Đáp án:
b) \(MaxB = \dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)A = {x^2} - 6x + 2 = {x^2} - 6x + 9 - 7\\
= {\left( {x - 3} \right)^2} - 7\\
Do:{\left( {x - 3} \right)^2} \ge 0\forall x\\
\to {\left( {x - 3} \right)^2} - 7 \ge - 7\\
\to Min = - 7\\
\Leftrightarrow x - 3 = 0 \to x = 3\\
b)B = 4{x^2} - x + 2 = 4{x^2} - 2.2x.\dfrac{1}{4} + \dfrac{1}{{16}} + \dfrac{{31}}{{16}}\\
= {\left( {2x - \dfrac{1}{4}} \right)^2} + \dfrac{{31}}{{16}}\\
Do:{\left( {2x - \dfrac{1}{4}} \right)^2} \ge 0\forall x\\
\to {\left( {2x - \dfrac{1}{4}} \right)^2} + \dfrac{{31}}{{16}} \ge \dfrac{{31}}{{16}}\\
\to Min = \dfrac{{31}}{{16}}\\
\Leftrightarrow 2x - \dfrac{1}{4} = 0 \Leftrightarrow x = \dfrac{1}{8}\\
b)A = - \left( {{x^2} - 5x - 3} \right)\\
= - \left( {{x^2} - 2.x.\dfrac{5}{2} + \dfrac{{25}}{4} - \dfrac{{37}}{4}} \right)\\
= - {\left( {x - \dfrac{5}{2}} \right)^2} + \dfrac{{37}}{4}\\
Do:{\left( {x - \dfrac{5}{2}} \right)^2} \ge 0\forall x\\
\to - {\left( {x - \dfrac{5}{2}} \right)^2} \le 0\\
\to - {\left( {x - \dfrac{5}{2}} \right)^2} + \dfrac{{37}}{4} \le \dfrac{{37}}{4}\\
\to Max = \dfrac{{37}}{4}\\
\Leftrightarrow x = \dfrac{5}{2}\\
B = \dfrac{1}{{{x^2} - 8x + 16 + 4}}\\
= \dfrac{1}{{{{\left( {x - 4} \right)}^2} + 4}}\\
Do:{\left( {x - 4} \right)^2} \ge 0\forall x\\
\to {\left( {x - 4} \right)^2} + 4 \ge 4\\
\to \dfrac{1}{{{{\left( {x - 4} \right)}^2} + 4}} \le \dfrac{1}{4}\\
\to Max = \dfrac{1}{4}\\
\Leftrightarrow x - 4 = 0 \to x = 4
\end{array}\)
