a: Thay m=-1 vào hệ, ta được:
\(\left\{{}\begin{matrix}3x-y=2\left(-1\right)+3=1\\x+2y=3\left(-1\right)+1=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x-2y=2\\x+2y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x-2y+x+2y=2-2=0\\x+2y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0\\2y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0\\y=-1\end{matrix}\right.\)
b: Vì \(\dfrac{3}{1}\ne\dfrac{1}{-2}\)
nên hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}3x-y=2m+3\\x+2y=3m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x-2y=4m+6\\x+2y=3m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x-2y+x+2y=4m+6+3m+1\\x+2y=3m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x=7m+7\\2y=3m+1-x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1\\2y=3m+1-m-1=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1\\y=m\end{matrix}\right.\)
\(x^2+y^2=5\)
=>\(\left(m+1\right)^2+m^2-5=0\)
=>\(2m^2+2m-4=0\)
=>\(m^2+m-2=0\)
=>(m+2)(m-1)=0
=>\(\left[{}\begin{matrix}m=-2\\m=1\end{matrix}\right.\)
