Ta có: \(A=x^2+5y^2-2xy+4y+3\)
\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2\)
\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(2y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow x=y=\dfrac{-1}{2}\)
Vậy \(MIN_A=2\) khi \(x=y=\dfrac{-1}{2}\)