Để hệ có nghiệm duy nhất thì \(\dfrac{2}{m+2}\ne\dfrac{m-2}{-2}\)
=>\(\left(m+2\right)\left(m-2\right)\ne-4\)
=>\(m^2-4\ne-4\)
=>\(m^2\ne0\)
=>\(m\ne0\)
\(\left\{{}\begin{matrix}2x+\left(m-2\right)y=m+1\\\left(m+2\right)x-2y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+\left(2m-4\right)y=2m+2\\\left(m^2-4\right)x-\left(2m-4\right)y=3\left(m-2\right)=3m-6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+\left(2m-4\right)y+\left(m^2-4\right)x-\left(2m-4\right)y=2m+2+3m-6\\\left(m+2\right)x-2y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}xm^2=5m-4\\2y=\left(m+2\right)x-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5m-4}{m^2}\\2y=\dfrac{\left(m+2\right)\left(5m-4\right)-3m^2}{m^2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5m-4}{m^2}\\y=\dfrac{5m^2-4m+10m-8-3m^2}{2m^2}=\dfrac{2m^2+6m-8}{2m^2}=\dfrac{m^2+3m-4}{m^2}\end{matrix}\right.\)
\(x+y=\dfrac{m^2+3m-4+5m-4}{m^2}=\dfrac{m^2+8m-8}{m^2}\)
\(=1+\dfrac{8}{m}-\dfrac{8}{m^2}\)
\(=-8\left(\dfrac{1}{m^2}-\dfrac{1}{m}-\dfrac{1}{8}\right)\)
\(=-8\left(\dfrac{1}{m^2}-2\cdot\dfrac{1}{m}\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{3}{8}\right)\)
\(=-8\left(\dfrac{1}{m}-\dfrac{1}{2}\right)^2+3< =3\forall m\ne0\)
Dấu '=' xảy ra khi m=2
