Đáp án:

\(MinA = \dfrac{1}{5}\)

Giải thích các bước giải:

\(\begin{array}{l}
A = 4{x^2} - 2.2x.3 + 9 + \dfrac{1}{5}\\
 = {\left( {2x - 3} \right)^2} + \dfrac{1}{5}\\
Do:{\left( {2x - 3} \right)^2} \ge 0\forall x\\
 \to {\left( {2x - 3} \right)^2} + \dfrac{1}{5} \ge \dfrac{1}{5}\\
 \to Min = \dfrac{1}{5}\\
 \Leftrightarrow 2x - 3 = 0\\
 \Leftrightarrow x = \dfrac{3}{2}\\
B = \dfrac{3}{{2{x^2} + 2.x\sqrt 2 .\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} + \dfrac{5}{2}}}\\
 = \dfrac{3}{{{{\left( {x\sqrt 2  + \dfrac{1}{{\sqrt 2 }}} \right)}^2} + \dfrac{5}{2}}}\\
Do:{\left( {x\sqrt 2  + \dfrac{1}{{\sqrt 2 }}} \right)^2} \ge 0\forall x\\
 \to {\left( {x\sqrt 2  + \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{5}{2} \ge \dfrac{5}{2}\\
 \to \dfrac{3}{{{{\left( {x\sqrt 2  + \dfrac{1}{{\sqrt 2 }}} \right)}^2} + \dfrac{5}{2}}} \le 3:\dfrac{5}{2}\\
 \to \dfrac{3}{{{{\left( {x\sqrt 2  + \dfrac{1}{{\sqrt 2 }}} \right)}^2} + \dfrac{5}{2}}} \le \dfrac{6}{5}\\
 \to Max = \dfrac{6}{5}\\
 \Leftrightarrow x\sqrt 2  + \dfrac{1}{{\sqrt 2 }} = 0\\
 \Leftrightarrow x =  - \dfrac{1}{2}
\end{array}\)

⇒ Biểu thức B không có GTNN