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