Đáp án:
$\begin{array}{l}
\sqrt {3x - 2} + \sqrt[3]{x} = 2\left( {dkxd:x \ge \frac{2}{3}} \right)\\
\Rightarrow \sqrt {3x - 2} - 1 + \sqrt[3]{x} - 1 = 0\\
\Rightarrow \frac{{\left( {\sqrt {3x - 2} - 1} \right)\left( {\sqrt {3x - 2} + 1} \right)}}{{\sqrt {3x - 2} + 1}} + \frac{{\left( {\sqrt[3]{x} - 1} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}}{{\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1}} = 0\\
\Rightarrow \frac{{3x - 2 - 1}}{{\sqrt {3x - 2} + 1}} + \frac{{x - 1}}{{\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1}} = 0\\
\Rightarrow \left( {x - 1} \right).\left( {\frac{3}{{\sqrt {3x - 2} + 1}} + \frac{1}{{\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1}}} \right) = 0\\
\Rightarrow x = 1\left( {tmdk} \right)\\
\left( {do:\frac{3}{{\sqrt {3x - 2} + 1}} + \frac{1}{{\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1}} > 0} \right)
\end{array}$
Vậy x=1
