Đặt \(A=x^2-4x+y^2-8y+6\)
\(\Leftrightarrow A=x^2-4x+4+y^2-8y+16-14\)
\(\Leftrightarrow A=\left(x-2\right)^2+\left(y-4\right)^2-14\)
Vì \(\left(x-2\right)^2\ge0;\left(y-4\right)^2\ge0\)
\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
Dấu = xảy ra khi \(\hept{\begin{cases}x-2=0\\y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\y=4\end{cases}}\)
Vậy Min A = -14 khi x=2;y=4
\(A=x^2-4x+y^2-8y+6=\left(x^2-2.x.2+2^2\right)+\left(y^2-2.y.4+4^2\right)+\left(6-4-16\right)\)
\(=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
Vậy \(MinA=-14\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}}\)
