Đáp án:
$\begin{array}{l}
1)Đkxđ:x > 0\\
y = {x^2}\ln x\\
\Rightarrow y' = 2x.\ln x + {x^2}.\frac{1}{x} = 2x.\ln + x = x\left( {2.\ln x + 1} \right)\\
y' = 0 \Rightarrow \left[ \begin{array}{l}
x = 0\left( {ktm} \right)\\
\ln x = \frac{{ - 1}}{2}
\end{array} \right. \Rightarrow x = {e^{\frac{{ - 1}}{2}}} = \frac{1}{{\sqrt e }}\\
\Rightarrow A\\
3)\sqrt {x\sqrt {x\sqrt {x\sqrt x } } } \left( {x > 0} \right)\\
= \sqrt {x\sqrt {x\sqrt {x.{x^{\frac{1}{2}}}} } } = \sqrt {x\sqrt {x\sqrt {{x^{\frac{3}{2}}}} } } = \sqrt {x\sqrt {x.{x^{\frac{3}{4}}}} } = \sqrt {x\sqrt {{x^{\frac{7}{4}}}} } = \sqrt {x.{x^{\frac{7}{8}}}} = \sqrt {{x^{\frac{{15}}{8}}}} = {x^{\frac{{15}}{{16}}}}\\
\Rightarrow C
\end{array}$
