Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{-1}\)
=>\(m^2\ne-1\)(luôn đúng)
=>Hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}x+my=1\\mx-y=-m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-my\\m\left(1-my\right)-y=-m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1-my\\m-m^2y-y=-m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1-my\\y\left(m^2+1\right)=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{2m}{m^2+1}\\x=1-my=1-\dfrac{2m^2}{m^2+1}=\dfrac{m^2+1-2m^2}{m^2+1}=\dfrac{-m^2+1}{m^2+1}\end{matrix}\right.\)
\(S=x+y=\dfrac{2m}{m^2+1}+\dfrac{-m^2+1}{m^2+1}=\dfrac{-m^2+2m-1+2}{m^2+1}\)
\(=\dfrac{-\left(m-1\right)^2+2}{m^2+1}< =\dfrac{2}{1}=2\forall m\)
Dấu '=' xảy ra khi m=0
