Đáp án:
\(Max = \dfrac{{25}}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
Xét:\Delta \ge 0\\
\to 9{m^2} + 6m + 1 - 8{m^2} - 4m + 4 \ge 0\\
\to {m^2} + 2m + 5 \ge 0\left( {ld} \right)\forall m\\
Có:A = {x_1}^2 + {x_2}^2 - 3{x_1}{x_2}\\
= {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 5{x_1}{x_2}\\
= {\left( {{x_1} + {x_2}} \right)^2} - 5{x_1}{x_2}\\
= {\left( {3m + 1} \right)^2} - 5\left( {2{m^2} + m - 1} \right)\\
= 9{m^2} + 6m + 1 - 10{m^2} - 5m + 5\\
= - {m^2} + m + 6\\
= - \left( {{m^2} - 2.m.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{{25}}{4}} \right)\\
= - {\left( {m - \dfrac{1}{2}} \right)^2} + \dfrac{{25}}{4}\\
Do:{\left( {m - \dfrac{1}{2}} \right)^2} \ge 0\forall m\\
\to - {\left( {m - \dfrac{1}{2}} \right)^2} \le 0\\
\to - {\left( {m - \dfrac{1}{2}} \right)^2} + \dfrac{{25}}{4} \le \dfrac{{25}}{4}\\
\to Max = \dfrac{{25}}{4}\\
\Leftrightarrow m - \dfrac{1}{2} = 0\\
\to m = \dfrac{1}{2}
\end{array}\)
