\(P=\dfrac{3x^2+6x+11}{x^2+2x+3}=\dfrac{3\left(x^2+2x+1\right)+9}{x^2+2x+3}=\dfrac{3.\left(x+1\right)^2+9}{\left(x+1\right)^2+2}=\dfrac{3.\left[\left(x+1\right)^2+2\right]+3}{\left(x+1\right)^2+2}=3+\dfrac{3}{\left(x+1\right)^2+2}\)=> \(P\le3+\dfrac{3}{2}=\dfrac{9}{2}\)

=> \(Max_P=\dfrac{9}{2}\Leftrightarrow x=-1\)

We have:P-4=\(\dfrac{3x^2+6x+11-4\left(x^2+2x+3\right)}{x^2+2x+3}\)

=\(\dfrac{3x^2+6x+11-4x^2-8x-12}{x^2+2x+3}\)\(=\dfrac{-x^2-2x-1}{x^2+2x+3}=\dfrac{-\left(x+1\right)^2}{x^2+2x+3}\le0\)

\(\Rightarrow P\le4\).So max_A=4 when x=-1