a: Để hệ có nghiệm duy nhất thì \(\dfrac{m}{1}\ne\dfrac{1}{m}\)
=>\(m^2\ne1\)
=>\(m\notin\left\{1;-1\right\}\)
Để hệ có vô số nghiệm thì \(\dfrac{m}{1}=\dfrac{1}{m}=\dfrac{3m-1}{m+1}\)
=>\(\left\{{}\begin{matrix}\dfrac{m}{1}=\dfrac{1}{m}\\\dfrac{1}{m}=\dfrac{3m-1}{m+1}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m^2=1\\3m^2-m=m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{1;-1\right\}\\3m^2-2m-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\in\left\{1;-1\right\}\\\left(m-1\right)\left(3m+1\right)=0\end{matrix}\right.\)
=>m=1
Để hệ vô nghiệm thì \(\dfrac{m}{1}=\dfrac{1}{m}\ne\dfrac{3m-1}{m+1}\)
=>\(\left\{{}\begin{matrix}\dfrac{m}{1}=\dfrac{1}{m}\\\dfrac{m}{1}\ne\dfrac{3m-1}{m+1}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m^2=1\\m^2+m\ne3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\in\left\{1;-1\right\}\\m^2-2m+1\ne0\end{matrix}\right.\)
=>m=-1
b: Để hệ có vô số nghiệm thì \(\dfrac{m}{1}=\dfrac{4}{m}=\dfrac{10-m}{4}\)
=>\(\left\{{}\begin{matrix}\dfrac{m}{1}=\dfrac{4}{m}\\\dfrac{4}{m}=\dfrac{10-m}{4}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m^2=4\\10m-m^2=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\m^2-10m+16=0\end{matrix}\right.\)
=>m=2
Để hệ vô nghiệm thì \(\dfrac{m}{1}=\dfrac{4}{m}\ne\dfrac{10-m}{4}\)
=>\(\left\{{}\begin{matrix}\dfrac{m}{1}=\dfrac{4}{m}\\\dfrac{m}{1}\ne\dfrac{10-m}{4}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m^2=4\\4m\ne10-m\end{matrix}\right.\Leftrightarrow m=-2\)
Để hệ có nghiệm duy nhất thì \(\dfrac{m}{1}\ne\dfrac{4}{m}\)
=>\(m^2\ne4\)
=>\(m\notin\left\{2;-2\right\}\)
