Đáp án:
\(\dfrac{1}{2} \ge m > - 16\)
Giải thích các bước giải:
Bài 7:
\(\begin{array}{l}
\left\{ \begin{array}{l}
mx - 4y = 1\\
x + my = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{m^2}x - 4my = m\\
4x + 4my = 16
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {{m^2} + 4} \right)x = m + 16\\
y = \dfrac{{mx - 1}}{4}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{m + 16}}{{{m^2} + 4}}\\
y = \dfrac{{m.\dfrac{{m + 16}}{{{m^2} + 4}} - 1}}{4} = \dfrac{{{m^2} + 16m - {m^2} - 4}}{{4\left( {{m^2} + 4} \right)}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{m + 16}}{{{m^2} + 4}}\\
y = \dfrac{{4m - 1}}{{{m^2} + 4}}
\end{array} \right.\\
Do:x > 0;y \le 0\\
\to \left\{ \begin{array}{l}
\dfrac{{m + 16}}{{{m^2} + 4}} > 0\\
\dfrac{{4m - 1}}{{{m^2} + 4}} \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m + 16 > 0\\
4m - 1 \le 0
\end{array} \right.\left( {do:{m^2} + 4 > 0\forall m} \right)\\
\to \dfrac{1}{2} \ge m > - 16
\end{array}\)
