a) Với \(m=-1\), hpt \(\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\x-y=-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{5}{2}\end{matrix}\right.\)
Vậy...
b) \(\left\{{}\begin{matrix}x+y=3\\-mx-y=2m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-y=x-3\\-mx+x-3=2m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(-m+1\right)=2m+3\\y=-x+3\end{matrix}\right.\)
+) Xét \(m=1\Leftrightarrow\left(x;y\right)\in\varnothing\)
+) Xét \(m\ne1\):
hpt \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-2m-3}{m-1}\\y=\frac{2m+3}{m-1}+3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-2m-3}{m-1}\\y=\frac{5m}{m-1}\end{matrix}\right.\)
Để \(y\) nguyên thì \(5m⋮\left(m-1\right)\)
\(\Leftrightarrow5\left(m-1\right)+5⋮\left(m-1\right)\)
\(\Leftrightarrow5⋮\left(m-1\right)\)
\(\Leftrightarrow\left(m-1\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Leftrightarrow m\in\left\{2;0;6;-4\right\}\)
Thử lại thay vào \(x\) rồi kết luận.