\(3^x+3^{-x}=2^{2023}\)
=>\(3^x+\dfrac{1}{3^x}=2^{2023}\)
=>\(3^{2x}+1=3^x\cdot2^{2023}\)
\(A=\dfrac{3^{6x}+3^{3x}+1}{3^{2x}\left(3^{2x}+1\right)}\)
\(=\dfrac{3^{3x}\left(3^{3x}+1\right)+1}{3^{2x}\cdot3^x\cdot2^{2023}}\)
\(=\dfrac{3^{3x}+2}{2^{2023}}\)