Đáp án:
\(Max = - \dfrac{6}{7}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{3}{{2{x^2} + 2.x\sqrt 2 .\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} - \dfrac{7}{2}}}\\
= \dfrac{3}{{{{\left( {x\sqrt 2 + \dfrac{1}{{\sqrt 2 }}} \right)}^2} - \dfrac{7}{2}}}\\
Do:{\left( {x\sqrt 2 + \dfrac{1}{{\sqrt 2 }}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt 2 + \dfrac{1}{{\sqrt 2 }}} \right)^2} - \dfrac{7}{2} \ge - \dfrac{7}{2}\\
\to \dfrac{3}{{{{\left( {x\sqrt 2 + \dfrac{1}{{\sqrt 2 }}} \right)}^2} - \dfrac{7}{2}}} \le 3:\left( { - \dfrac{7}{2}} \right)\\
\to \dfrac{3}{{{{\left( {x\sqrt 2 + \dfrac{1}{{\sqrt 2 }}} \right)}^2} - \dfrac{7}{2}}} \le - \dfrac{6}{7}\\
\to Max = - \dfrac{6}{7}\\
\Leftrightarrow x\sqrt 2 + \dfrac{1}{{\sqrt 2 }} = 0\\
\to x = - \dfrac{1}{2}
\end{array}\)
