\(\left\{{}\begin{matrix}x+my=1\left(1\right)\\-mx+y=m\left(2\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-my\\-m\left(1-my\right)+y-m=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1-my\\-m+m^2y+y-m=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-my\\\left(m^2+1\right)y-2m=0\end{matrix}\right.\)
Hệ pt có nghiệm duy nhất khi m2+1\(\ne\)0 ( luôn đúng)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1-my\\y=\dfrac{2m}{m^2+1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-\dfrac{2m^2}{m^2+1}=\dfrac{1-m^2}{m^2+1}\\y=\dfrac{2m}{m^2+1}\end{matrix}\right.\)
\(x+y=\dfrac{1-m^2}{m^2+1}+\dfrac{2m}{m^2+1}=\dfrac{-m^2+2m+1}{m^2+1}\)
Để x+y>0 thì -m2+2m+1>0
\(\Leftrightarrow-\left(m^2-2m+1\right)+2>0\Leftrightarrow2-\left(m-1\right)^2>0\)
mà \(2-\left(m-1\right)^2\le2\)\(\Rightarrow0< 2-\left(m-1\right)^2\le2\)
