Đáp án:
c) Min=-67
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:m = 4\\
Pt \to 2{x^2} - 5x + 3 = 0\\
\to \left( {2m - 3} \right)\left( {m - 1} \right) = 0\\
\to \left[ \begin{array}{l}
m = \dfrac{3}{2}\\
m = 1
\end{array} \right.
\end{array}\)
b) Để phương trình có 2 nghiệm
\(\begin{array}{l}
\to \Delta \ge 0\\
\to {\left( {m + 1} \right)^2} - 4.2.3 \ge 0\\
\to {m^2} + 2m - 23 \ge 0\\
\to {\left( {m + 1} \right)^2} \ge 24\\
\to \left[ \begin{array}{l}
m + 1 \ge 2\sqrt 6 \\
m + 1 \le - 2\sqrt 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
m \ge - 1 + 2\sqrt 6 \\
m \le - 1 - 2\sqrt 6
\end{array} \right.\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{m + 1}}{2}\\
{x_1}{x_2} = \dfrac{3}{2}
\end{array} \right.\\
Có:{x_1} + {x_2} + {x_1}{x_2} = 2019\\
\to \dfrac{{m + 1}}{2} + \dfrac{3}{2} = 2019\\
\to m + 4 = 4038\\
\to m = 4034\\
c)M = {x_1}^2 + {x_2}^2 - 16\left( {{x_1} + {x_2}} \right)\\
= \left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}} \right) - 2{x_1}{x_2} - 16\left( {{x_1} + {x_2}} \right)\\
= {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} - 16\left( {{x_1} + {x_2}} \right)\\
= \dfrac{{{m^2} + 2m + 1}}{4} - 2.\dfrac{3}{2} - 16.\dfrac{{m + 1}}{2}\\
= \dfrac{{{m^2} + 2m + 1 - 12 - 32m - 32}}{4}\\
= \dfrac{{{m^2} - 30m - 43}}{4}\\
= \dfrac{{{{\left( {m - 15} \right)}^2} - 268}}{4}\\
Do:{\left( {m - 15} \right)^2} \ge 0\forall m\\
\to {\left( {m - 15} \right)^2} - 268 \ge - 268\\
\to \dfrac{{{{\left( {m - 15} \right)}^2} - 268}}{4} \ge - 67\\
\to Min = - 67\\
\Leftrightarrow m = 15
\end{array}\)
