\(A=x^2-4x+y^2-8y+8\)
\(=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-12\)
\(=\left(x-2\right)^2+\left(y-4\right)^2-12\ge-12\)
Dấu "=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\y-4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)