\(A=\frac{6}{2x^2-6x+7}\)
Ta có \(2x^2-6x+7=2\left(x^2-3x+\frac{9}{4}\right)+\frac{5}{2}=2\left(x-\frac{3}{2}\right)^2+\frac{5}{2}\ge\frac{5}{2}\)
\(\Rightarrow A=\frac{6}{2x^2-6x+7}\le\frac{6}{\frac{5}{2}}=\frac{12}{5}\)
\(Max_A=\frac{12}{5}\Leftrightarrow x=\frac{3}{2}\)