\(\left\{{}\begin{matrix}x+my=m+1\\mx+y=3m-1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=m+1-my\\m\left(m+1-my\right)+y=3m-1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=m+1-my\\m^2+m-m^2y+y=3m-1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=m+1-my\\y\left(m^2-1\right)=m^2-2m+1\end{matrix}\right.\)
Với m = 1 ta có: \(\left\{{}\begin{matrix}x=2-y\\0y=0\left(VSN\right)\end{matrix}\right.\)
\(\Rightarrow\) Hpt vô số nghiệm
Với m = -1 ta có: \(\left\{{}\begin{matrix}x=y\\0y=4\left(VN\right)\end{matrix}\right.\)
\(\Rightarrow\) Hpt vô nghiệm
Với m \(\ne\) \(\pm\)1 ta có: \(\left\{{}\begin{matrix}x=m+1-my\\y=\dfrac{m^2-2m+1}{m^2-1}\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=m+1-\dfrac{m\left(m-1\right)^2}{\left(m-1\right)\left(m+1\right)}=m+1-\dfrac{m\left(m-1\right)}{m+1}=m+1-\dfrac{m^2-m}{m+1}\\y=\dfrac{m^2-2m+1}{m^2-1}=\dfrac{\left(m-1\right)^2}{\left(m-1\right)\left(m+1\right)}=\dfrac{m-1}{m+1}\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{m^2+2m+1-m^2+m}{m+1}=\dfrac{3m+1}{m+1}\\y=\dfrac{m-1}{m+1}\end{matrix}\right.\)
Vậy hpt có nghiệm duy nhất x = ..; y = ... với x \(\ne\) \(\pm\) 1
Ta có: x = |y|
\(\Leftrightarrow\) \(\dfrac{3m+1}{m+1}=\left|\dfrac{m-1}{m+1}\right|\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}\dfrac{3m+1}{m+1}=\dfrac{m-1}{m+1}\\\dfrac{3m+1}{m+1}=\dfrac{1-m}{m+1}\end{matrix}\right.\)
\(\Rightarrow\) \(\left[{}\begin{matrix}3m+1=m-1\\3m+1=1-m\end{matrix}\right.\) (Vì m \(\ne\) -1)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2m=-2\\4m=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}m=-1\\m=0\end{matrix}\right.\)
Vì m \(\ne\) -1 nên m = -1 KTM
\(\Rightarrow\) m = 0 thỏa mãn đk
Vậy m = 0
Chúc bn học tốt!