\(x^2+y^2-2x+4y+8=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy GTNN là 3 khi x=1 và y=-2
=>Chọn B
\(x^2+y^2-2x+4y+8=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3=\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\)
Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)