\(A=x^2-4x+y^2-8y+6\\ =x^2-4x+4+y^2-8y+16-14\\ =\left(x-2\right)^2+\left(y-4\right)^2-14\\ \left(x-2\right)^2\ge0\forall x\\ \left(y-4\right)^2\ge0\forall y\\ \Rightarrow\left(x-2\right)^2+\left(y-4\right)^2\ge0\forall x,y\\ \Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\forall x,y\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2=0\\y-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
Vậy \(Min_A=-14\) khi \(x=2;y=4\)
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