Question

\(\left\{{}\begin{matrix}2x+y=5m-1\\x-2y=2\end{matrix}\right.\)

tìm m để hệ có nghiệm thỏa mãn `x^2 -2y^2 =-2`

Answer 1

Vì \(\dfrac{2}{1}\ne\dfrac{1}{-2}\)

nên hệ  luôn có nghiệm duy nhất

\(\left\{{}\begin{matrix}2x+y=5m-1\\x-2y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+2y=10m-2\\x-2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=10m\\x-2y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2m\\2y=x-2=2m-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2m\\y=m-1\end{matrix}\right.\)

\(x^2-2y^2=-2\)

=>\(\left(2m\right)^2-2\left(m-1\right)^2=-2\)

=>\(4m^2-2\left(m^2-2m+1\right)=-2\)

=>\(4m^2-2m^2+4m=0\)

=>\(2m^2+4m=0\)

=>\(m^2+2m=0\)

=>m(m+2)=0

=>\(\left[{}\begin{matrix}m=0\\m=-2\end{matrix}\right.\)