\(A=\dfrac{2024}{x^2-2x+8}\)
Ta có :
\(x^2-2x+8=x^2-2x+1+7=\left(x-1\right)^2+7\ge7,\forall x\inℝ\)
\(\Rightarrow\dfrac{1}{x^2-2x+8}\le\dfrac{1}{7}\)
\(\Rightarrow A=\dfrac{2024}{x^2-2x+8}\le\dfrac{2024}{7}\)
\(\Rightarrow Max\left(A\right)=\dfrac{2024}{7}\left(tạix=1\right)\)