Đáp án:
\(MinA = \frac{1}{5}\)
Giải thích các bước giải:
\(\begin{array}{l}
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 3m - 1\\
{x_1}{x_2} = 2{m^2} - m
\end{array} \right.\\
A = {x_1}^2 + {x_2}^2\\
= {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}\\
= {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2}\\
= {\left( {3m - 1} \right)^2} - 2\left( {2{m^2} - m} \right)\\
= 9{m^2} - 6m + 1 - 4{m^2} + 2m\\
= 5{m^2} - 4m + 1\\
= {\left( {m\sqrt 5 } \right)^2} - 2.\left( {m\sqrt 5 } \right).\frac{2}{{\sqrt 5 }} + {\left( {\frac{2}{{\sqrt 5 }}} \right)^2} + \frac{1}{5}\\
= {\left( {m\sqrt 5 - \frac{2}{{\sqrt 5 }}} \right)^2} + \frac{1}{5}\\
Do:{\left( {m\sqrt 5 - \frac{2}{{\sqrt 5 }}} \right)^2} \ge 0\forall m \in R\\
\to {\left( {m\sqrt 5 - \frac{2}{{\sqrt 5 }}} \right)^2} + \frac{1}{5} \ge \frac{1}{5}\\
\to A \ge \frac{1}{5}\\
\to MinA = \frac{1}{5}\\
\Leftrightarrow m\sqrt 5 - \frac{2}{{\sqrt 5 }} = 0\\
\Leftrightarrow m\sqrt 5 = \frac{2}{{\sqrt 5 }}\\
\Leftrightarrow 5m = 2\\
\Leftrightarrow m = \frac{2}{5}
\end{array}\)
