Để phương trình có nghiệm duy nhất thì \(\dfrac{m-1}{2}\ne\dfrac{-m}{-1}=m\)
=>\(m-1\ne2m\)
=>\(m\ne-1\)
\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\2x-y=m+5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\y=2x-m-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x-m-5\\\left(m-1\right)x-m\left(2x-m-5\right)=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x-m-5\\\left(m-1\right)x-2xm+m^2+5m=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x-m-5\\x\left(m-1-2m\right)=-m^2-5m+3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x-m-5\\x\left(-m-1\right)=-m^2-2m-1=-\left(m+1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x-m-5\\x\cdot\left(-1\right)\cdot\left(m+1\right)=-\left(m+1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1\\y=2\left(m+1\right)-m-5=2m+2-m-5=m-3\end{matrix}\right.\)
\(x^2-y^2=24\)
=>\(\left(m+1\right)^2-\left(m-3\right)^2=24\)
=>\(m^2+2m+1-m^2+6m-9=24\)
=>8m-8=24
=>m=4(nhận)
