\(A=\frac{13}{\left(3x-2\right)^2+11}\)
Vì \(\left(3x-2\right)^2\ge0;\forall x\)
\(\Rightarrow\left(3x-2\right)^2+11\ge0+11;\forall x\)
\(\Rightarrow\frac{13}{\left(3x-2\right)^2+11}\le\frac{13}{11};\forall x\)
Dấu"="xảy ra \(\Leftrightarrow\left(3x-2\right)^2=0\)
\(\Leftrightarrow x=\frac{2}{3}\)
Vậy Max\(A=\frac{13}{11}\)\(\Leftrightarrow x=\frac{2}{3}\)