Đáp án:
\(\begin{array}{l}
{A_{\min }} = \frac{3}{4} \Leftrightarrow x = \frac{5}{4}\\
{B_{\max }} = 4 \Leftrightarrow x = 4
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = 4{x^2} - 10x + 7\\
A = {\left( {2x} \right)^2} - 2.2x.\frac{5}{2} + \frac{{25}}{4} - \frac{{25}}{4} + 7\\
A = {\left( {2x - \frac{5}{2}} \right)^2} + \frac{3}{4}\\
Do\,\,{\left( {2x - \frac{5}{2}} \right)^2} \ge 0\,\,\forall x \in R \Rightarrow {\left( {2x - \frac{5}{2}} \right)^2} + \frac{3}{4} \ge \frac{3}{4}\,\,\forall x \in R\\
\Rightarrow A \ge \frac{3}{4}\,\,\forall x \in R \Rightarrow {A_{\min }} = \frac{3}{4} \Leftrightarrow 2x = \frac{5}{2} \Leftrightarrow x = \frac{5}{4}\\
B = - {x^2} + 8x - 12\\
B = - \left( {{x^2} - 8x} \right) - 12\\
B = - \left( {{x^2} - 2.x.4 + 16} \right) + 16 - 12\\
B = - {\left( {x - 4} \right)^2} + 4\\
Do\,\,{\left( {x - 4} \right)^2} \ge 0\,\,\forall x \in R \Rightarrow - {\left( {x - 4} \right)^2} \le 0\,\,\forall x \in R\\
\Rightarrow - {\left( {x - 4} \right)^2} + 4 \le 4\,\,\forall x \in R\\
\Rightarrow B \le 4\,\,\,\forall x \in R\\
\Rightarrow {B_{\max }} = 4 \Leftrightarrow x = 4
\end{array}\)
