a: Thay m=1 vào hệ, ta được:
\(\left\{{}\begin{matrix}2x-y=1-2=-1\\x+2y=3\cdot1+4=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x-2y=-2\\x+2y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=5\\x+2y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\2y=7-x=7-1=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
b: Vì \(\dfrac{2}{1}\ne\dfrac{-1}{2}\)
nên hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}2x-y=m-2\\x+2y=3m+4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x-2y=2m-4\\x+2y=3m+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=5m\\2x-y=m-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m\\y=2x-m+2=2m-m+2=m+2\end{matrix}\right.\)
\(x^2+y^2=10\)
=>\(m^2+\left(m+2\right)^2=10\)
=>\(2m^2+4m-6=0\)
=>\(m^2+2m-3=0\)
=>(m+3)(m-1)=0
=>\(\left[{}\begin{matrix}m=-3\\m=1\end{matrix}\right.\)