\(\left\{{}\begin{matrix}x+my=m+1\\mx+y=3m-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}mx+m^2y=m^2+m\\mx+y=3m-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2-1\right)y=m^2-2m+1\\x+my=m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-1}{m+1}\\x=m+1-my\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-1}{m+1}\\x=m+1-\dfrac{m^2-m}{m+1}=\dfrac{m^2+2m+1-m^2+m}{m+1}=\dfrac{3m+1}{m+1}\end{matrix}\right.\)
Thay vào bth ta được \(\left(\dfrac{3m+1}{m+1}\right)^2-\left(\dfrac{m-1}{m+1}\right)^2=4\)đk m khác -1
\(\Leftrightarrow\dfrac{3m+1-m+1}{m+1}.\dfrac{3m+1+m-1}{m+1}=4\)
\(\Leftrightarrow2.\dfrac{4m}{m+1}=4\Leftrightarrow4m=2m+2\Leftrightarrow m=1\)(tm)
Vậy có 1 giá trị tham số m
