a: Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{-1}\)
=>\(m^2\ne-1\)(luôn đúng)
=>HPT luôn có nghiệm duy nhất
b: \(\left\{{}\begin{matrix}x+my=1\\mx-y=-m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+my=1\\m^2x-my=-m^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(m^2+1\right)=-m^2+1\\mx-y=-m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-m^2+1}{m^2+1}\\y=mx+m=\dfrac{-m^3+m+m^3+m}{m^2+1}=\dfrac{2m}{m^2+1}\end{matrix}\right.\)
Để x<1 và y<1 thì \(\left\{{}\begin{matrix}\dfrac{-m^2+1}{m^2+1}-1< 0\\\dfrac{2m}{m^2+1}-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-m^2+1-m^2-1}{m^2+1}< 0\\\dfrac{2m-m^2-1}{m^2+1}< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2m^2< 0\\-m^2+2m-1< 0\end{matrix}\right.\Leftrightarrow m\in\varnothing\)
c: Để x,y nguyên thì \(\left\{{}\begin{matrix}-m^2+1⋮m^2+1\\2m⋮m^2+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-m^2-1+2⋮m^2+1\\2m⋮m^2+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2⋮m^2+1\\2m⋮m^2+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m^2+1\in\left\{1;2\right\}\\2m⋮m^2+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{0;1;-1\right\}\\2m⋮m^2+1\end{matrix}\right.\Leftrightarrow m\in\left\{0;1;-1\right\}\)
