\(A=\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(y^2-6y+9\right)+2006\)\(=\left(x+y\right)^2+2\left(x+y\right)+1+\left(y-3\right)^2+2006\)
\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2006\)
Ta có: \(\left(x+y+1\right)^2+\left(y-3\right)^2\ge0\left(\forall x;y\right)\)
\(\Rightarrow A\ge2006\).
Vậy MIN A = 2006 \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)
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