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