Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 1\\
\sqrt {3x - 2} + \sqrt {x - 1} = t\\
\Rightarrow 4x - 3 + 2\sqrt {3{x^2} - 5x + 2} = {t^2}\\
\Rightarrow t = {t^2} - 6\\
\Rightarrow {t^2} - t - 6 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = - 2\left( L \right)\\
t = 3\left( N \right)
\end{array} \right.\\
\Rightarrow \sqrt {3x - 2} + \sqrt {x - 1} = 3\\
\Rightarrow 4x - 3 + 2\sqrt {3{x^2} - 5x + 2} = 9\\
\Leftrightarrow \sqrt {3{x^2} - 5x + 2} = 6 - x\left( {x \le 6} \right)\\
\Rightarrow 3{x^2} - 5x + 2 = 36 - 12x + {x^2}\\
\Leftrightarrow 2{x^2} + 7x - 34 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 7 + \sqrt {321} }}{4}\left( N \right)\\
x = \frac{{ - 7 - \sqrt {321} }}{4}\left( L \right)
\end{array} \right.\\
2a)\left\{ \begin{array}{l}
\left( {{x^2} + 5x} \right)\left( {5x + y} \right) = 36\\
{x^2} + 5x + 5x + y = 12
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
5x + y = 12 - \left( {{x^2} + 5x} \right)\\
\left( {{x^2} + 5x} \right)\left( {12 - \left( {{x^2} + 5x} \right)} \right) = 36
\end{array} \right.\\
\Rightarrow {\left( {{x^2} + 5x} \right)^2} - 12\left( {{x^2} + 5x} \right) + 36 = 0\\
\Leftrightarrow {x^2} + 5x = 6\\
\Leftrightarrow {x^2} + 5x - 6 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1 \Rightarrow y = 1\\
x = - 6 \Rightarrow y = 36
\end{array} \right.
\end{array}\)
