Để hệ có nghiệm duy nhất thì \(\dfrac{m}{3}\ne\dfrac{-1}{m}\)
=>\(m^2\ne-3\)(luôn đúng)
=>hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}mx-y=3\\3x+my=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=mx-3\\3x+m\left(mx-3\right)=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=mx-3\\x\left(m^2+3\right)=3m+5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3m+5}{m^2+3}\\y=\dfrac{m\left(3m+5\right)}{m^2+3}-3=\dfrac{3m^2+5m-3m^2-9}{m^2+3}=\dfrac{5m-9}{m^2+3}\end{matrix}\right.\)
\(x+y-\dfrac{7\left(m-1\right)}{m^2+3}=1\)
=>\(\dfrac{3m+5+5m-9-7m+7}{m^2+3}=1\)
=>\(m^2+3=m+3\)
=>\(m^2-m=0\)
=>\(m\left(m-1\right)=0\)
=>\(\left[{}\begin{matrix}m=0\\m=1\end{matrix}\right.\)
