\(\left\{{}\begin{matrix}2x+my=1\\mx+2y=1\end{matrix}\right.\)
+) Xét \(m=0\Rightarrow\left\{{}\begin{matrix}2x=1\\2y=1\end{matrix}\right.\)\(\Leftrightarrow x=y=\frac{1}{2}\) ( loại vì x, y nguyên )
+) Xét \(m\ne0\)
Hpt \(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{1-2x}{m}\left(1\right)\\mx+\frac{2\left(1-2x\right)}{m}=1\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\frac{xm^2+2-4x}{m}=1\)
\(\Leftrightarrow x\left(m^2-4\right)+2=m\)
\(\Leftrightarrow x=\frac{m-2}{m^2-4}=\frac{m-2}{\left(m-2\right)\left(m+2\right)}=\frac{1}{m+2}\) ( \(m\ne\pm2\) )
Vì \(x\) nguyên nên \(\frac{1}{m+2}\) nguyên \(\Rightarrow1⋮\left(m+2\right)\)
\(\Rightarrow\left(m+2\right)\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Rightarrow m\in\left\{-1;-3\right\}\)( thỏa )
Thử lại:
+) \(m=-1\Leftrightarrow\)\(\left\{{}\begin{matrix}2x-y=1\\2y-x=1\end{matrix}\right.\)\(\Leftrightarrow x=y=1\) ( thỏa mãn )
+) \(m=-3\Leftrightarrow\)\(\left\{{}\begin{matrix}2x-3y=1\\2y-3x=1\end{matrix}\right.\)\(\Leftrightarrow x=y=-1\) ( thỏa mãn )
Vậy...
