Đáp án:
\(m \ne \pm \dfrac{5}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
2mx - 5y = - 2\\
5x - 2my = 4 - 4m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{2mx + 2}}{5}\\
5x - 2m.\dfrac{{2mx + 2}}{5} = 4 - 4m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{2mx + 2}}{5}\\
25x - 4{m^2}x - 4m = 20 - 20m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{2mx + 2}}{5}\\
\left( {25 - 4{m^2}} \right)x = 20 - 16m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{2mx + 2}}{5} = \dfrac{{2m.\dfrac{{20 - 16m}}{{25 - 4{m^2}}} + 2}}{5}\\
x = \dfrac{{20 - 16m}}{{25 - 4{m^2}}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{20 - 16m}}{{25 - 4{m^2}}}\\
y = \dfrac{{40m - 32{m^2} + 50 - 8{m^2}}}{{5\left( {25 - 4{m^2}} \right)}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{20 - 16m}}{{25 - 4{m^2}}}\\
y = \dfrac{{ - 40{m^2} + 90}}{{5\left( {25 - 4{m^2}} \right)}}
\end{array} \right.\\
DK:25 - 4{m^2} \ne 0\\
\to m \ne \pm \dfrac{5}{2}
\end{array}\)