Đáp án:
Giải thích các bước giải:
$\text{ta có :}\\D = \left[\sqrt{2} ; 0\right] \cup \left[\sqrt{6} ; +\infty\right]\\y' = \dfrac{3x^2 - 2}{2\sqrt{x^3 - 2x}}\\y' = 0\\\Leftrightarrow x = \pm \dfrac{\sqrt{6}}3\\\Leftrightarrow y = -\dfrac{\sqrt{6}}3\\\text{ta có BBT :}\\\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\sqrt{2}$}&\text{}&\text{}\dfrac{-\sqrt{6}}{3}&\text{}&\text{}0&\text{}\sqrt{2}&\text{$$}&+\infty\text{}\\\hline \text{$y'$}&∥\text{}&+\text{}&\text{0}&-\text{}&∥\text{}&∥\text{}&&+\text{}\\\hline \text{$y$}&\text{}&\text{}&\text{}\dfrac{2}{3} . \sqrt[4]{6}&\text{}&\text{}&\text{}&&+\infty\\&\text{}&\text{$\nearrow$}&\text{}&\text{}\searrow&\text{}&\text{}&\nearrow\\&\text{$$}+\infty&\text{}&\text{}&\text{}&\text{}0 &0\text{}&\text{}\\\hline \end{array}\\\mathscr{X_{\textit{in hay nhất}}}$$