Đặt \(K=4x^2+2y^2+4xy-16x-12y+5\)
\(K=\left(4x^2+4xy+y^2\right)+y^2-16x-12y+5\)
\(K=\left[\left(2x+y\right)^2-2\left(2x+y\right).4+16\right]+\left(y^2-4y+4\right)-15\)
\(K=\left(2x+y-4\right)^2+\left(y-2\right)^2-15\)
Mà \(\left(2x+y-4\right)^2\ge0\forall x;y\)
\(\left(y-2\right)^2\ge0\forall y\)
\(\Rightarrow K\ge-15\)
Dấu "=" xảy ra khi : \(\hept{\begin{cases}2x+y-4=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)
Vậy \(K_{Min}=-15\Leftrightarrow\left(x;y\right)=\left(1;2\right)\)
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