Đáp án:
a) \(\dfrac{1}{2} > m > - 4\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x = 2 - my\\
m\left( {2 - my} \right) - 2y = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 2 - my\\
2m - {m^2}y - 2y = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 2 - my\\
- \left( {{m^2} + 2} \right)y = 1 - 2m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = - \dfrac{{1 - 2m}}{{{m^2} + 2}} = \dfrac{{2m - 1}}{{{m^2} + 2}}\\
x = 2 - m.\dfrac{{2m - 1}}{{{m^2} + 2}} = \dfrac{{2{m^2} + 4 - 2{m^2} + m}}{{{m^2} + 2}} = \dfrac{{m + 4}}{{{m^2} + 2}}
\end{array} \right.\\
a)Do:x > 0;y < 0\\
\to \left\{ \begin{array}{l}
\dfrac{{m + 4}}{{{m^2} + 2}} > 0\\
\dfrac{{2m - 1}}{{{m^2} + 2}} < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m + 4 > 0\\
2m - 1 < 0
\end{array} \right.\left( {do:{m^2} + 2 > 0\forall m} \right)\\
\to \dfrac{1}{2} > m > - 4\\
b)S = x - y = \dfrac{{m + 4}}{{{m^2} + 2}} - \dfrac{{2m - 1}}{{{m^2} + 2}}\\
= \dfrac{{5 - m}}{{{m^2} + 2}}
\end{array}\)
