A=\(x^2+2y^2+7x+7y+12=x^2+2xy+y^2+7\left(x+y\right)+12+y^2\)
\(=\left(x+y\right)^2+2\dfrac{7}{2}\left(x+y\right)+\left(\dfrac{7}{2}\right)^2-\dfrac{1}{4}+y^2\)
\(=\left(x+y+\dfrac{7}{2}\right)^2+y^2-\dfrac{1}{4}\ge\dfrac{-1}{4}\)
Vậy Min A =\(\dfrac{-1}{4}\) .Dấu = xảy ra\(\left\{{}\begin{matrix}x+y+\dfrac{7}{2}=0\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\y=0\end{matrix}\right.\)
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