\(A=4x^2-6xy+3y^2+8x-5\)
\(=3\left(x^2-2xy+y^2\right)+\left(x^2+8x+16\right)-21\)
\(=3\left(x-y\right)^2+\left(x+4\right)^2-21\le-21\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x+4=0\end{matrix}\right.\Leftrightarrow x=y=-4\)
Vậy \(MinA=-21\)