Ta có: \(P=\dfrac{2023}{x^2-4x+11}=\dfrac{2023}{ \left(x^2-4x+4\right)+7}=\dfrac{2023}{\left(x-2\right)^2+7}\)
Ta thấy: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2+7\ge7\forall x\)
\(\Rightarrow\dfrac{1}{\left(x-2\right)^2+7}\le\dfrac{1}{7}\forall x\)
\(\Rightarrow P=\dfrac{2023}{\left(x-2\right)^2+7}\le\dfrac{2023}{7}=289\forall x\)
Dấu \("="\) xảy ra khi: \(x-2=0\Leftrightarrow x=2\)
Vậy \(P_{max}=289\) tại \(x=2\)
\(\rightarrow\) Chọn D. 289