Đáp án:
\(\begin{array}{l}
{A_{\min }} = \frac{2}{3} \Leftrightarrow x = \frac{2}{3}\\
{B_{\max }} = \frac{{153}}{{64}} \Leftrightarrow x = \frac{5}{{32}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = 3{x^2} - 4x + 2\\
A = 3\left( {{x^2} - \frac{4}{3}x} \right) + 2\\
A = 3\left( {{x^2} - 2.x.\frac{2}{3} + \frac{4}{9}} \right) - \frac{4}{3} + 2\\
A = 3{\left( {x - \frac{2}{3}} \right)^2} + \frac{2}{3}\\
Do\,\,{\left( {x - \frac{2}{3}} \right)^2} \ge 0\,\,\forall x \Rightarrow 3{\left( {x - \frac{2}{3}} \right)^2} \ge 0\,\,\forall x\\
\Rightarrow 3{\left( {x - \frac{2}{3}} \right)^2} + \frac{2}{3} \ge \frac{2}{3}\,\,\forall x \Rightarrow A \ge \frac{2}{3}\,\,\forall x\\
\Rightarrow {A_{\min }} = \frac{2}{3} \Leftrightarrow x = \frac{2}{3}\\
B = - 16{x^2} + 5x + 2\\
B = - \left( {16{x^2} - 5x} \right) + 2\\
B = - \left[ {{{\left( {4x} \right)}^2} - 2.4x.\frac{5}{8} + \frac{{25}}{{64}}} \right] + \frac{{25}}{{64}} + 2\\
B = - {\left( {4x - \frac{5}{8}} \right)^2} + \frac{{153}}{{64}}\\
Do\,\,{\left( {4x - \frac{5}{8}} \right)^2} \ge 0\,\,\forall x \Rightarrow - {\left( {4x - \frac{5}{8}} \right)^2} \le 0\,\,\forall x\\
\Rightarrow - {\left( {4x - \frac{5}{8}} \right)^2} + \frac{{153}}{{64}} \le \frac{{153}}{{64}}\,\,\forall x \Rightarrow B \le \frac{{153}}{{64}}\,\,\forall x\\
\Rightarrow {B_{\max }} = \frac{{153}}{{64}} \Leftrightarrow x = \frac{5}{{32}}
\end{array}\)
