2 câu trả lời
$\begin{array}{l}
P = x\left( {1 - 2x} \right)\\
= - 2{x^2} + x\\
= - 2\left( {{x^2} - \dfrac{1}{2}x + \dfrac{1}{{16}}} \right) + \dfrac{1}{8}\\
= - 2{\left( {x - \dfrac{1}{4}} \right)^2} + \dfrac{1}{8} \le \dfrac{1}{8}\\
\Rightarrow \max P = \dfrac{1}{8}\\
' = ' \Leftrightarrow x = \dfrac{1}{4}
\end{array}$
