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 = \(\dfrac{7}{3}\)
\(A=2x^2+9y^2-6xy-6x-12y+2004\)
\(=\left(9y^2-6xy-12y\right)+2x^2-6x+2004\)
\(=\left[9y^2-6y\left(x+2\right)+\left(x+2\right)^2\right]+2x^2-6x+2004\)\(=\left(3y-x-2\right)^2+2x^2-6x+2004-x^2-4x-4\)\(=\left(3y-x-2\right)^2+\left(x^2-10x+25\right)+1979\)
\(=\left(3y-x-2\right)^2+\left(x-5\right)^2+1979\)
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 \(\left\{{}\begin{matrix}3y-x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y-5-2=0\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3y=7\\x=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{3}\\x=5\end{matrix}\right.\)