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