Đáp án:
a) m<-3
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + my = 2m + 3\\
x - y = m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
my + y = 2m + 3 - m\\
x = m + y
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {m + 1} \right)y = m + 3\\
x = m + y
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{m + 3}}{{m + 1}}\\
x = m + \dfrac{{m + 3}}{{m + 1}} = \dfrac{{{m^2} + m + m + 3}}{{m + 1}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{m + 3}}{{m + 1}}\\
x = \dfrac{{{m^2} + 2m + 3}}{{m + 1}}
\end{array} \right.\\
DK:m \ne - 1\\
a)Do:xy < 0\\
\to \dfrac{{m + 3}}{{m + 1}}.\dfrac{{{m^2} + 2m + 3}}{{m + 1}} < 0\\
\to \dfrac{{\left( {m + 3} \right)\left( {{m^2} + 2m + 3} \right)}}{{{{\left( {m + 1} \right)}^2}}} < 0\\
Do:\left\{ \begin{array}{l}
{m^2} + 2m + 3 > 0\forall m \ne - 1\\
{\left( {m + 1} \right)^2} > 0\forall m \ne - 1
\end{array} \right.\\
\to m + 3 < 0\\
\to m < - 3\\
b)\left\{ \begin{array}{l}
x = \dfrac{{m + 3}}{{m + 1}} = \dfrac{{m + 1 + 2}}{{m + 1}} = 1 + \dfrac{2}{{m + 1}}\\
y = \dfrac{{{m^2} + 2m + 3}}{{m + 1}} = \dfrac{{\left( {{m^2} + 2m + 1} \right) + 2}}{{m + 1}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 + \dfrac{2}{{m + 1}}\\
y = \dfrac{{{{\left( {m + 1} \right)}^2} + 2}}{{m + 1}} = m + 1 + \dfrac{2}{{m + 1}}
\end{array} \right.\\
Do:x \in Z;y \in Z\\
\to \dfrac{2}{{m + 1}} \in Z\\
\to m + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
m + 1 = 2\\
m + 1 = - 2\\
m + 1 = 1\\
m + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
m = 1\\
m = - 3\\
m = 0\\
m = - 2
\end{array} \right.
\end{array}\)
