Đáp án:
\(MaxA = \dfrac{6}{5}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{3}{{2{x^2} + 2x + 3}} = \dfrac{3}{{2{x^2} + 2.x\sqrt 2 .\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} + \dfrac{5}{2}}}\\
= \dfrac{3}{{{{\left( {x\sqrt 2 + \dfrac{1}{{\sqrt 2 }}} \right)}^2} + \dfrac{5}{2}}}\\
Do:{\left( {x\sqrt 2 + \dfrac{1}{{\sqrt 2 }}} \right)^2} \ge 0\forall x \in R\\
\to {\left( {x\sqrt 2 + \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{5}{2} \ge \dfrac{5}{2}\\
\to \dfrac{3}{{{{\left( {x\sqrt 2 + \dfrac{1}{{\sqrt 2 }}} \right)}^2} + \dfrac{5}{2}}} \le 3:\dfrac{5}{2}\\
\to \dfrac{3}{{{{\left( {x\sqrt 2 + \dfrac{1}{{\sqrt 2 }}} \right)}^2} + \dfrac{5}{2}}} \le \dfrac{6}{5}\\
\to MaxA = \dfrac{6}{5}\\
\Leftrightarrow x\sqrt 2 + \dfrac{1}{{\sqrt 2 }} = 0\\
\Leftrightarrow x = - \dfrac{1}{2}
\end{array}\)
