A = 2x² + 9y² - 6xy - 6x - 12y + 2004

= (x²-6xy+9y²) + 4(x-3y) + 4 + (x²-10x+25) + 1975

= (x-3y)² + 4(x-3y) + 4 + (x-5)² + 1975

= (x-3y+2)² + (x-5)² + 1975 \(\ge\) 1975

Vậy MinA = 1975

Dấu "=" xảy ra khi x = 5; y = \(\dfrac{7}{3}\)

Sao lai y = 7/3

\(\left\{{}\begin{matrix}x-5=0\\x-3y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\5-3y+2=0\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=5\\y=\dfrac{7}{3}\end{matrix}\right.\)

\(A=2x^2+9y^2-6xy-6x-12y+2004\)

\(=\left(9y^2-6xy-12y\right)+2x^2-6x+2004\)

\(=\left[9y^2-6y\left(x+2\right)+\left(x+2\right)^2\right]+2x^2-6x+2004\)\(=\left(3y-x-2\right)^2+2x^2-6x+2004-x^2-4x-4\)\(=\left(3y-x-2\right)^2+\left(x^2-10x+25\right)+1979\)

\(=\left(3y-x-2\right)^2+\left(x-5\right)^2+1979\)

Với mọi giá trị của x;y ta có:

\(\left(3y-x-2\right)^2\ge0;\left(x-5\right)^2\ge0\)

\(\Rightarrow\left(3y-x-2\right)^2+\left(x-5\right)^2+1979\ge1979\)

Vậy Min A = 1979

Để A = 1979 thì \(\left\{{}\begin{matrix}3y-x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y-5-2=0\\x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3y=7\\x=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{3}\\x=5\end{matrix}\right.\)