Vì \(\dfrac{2}{1}\ne\dfrac{1}{-2}\)
nên hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}2x+y=5m-1\\x-2y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+2y=10m-2\\x-2y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+2y+x-2y=10m-2+2\\x-2y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5x=10m\\2y=x-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2m\\2y=2m-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2m\\y=m-1\end{matrix}\right.\)
\(x^2+2y^2=2\)
=>\(\left(2m\right)^2+2\left(m-1\right)^2=2\)
=>\(4m^2+2\left(m^2-2m+1\right)-2=0\)
=>\(6m^2-4m=0\)
=>\(m\left(6m-4\right)=0\)
=>\(\left[{}\begin{matrix}m=0\\m=\dfrac{2}{3}\end{matrix}\right.\)
