Đáp án:
\(MinP = \dfrac{{39}}{8}\)
Giải thích các bước giải:
\(\begin{array}{l}
P = {\left( {x\sqrt 2 } \right)^2} - 2.x\sqrt 2 .\dfrac{1}{{2\sqrt 2 }} + {\left( {\dfrac{1}{{2\sqrt 2 }}} \right)^2} + \dfrac{{39}}{8}\\
= {\left( {x\sqrt 2 - \dfrac{1}{{2\sqrt 2 }}} \right)^2} + \dfrac{{39}}{8}\\
Do:{\left( {x\sqrt 2 - \dfrac{1}{{2\sqrt 2 }}} \right)^2} \ge 0\forall x \in R\\
\to {\left( {x\sqrt 2 - \dfrac{1}{{2\sqrt 2 }}} \right)^2} + \dfrac{{39}}{8} \ge \dfrac{{39}}{8}\\
\to P \ge \dfrac{{39}}{8}\\
\to MinP = \dfrac{{39}}{8}\\
\Leftrightarrow x\sqrt 2 - \dfrac{1}{{2\sqrt 2 }} = 0\\
\Leftrightarrow x = \dfrac{1}{4}
\end{array}\)
