Giải thích các bước giải:
6,
ĐKXĐ: \(x \ge \frac{1}{2}\)
Ta có:
\(\begin{array}{l}
\sqrt {2x - 1} + {x^2} - 3x + 1 = 0\\
\Leftrightarrow \left( {\sqrt {2x - 1} - x} \right) + \left( {{x^2} - 2x + 1} \right) = 0\\
\Leftrightarrow \left( {\frac{{2x - 1 - {x^2}}}{{\sqrt {2x - 1} + x}}} \right) + {\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow \frac{{ - {{\left( {x - 1} \right)}^2}}}{{\sqrt {2x - 1} + x}} + {\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2}\left( {1 - \frac{1}{{\sqrt {2x - 1} + x}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\sqrt {2x - 1} + x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\sqrt {2x - 1} = 1 - x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\left\{ \begin{array}{l}
x \le 1\\
2x - 1 = {x^2} - 2x + 1
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\left\{ \begin{array}{l}
x \le 1\\
{x^2} - 4x + 2 = 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2 - \sqrt 2
\end{array} \right.\left( {t/m} \right)
\end{array}\)
