\(A=2x^2+9y^2-6xy-12y+2004\)
\(=\left(9y^2-6xy-12y\right)+2x^2+2004\)
\(=\left[9y^2-2.3y\left(x+2\right)+\left(x+2\right)^2\right]+2x^2-\left(x+2\right)^2+2004\)\(=\left(3y-x-2\right)^2+2x^2-x^2-4x-4+2004\)
\(=\left(3y-x-2\right)^2+x^2-4x+4+1996\)
\(=\left(3y-x-2\right)^2+\left(x-2\right)^2+1996\)
Với mọi giá trị của x;y ta có:
\(\left(3y-x-2\right)^2\ge0;\left(x-2\right)^2\ge0\)
\(\Rightarrow\left(3y-x-2\right)^2+\left(x-2\right)^2+1996\ge1996\)
Vậy Min A = 1996
Để A = 1996 thì \(\left\{{}\begin{matrix}3y-x-2=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3y-4=0\\x=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
