Đáp án:
b) m=16
Giải thích các bước giải:
Để phương trình có 2 nghiệm
\(\begin{array}{l}
\to \Delta ' \ge 0\\
\to 1 + 2m - 8 \ge 0\\
\to m \ge \dfrac{7}{2}\\
\to \left[ \begin{array}{l}
x = - 1 + \sqrt {2m - 7} \\
x = - 1 - \sqrt {2m - 7}
\end{array} \right.\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - 2\\
{x_1}{x_2} = - 2m + 8
\end{array} \right.\\
a){x_1} = 2{x_2}\\
\to {x_1} + {x_2} = 3{x_2}\\
\to \left[ \begin{array}{l}
- 2 = 3\left( { - 1 + \sqrt {2m - 7} } \right)\\
- 2 = 3\left( { - 1 - \sqrt {2m - 7} } \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
3\sqrt {2m - 7} = 1\\
3\sqrt {2m - 7} = - 1\left( l \right)
\end{array} \right.\\
\to 2m - 7 = \dfrac{1}{9}\\
\to m = \dfrac{{32}}{9}\left( {TM} \right)\\
b)2{x_1} + 3{x_2} = 0\\
\to 2\left( {{x_1} + {x_2}} \right) + {x_2} = 0\\
\to \left[ \begin{array}{l}
2.\left( { - 2} \right) - 1 - \sqrt {2m - 7} = 0\\
2.\left( { - 2} \right) - 1 + \sqrt {2m - 7} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt {2m - 7} = - 5\left( l \right)\\
\sqrt {2m - 7} = 5
\end{array} \right.\\
\to 2m - 7 = 25\\
\to m = 16\left( {TM} \right)\\
c){x_1}^2 + {x_2}^2 + 3{x_1}{x_2} = 20\\
\to {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} + {x_1}{x_2} = 20\\
\to {\left( {{x_1} + {x_2}} \right)^2} + {x_1}{x_2} = 20\\
\to 4 - 2m + 8 = 20\\
\to 2m = - 8\\
\to m = - 4\left( {KTM} \right)\\
\to m \in \emptyset \\
d){x_1} < 2 < {x_2}\\
\to \left\{ \begin{array}{l}
{x_1} - 2 < 0\\
{x_2} - 2 > 0
\end{array} \right.\\
\to \left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right) < 0\\
\to {x_1}{x_2} - 2\left( {{x_1} + {x_2}} \right) + 4 < 0\\
\to - 2m + 8 - 2.\left( { - 2} \right) + 4 < 0\\
\to 2m > 16\\
\to m > 8
\end{array}\)
