Tìm GTLN, GTNN của biểu thức sau A = 3x mũ 2 -4x +2 (GTNN) B = -16x mũ 2 +5x+2 (GTLN)

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