Đáp án:
\(\left[ \begin{array}{l}
x = 1\\
x = 2 + \sqrt 2 \\
x = 2 - \sqrt 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
{x^2} - 3x + 1 + \sqrt {2x - 1} = 0\\
\to \sqrt {2x - 1} = - 1 + 3x - {x^2}\\
\to 2x - 1 = 1 + 9{x^2} + {x^4} - 6x - 6{x^3} + 2{x^2}\\
\to {x^4} - 6{x^3} + 11{x^2} - 8x + 2 = 0\\
\to {x^4} - {x^3} - 5{x^3} + 5{x^2} + 6{x^2} - 6x - 2x + 2 = 0\\
\to {x^3}\left( {x - 1} \right) - 5{x^2}\left( {x - 1} \right) + 6x\left( {x - 1} \right) - 2\left( {x - 1} \right) = 0\\
\to \left( {x - 1} \right)\left( {{x^3} - 5{x^2} + 6x - 2} \right) = 0\\
\to \left( {x - 1} \right)\left( {{x^3} - {x^2} - 4{x^2} + 4x + 2x - 2} \right) = 0\\
\to \left( {x - 1} \right)\left( {x - 1} \right)\left( {{x^2} - 4x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
{x^2} - 4x + 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = 2 + \sqrt 2 \\
x = 2 - \sqrt 2
\end{array} \right.
\end{array}\)
