Đáp án:
\(m > \dfrac{3}{4};m \ne \dfrac{5}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x = 8 - 2y\\
m\left( {8 - 2y} \right) + 5y = 6
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 8 - 2y\\
8m - 2my + 5y = 6
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 8 - 2y\\
\left( {5 - 2m} \right)y = 6 - 8m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 8 - 2y\\
y = \dfrac{{6 - 8m}}{{5 - 2m}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 8 - 2.\dfrac{{6 - 8m}}{{5 - 2m}}\\
y = \dfrac{{6 - 8m}}{{5 - 2m}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{40 - 16m - 12 + 16m}}{{5 - 2m}}\\
y = \dfrac{{6 - 8m}}{{5 - 2m}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{28}}{{5 - 2m}}\\
y = \dfrac{{6 - 8m}}{{5 - 2m}}
\end{array} \right.\left( {DK:m \ne \dfrac{5}{2}} \right)\\
Do:xy < 0\\
\to \dfrac{{28}}{{5 - 2m}}.\dfrac{{6 - 8m}}{{5 - 2m}} < 0\\
\to \dfrac{{6 - 8m}}{{{{\left( {5 - 2m} \right)}^2}}} < 0\left( {do:{{\left( {5 - 2m} \right)}^2} > 0\forall m \ne \dfrac{5}{2}} \right)\\
\to 6 - 8m < 0\\
\to m > \dfrac{3}{4};m \ne \dfrac{5}{2}
\end{array}\)
