Đáp án:
Min=9
Giải thích các bước giải:
\(\begin{array}{l}
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m - 2\\
{x_1}{x_2} = - m - 3
\end{array} \right.\\
A = \dfrac{{2\left( {{x_1}^2 + 2{x_1}{x_2} + {x_2}^2} \right) - 2{x_1}{x_2}}}{{{x_1} + {x_2}}}\\
= \dfrac{{2{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}}}{{{x_1} + {x_2}}}\\
= 2\left( {{x_1} + {x_2}} \right) - \dfrac{{2{x_1}{x_2}}}{{{x_1} + {x_2}}}\\
= 2\left( {2m - 2} \right) - \dfrac{{2\left( { - m - 3} \right)}}{{2m - 2}}\\
= 4m - 4 + \dfrac{{2m + 6}}{{2m - 2}}\\
= 4m - 4 + \dfrac{{2m - 2 + 8}}{{2m - 2}}\\
= 4m - 4 + 1 + \dfrac{8}{{2m - 2}}\\
= 2\left( {2m - 2} \right) + \dfrac{8}{{2m - 2}} + 1\\
Do:m > 1\\
BDT:Co - si:2\left( {2m - 2} \right) + \dfrac{8}{{2m - 2}} \ge 2\sqrt {2\left( {2m - 2} \right).\dfrac{8}{{2m - 2}}} = 2.4 = 8\\
\to 2\left( {2m - 2} \right) + \dfrac{8}{{2m - 2}} + 1 \ge 9\\
\to Min = 9\\
\Leftrightarrow 2\left( {2m - 2} \right) = \dfrac{8}{{2m - 2}}\\
\to {\left( {2m - 2} \right)^2} = 4\\
\to \left[ \begin{array}{l}
2m - 2 = 2\\
2m - 2 = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = 2\\
m = 0\left( l \right)
\end{array} \right.
\end{array}\)
