 4} \right)\\
\Leftrightarrow {x^4} - 12{x^3} + 40{x^2} - 24x + 4 = 8{x^3} - 36{x^2} + 48x - 16\\
\Leftrightarrow {x^4} - 20{x^3} + 76{x^2} - 72x + 20 = 0\\
\Leftrightarrow \left( {{x^4} - 4{x^3} + 2{x^2}} \right) - \left( {16{x^3} - 64{x^2} + 32x} \right) + \left( {10{x^2} - 40x + 20} \right) = 0\\
\Leftrightarrow {x^2}\left( {{x^2} - 4x + 2} \right) - 16x\left( {{x^2} - 4x + 2} \right) + 10\left( {{x^2} - 4x + 2} \right) = 0\\
\Leftrightarrow \left( {{x^2} - 4x + 2} \right)\left( {{x^2} - 16x + 10} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 4x + 2 = 0\\
{x^2} - 16x + 10 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2 \pm \sqrt 2 \left( {t/m} \right)\\
x = 8 \pm 3\sqrt 6 \left( {t/m} \right)
\end{array} \right.
\end{array}\)
