Đáp án:
\(4 > m > - 2;x \ne 2\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + my = 3\\
mx + 4y = 6
\end{array} \right.\\
\to \left\{ \begin{array}{l}
mx + {m^2}y = 3m\\
mx + 4y = 6
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {{m^2} - 4} \right)y = 3m - 6\\
x = 3 - my
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {m - 2} \right)\left( {m + 2} \right)y = 3\left( {m - 2} \right)\\
x = 3 - my
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{3}{{m + 2}}\left( {DK:m \ne \pm 2} \right)\\
x = 3 - m.\dfrac{3}{{m + 2}} = \dfrac{{3m + 6 - 3m}}{{m + 2}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{3}{{m + 2}}\\
x = \dfrac{6}{{m + 2}}
\end{array} \right.\\
Do:x > 1;y > 0\\
\to \left\{ \begin{array}{l}
\dfrac{6}{{m + 2}} > 1\\
\dfrac{3}{{m + 2}} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{6 - m - 2}}{{m + 2}} > 0\\
m + 2 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4 - m > 0\\
m > - 2
\end{array} \right.\\
\to 4 > m > - 2;x \ne 2
\end{array}\)
