a)\(A=x^2+3x+7\)

\(=x^2+3x+\dfrac{9}{4}+\dfrac{19}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

When \(x=-\dfrac{3}{2}\)

B = 2x² + 9y² - 6xy - 6x - 12y + 2004

   = (x² + 9y² - 6xy) + (4x - 12y) + x² - 10x + 2004

   = [x² + (3y)² - 2x.3y] + 4(x - 3y) + (x² - 10x) + 2004

   = (x - 3y)² + 2.(x - 3y).2 + 4 + (x² - 2x.5 + 25) + 2004 - 4 - 25

   = (x - 3y + 2)² + (x - 5)² + 1975

We have: (x - 3y + 2)² \(\ge0\forall x;y\);  \(\left(x-5\right)^2\ge0\forall x\)

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

\(\Rightarrow\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\ge1975\forall x;y\)

or \(B\ge1975\forall x;y\)

Equal sign occurs when and only if \(\left\{{}\begin{matrix}x-3y+2=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+2=3y\\x=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{3}\\x=5\end{matrix}\right.\)