\(A=8x^2+3y^2-16x+6y+100\)
\(=\left(8x^2-16x+8\right)+\left(3y^2+6y+3\right)+89\)
\(=2\left(4x^2-8x+4\right)+3\left(y^2+2y+1\right)+89\)
\(=2\left[\left(2x\right)^2-2.2x.2+4\right]+3\left(y+1\right)^2+89\)
\(=2\left(2x-2\right)^2+3\left(y+1\right)^2+89\ge89\forall x;y\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2\\y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Vậy Min A là : \(89\Leftrightarrow x=1;y=-1\)
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A=8.(x^2-2x+1)+3.(y^2+2x+1)+100-8-3
A=8.(x-1)^2+3.(y+1)^2+89 >= 89
dấu "=" xảy ra khi
x=1 và y=-1
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