Đáp án:
\(MinE = \sqrt {8136578} \)
Giải thích các bước giải:
Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to \Delta > 0\\
\to {m^2} + 6m + 9 - 4.2.m > 0\\
\to {m^2} - 2m + 9 > 0\left( {ld} \right)\forall m\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{m + 3}}{2}\\
{x_1}{x_2} = \dfrac{m}{2}
\end{array} \right.\\
E = \left| {2017{x_1} - 2017{x_2}} \right|\\
= 2017\left| {{x_1} - {x_2}} \right|\\
\to {E^2} = {2017^2}\left( {{x_1}^2 - 2{x_1}{x_2} + {x_2}^2} \right)\\
= {2017^2}\left( {{x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - 4{x_1}{x_2}} \right)\\
= {2017^2}.\left( {{{\left( {{x_1} + {x_2}} \right)}^2} - 4{x_1}{x_2}} \right)\\
= {2017^2}.\left( {\dfrac{{{m^2} + 6m + 9}}{4} - 4.\dfrac{m}{2}} \right)\\
= {2017^2}.\dfrac{{{m^2} + 6m + 9 - 8m}}{4}\\
= \dfrac{{{{2017}^2}}}{4}.\left( {{m^2} - 2m + 9} \right)\\
= \dfrac{{{{2017}^2}}}{4}.\left[ {{{\left( {m - 1} \right)}^2} + 8} \right]\\
Do:{\left( {m - 1} \right)^2} \ge 0\forall m\\
\to {\left( {m - 1} \right)^2} + 8 \ge 8\\
\to \dfrac{{{{2017}^2}}}{4}.\left[ {{{\left( {m - 1} \right)}^2} + 8} \right] \ge 8136578\\
\to E \ge \sqrt {8136578} \\
\to MinE = \sqrt {8136578} \\
\Leftrightarrow m = 1
\end{array}\)
