Đáp án:
\(\dfrac{1}{2} > m > - 2\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
y = mx - 1\\
x + m\left( {mx - 1} \right) = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = mx - 1\\
x + {m^2}x - m = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = mx - 1\\
\left( {{m^2} + 1} \right)x = m + 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{m + 2}}{{{m^2} + 1}}\\
y = m.\dfrac{{m + 2}}{{{m^2} + 1}} - 1 = \dfrac{{{m^2} + 2m - {m^2} - 1}}{{{m^2} + 1}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{m + 2}}{{{m^2} + 1}}\\
y = \dfrac{{2m - 1}}{{{m^2} + 1}}
\end{array} \right.\\
Do:\left\{ \begin{array}{l}
x > 0\\
y < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{m + 2}}{{{m^2} + 1}} > 0\\
\dfrac{{2m - 1}}{{{m^2} + 1}} < 0
\end{array} \right. \to \left\{ \begin{array}{l}
m + 2 > 0\\
2m - 1 < 0
\end{array} \right.\left( {do:{m^2} + 1 > 0\forall x} \right)\\
\to \dfrac{1}{2} > m > - 2\\
S = 2x + y = 2.\dfrac{{m + 2}}{{{m^2} + 1}} + \dfrac{{2m - 1}}{{{m^2} + 1}}\\
= \dfrac{{2m + 4 + 2m - 1}}{{{m^2} + 1}}\\
= \dfrac{{4m + 3}}{{{m^2} + 1}}
\end{array}\)
