Đáp án:
\(1 - \sqrt 3 < m < 1 + \sqrt 3 \)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + my = 3m\\
{m^2}x - my = {m^3} - 2m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {{m^2} + 1} \right)x = {m^3} + m\\
y = \frac{{3m - x}}{m}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \frac{{m\left( {{m^2} + 1} \right)}}{{{m^2} + 1}}\\
y = \frac{{3m - x}}{m}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = m\\
y = \frac{{3m - m}}{m} = 2
\end{array} \right.\\
Có:{x^2} - 2x - y > 0\\
\to {m^2} - 2m - 2 > 0\\
Xét:{m^2} - 2m - 2 = 0\\
\to \left[ \begin{array}{l}
m = 1 + \sqrt 3 \\
m = 1 - \sqrt 3
\end{array} \right.\\
Bpt:{m^2} - 2m - 2 > 0\\
\to \left( {m - 1 - \sqrt 3 } \right)\left( {m - 1 + \sqrt 3 } \right) > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m - 1 - \sqrt 3 > 0\\
m - 1 + \sqrt 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m - 1 - \sqrt 3 < 0\\
m - 1 + \sqrt 3 < 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m > 1 + \sqrt 3 \\
m < 1 - \sqrt 3
\end{array} \right.\left( l \right)\\
\left\{ \begin{array}{l}
m < 1 + \sqrt 3 \\
m > 1 - \sqrt 3
\end{array} \right.
\end{array} \right.\\
KL:1 - \sqrt 3 < m < 1 + \sqrt 3
\end{array}\)
