\(A=4x^2+y^2-12x+8y+28\)
\(=\left(4x^2-12x+9\right)+\left(y^2+8y+16\right)+3\)
\(=\left[\left(2x\right)^2+2.2x.3+3^2\right]+\left(y^2+2.y.4+4^2\right)+3\)
\(=\left(2x+3\right)^2+\left(y+4\right)^2+3\)
Ta có :
\(\left(2x+3\right)^2\ge0\forall x\)
\(\left(y+4\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x+3\right)^2+\left(y+4\right)^2+3\ge3\forall x\)
Dấu = xảy ra khi
\(\left\{{}\begin{matrix}\left(2x+3\right)^2=0\\\left(y+4\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x+3=0\\y+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-4\end{matrix}\right.\)
Vậy \(Min_A=3\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=-4\end{matrix}\right.\)
