Để hệ có nghiệm duy nhất thì \(\frac{3}{m-1}<>\frac{m-1}{12}\)
=>\(\left(m-1\right)^2<>36\)
=>m-1∉{6;-6}
=>m∉{7;-5}
\(\begin{cases}3x+\left(m-1\right)y=12\\ \left(m-1\right)x+12y=24\end{cases}\Rightarrow\begin{cases}3\left(m-1\right)x+\left(m-1\right)^2y=12\left(m-1\right)\\ 3\left(m-1\right)x+36y=72\end{cases}\)
=>\(\begin{cases}3\left(m-1\right)x+\left(m-1\right)^2\cdot y-3\left(m-1\right)x-36y=12\left(m-1\right)-72\\ 3x+\left(m-1\right)y=12\end{cases}\)
=>\(\begin{cases}y\left\lbrack\left(m-1\right)^2-36\right\rbrack=12\left(m-1-6\right)\\ 3x=12-\left(m-1\right)y\end{cases}\Rightarrow\begin{cases}y\left(m-1+6\right)=12\\ 3x=12-\left(m-1\right)y\end{cases}\)
=>\(\begin{cases}y=\frac{12}{m+5}\\ x=4-y\cdot\frac{m-1}{3}=4-\frac{m-1}{3}\cdot\frac{12}{m+5}=4-\frac{4\left(m-1\right)}{m+5}=\frac{4m+20-4m+4}{m+5}=\frac{24}{m+5}\end{cases}\)
x+y=-1
=>\(\frac{12}{m+5}+\frac{24}{m+5}=-1\)
=>\(\frac{36}{m+5}=-1\)
=>m+5=-36
=>m=-41(nhận)
