Ta có: \(x^2+2x+3=x^2+2x+1+2\)
\(=\left(x+1\right)^2+2\ge2\)
\(\Rightarrow\dfrac{1}{x^2+2x+3}=\dfrac{1}{\left(x+1\right)^2+2}\le\dfrac{1}{2}\)
\(\Rightarrow A=\dfrac{14}{\left(x+1\right)^2+2}\le\dfrac{14}{2}=7\)
Đẳng thức xảy ra khi \(\left(x+1\right)^2=0\Rightarrow x=-1\)