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