Đáp án:
\(\left[ \begin{array}{l}
x = 17\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 1\\
\sqrt {3x - 2} + \sqrt {x - 1} = 3\\
\to 3x - 2 + 2\sqrt {\left( {3x - 2} \right)\left( {x - 1} \right)} + x - 1 = 9\\
\to 4x - 3 + 2\sqrt {3{x^2} - 5x + 2} = 9\\
\to 2\sqrt {3{x^2} - 5x + 2} = 12 - 4x\\
\to \sqrt {3{x^2} - 5x + 2} = 6 - 2x\\
\to 3{x^2} - 5x + 2 = 36 - 24x + 4{x^2}\\
\to {x^2} - 19x + 34 = 0\\
\to \left( {x - 2} \right)\left( {x - 17} \right) = 0\\
\to \left[ \begin{array}{l}
x = 17\\
x = 2
\end{array} \right.
\end{array}\)
