Câu 2: Để hệ có nghiệm duy nhất thì \(\frac{2}{m}<>\frac{m-1}{3}\)
=>\(m\left(m-1\right)<>6\)
=>\(m^2-m-6<>0\)
=>(m-3)(m+2)<>0
=>m∉{3;-2}
\(\begin{cases}2x+\left(m-1\right)y=4\\ mx+3y=m+3\end{cases}\Rightarrow\begin{cases}2mx+\left(m^2-m\right)y=4m\\ 2mx+6y=2m+6\end{cases}\)
=>\(\begin{cases}2mx+\left(m^2-m\right)y-2mx-6y=4m-2m-6\\ mx+3y=m+3\end{cases}\)
=>\(\begin{cases}y\left(m^2-m-6\right)=2m-6\\ mx+3y=m+3\end{cases}\Rightarrow\begin{cases}y=\frac{2\left(m-3\right)}{\left(m-3\right)\left(m+2\right)}=\frac{2}{m+2}\\ mx=m+3-3y=m+3-\frac{6}{m+2}=\frac{m^2+5m+6-6}{m+2}=\frac{m^2+5m}{m+2}\end{cases}\)
=>\(\begin{cases}y=\frac{2}{m+2}\\ x=\frac{m+5}{m+2}\end{cases}\)
x-3y=2
=>\(\frac{m+5}{m+2}-\frac{6}{m+2}=2\)
=>\(\frac{m-1}{m+2}=2\)
=>2(m+2)=m-1
=>2m+4=m-1
=>m=-5(nhận)
