\(2x^2+9y^2-6xy-6x-12y+2004\)
\(=x^2-10x+25+x^2+9y^2+4-6xy+4x-12y+1975\)
\(=\left(x-5\right)^2+\left(x-3y+2\right)^2+1975\ge1975\)
Dấu \(=\)khi \(\hept{\begin{cases}x-5=0\\x-3y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{7}{3}\end{cases}}\).