Đáp án:
\[{A_{\min }} = \dfrac{1}{8} \Leftrightarrow x = \dfrac{1}{4}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = 2{x^2} - x + \dfrac{1}{4}\\
= 2.\left( {{x^2} - \dfrac{1}{2}x + \dfrac{1}{{16}}} \right) + \dfrac{1}{8}\\
= 2.\left[ {{x^2} - 2.x.\dfrac{1}{4} + {{\left( {\dfrac{1}{4}} \right)}^2}} \right] + \dfrac{1}{8}\\
= 2.{\left( {x - \dfrac{1}{4}} \right)^2} + \dfrac{1}{8}\\
{\left( {x - \dfrac{1}{4}} \right)^2} \ge 0,\,\,\,\,\forall x \Rightarrow 2.{\left( {x - \dfrac{1}{4}} \right)^2} + \dfrac{1}{8} \ge \dfrac{1}{8},\,\,\,\,\forall x\\
\Rightarrow A \ge \dfrac{1}{8},\,\,\,\forall x\\
\Rightarrow {A_{\min }} = \dfrac{1}{8} \Leftrightarrow {\left( {x - \dfrac{1}{4}} \right)^2} = 0 \Leftrightarrow x = \dfrac{1}{4}
\end{array}\)
Vậy \({A_{\min }} = \dfrac{1}{8} \Leftrightarrow x = \dfrac{1}{4}\)
