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