\(A=4x^2+y^2+4x+2y+7\)
\(=\left(4x^2+4x+1\right)+\left(y^2+2y+1\right)+5\)
\(=\left(2x+1\right)^2+\left(y+1\right)^2+5\ge5\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-1\end{matrix}\right.\)
Vậy \(Min_A=5\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-1\end{matrix}\right.\)
\(B=6x+3x^2+4\)
\(=3\left(x^2+2x+1\right)+1\)
\(=3\left(x+1\right)^2+1\ge1\)
Dấu = xảy ra \(\Leftrightarrow x=-1\)
Vậy \(Min_B=1\Leftrightarrow x=-1\)
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