Đáp án:
b) \(m > - \dfrac{5}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left\{ \begin{array}{l}
x - y = 1\\
2x - y = m - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
3x = m\\
y = x - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{m}{3}\\
y = \dfrac{m}{3} - 1
\end{array} \right.\\
Do:x \in Z;y \in Z\\
\to \dfrac{m}{3} \in Z\\
\Leftrightarrow m \in B\left( 3 \right)\\
\to \left[ \begin{array}{l}
m = 3\\
m = 6\\
m = 9\\
m = 12\\
...
\end{array} \right.\\
b)\left\{ \begin{array}{l}
x - y = 2\\
3x + my = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = y + 2\\
3\left( {y + 2} \right) + my = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = y + 2\\
3y + 6 + my = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = y + 2\\
\left( {m + 3} \right)y = - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = - \dfrac{1}{{m + 3}}\\
x = - \dfrac{1}{{m + 3}} + 2 = \dfrac{{ - 1 + 2m + 6}}{{m + 3}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = - \dfrac{1}{{m + 3}}\\
x = \dfrac{{2m + 5}}{{m + 3}}
\end{array} \right.\\
Do:x \in \left( {IV} \right);y \in \left( {IV} \right)\\
\to \left\{ \begin{array}{l}
x > 0\\
y < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{2m + 5}}{{m + 3}} > 0\\
- \dfrac{1}{{m + 3}} < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m + 3 > 0\\
2m + 5 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - 3\\
m > - \dfrac{5}{2}
\end{array} \right.
\end{array}\)
\(KL:m > - \dfrac{5}{2}\)
