Đáp án:
\(m \in \left( { - \infty ;0} \right) \cup \left( {0;1} \right) \cup \left( {\frac{8}{5}; + \infty } \right)\)
Giải thích các bước giải:
Để phương trình có hai nghiệm phân biệt
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
4\left( {{m^2} - 2m + 1} \right) - m\left( {m - 5} \right) > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
4{m^2} - 8m + 4 - {m^2} + 5m > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
3{m^2} - 3m + 4 > 0\left( {ld} \right)
\end{array} \right.\\
\to m \ne 0\left( 1 \right)\\
Có:{x_1}^2 + {x_2}^2 + {x_1} + {x_2} - 8 > 0\\
\to {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} + {x_1} + {x_2} - 8 > 0\\
\to {\left( {\frac{{4m - 4}}{m}} \right)^2} - 2.\frac{{m - 5}}{m} + \frac{{4m - 4}}{m} - 8 > 0\\
\to \frac{{16{m^2} - 32m + 16}}{{{m^2}}} - \frac{{2m - 10}}{m} + \frac{{4m - 4}}{m} - 8 > 0\\
\to \frac{{16{m^2} - 32m + 16 - 2{m^2} + 10m + 4{m^2} - 4m - 8{m^2}}}{{{m^2}}} > 0\\
\to 10{m^2} - 26m + 16 > 0\left( {do:{m^2} > 0\forall m \ne 0} \right)\\
\to \left( {5m - 8} \right)\left( {2m - 2} \right) > 0\\
\to m \in \left( { - \infty ;1} \right) \cup \left( {\frac{8}{5}; + \infty } \right)\left( 2 \right)\\
\left( 1 \right);\left( 2 \right) \to m \in \left( { - \infty ;0} \right) \cup \left( {0;1} \right) \cup \left( {\frac{8}{5}; + \infty } \right)
\end{array}\)
