a: Để hệ có nghiệm duy nhất thì \(\dfrac{m-1}{1}\ne\dfrac{1}{m-1}\)
=>\(\left(m-1\right)^2\ne1\)
=>\(m-1\notin\left\{1;-1\right\}\)
=>\(m\notin\left\{2;0\right\}\)
\(\left\{{}\begin{matrix}\left(m-1\right)x+y=m\\x+\left(m-1\right)y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=m-\left(m-1\right)x\\x+\left(m-1\right)\left[m-\left(m-1\right)x\right]=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=m-\left(m-1\right)x\\x+m\left(m-1\right)-x\left(m-1\right)^2=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=m-\left(m-1\right)x\\x\left[1-\left(m-1\right)^2\right]=2-m\left(m-1\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=m-\left(m-1\right)x\\x\left[\left(m-1\right)^2-1\right]=m\left(m-1\right)-2=\left(m-2\right)\left(m+1\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=m-\left(m-1\right)x\\x\left(m-2\right)\cdot m=\left(m-2\right)\left(m+1\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=m-\left(m-1\right)x\\x=\dfrac{m+1}{m}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=m-\dfrac{\left(m-1\right)\left(m+1\right)}{m}=\dfrac{m^2-m^2+1}{m}=\dfrac{1}{m}\\x=\dfrac{m+1}{m}\end{matrix}\right.\)
\(x-y=\dfrac{m+1}{m}-\dfrac{1}{m}=\dfrac{m}{m}=1\)
=>x-y là hệ thức không phụ thuộc vào m
b: \(2x^2-7y=1\)
=>\(2\cdot\left(\dfrac{m+1}{m}\right)^2-\dfrac{7}{m}=1\)
=>\(\dfrac{2\left(m+1\right)^2-7m}{m^2}=1\)
=>\(2\left(m^2+2m+1\right)-7m=m^2\)
=>\(2m^2+4m+2-7m-m^2=0\)
=>\(m^2-3m+2=0\)
=>(m-2)(m-1)=0
=>\(\left[{}\begin{matrix}m=2\left(loại\right)\\m=1\left(nhận\right)\end{matrix}\right.\)
c: Đặt \(A=\dfrac{2x-3y}{x+y}=\dfrac{2\cdot\dfrac{m+1}{m}-\dfrac{3}{m}}{\dfrac{m+1}{m}+\dfrac{1}{m}}=\dfrac{2m+2-3}{m}:\dfrac{2m+1}{m}=\dfrac{2m-1}{2m+1}\)
Để A là số nguyên thì \(2m-1⋮2m+1\)
=>\(2m+1-2⋮2m+1\)
=>\(-2⋮2m+1\)
=>\(2m+1\in\left\{1;-1;2;-2\right\}\)
=>\(m\in\left\{0;-1;\dfrac{1}{2};-\dfrac{3}{2}\right\}\)
mà m nguyên và m<>0; m<>2
nên m=-1