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
