hệ phương trình có nghiệm duy nhất khi:
\(\frac{1}{m}\ne m\Leftrightarrow m^2\ne1\Leftrightarrow m\ne\pm1\)
ta có:
\(\left\{{}\begin{matrix}x+my=m+1\\mx+y=3m-1\end{matrix}\right.\) \(\Leftrightarrow x+mx+my+y=4m\)
\(\Leftrightarrow\left(x+y\right)\left(m+1\right)=4m\) \(\Leftrightarrow x+y=\frac{4m}{m+1}\)
\(\frac{\Leftrightarrow4m}{m+1}< 0\)(do x+y<0)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4m< 0\\m+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}4m>0\\m+1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m< 0\\m>-1\end{matrix}\right.\\\left\{{}\begin{matrix}m>0\\m< -1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-1< m< 0\\0< m< -1\left(vl\right)\end{matrix}\right.\)
kết hợp với đk => \(-1< m< 0\)
vậy...
