Đáp án:
\(MaxA = \frac{{8065}}{4}\)
Giải thích các bước giải:
Có:
\(\begin{array}{l}
A = 2020 - {x_1}^2 - {x_2}^2 + 6{x_1}{x_2}\\
= 2020 - \left( {{x_1}^2 + {x_2}^2} \right) + 6{x_1}{x_2}\\
= 2020 - \left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right) + 6{x_1}{x_2}\\
= 2020 - {\left( {{x_1} + {x_2}} \right)^2} + 10{x_1}{x_2}\\
Theo:Vi - et\\
\to A = 2020 - {\left( { - 1 - 2m} \right)^2} + 10\left( {m - 5} \right)\\
= 2020 - \left( {1 + 4m + 4{m^2}} \right) + 10m - 50\\
= 2015 - 1 - 4m - 4{m^2} + 10m\\
= 2014 + 6m - 4{m^2}\\
= - \left( {4{m^2} - 6m - 2014} \right)\\
= - \left( {4{m^2} - 2.\frac{3}{2}.2m + \frac{9}{4} - \frac{{8065}}{4}} \right)\\
= - \left( {4{m^2} - 2.\frac{3}{2}.2m + \frac{9}{4}} \right) + \frac{{8065}}{4}\\
= - {\left( {2m - \frac{3}{2}} \right)^2} + \frac{{8065}}{4}\\
Do:{\left( {2m - \frac{3}{2}} \right)^2} \ge 0\forall m \in R\\
\to - {\left( {2m - \frac{3}{2}} \right)^2} \le 0\\
\to - {\left( {2m - \frac{3}{2}} \right)^2} + \frac{{8065}}{4} \le \frac{{8065}}{4}\\
\to A \le \frac{{8065}}{4}\\
\to MaxA = \frac{{8065}}{4}\\
\Leftrightarrow 2m - \frac{3}{2} = 0\\
\Leftrightarrow m = \frac{3}{4}
\end{array}\)
