\(A=\frac{4}{4x^2-4x+7}\)
Xét \(4x^2-4x+7=4x^2-4x+1+6=\left(2x-1\right)^2+6\)
Vì \(\left(2x-1\right)^2\ge0\forall x\in R\)
\(\Leftrightarrow\left(2x-1\right)^2+6\ge6\forall x\in R\)
\(\Leftrightarrow\frac{1}{\left(2x-1\right)^2+6}\le\frac{1}{6}\alpha\forall x\in R\)
\(\Leftrightarrow\frac{4}{\left(2x-1\right)^2+6}\le\frac{2}{3}\alpha\forall x\in R\)
\(\Leftrightarrow A\le\frac{2}{3}\forall x\in R\)
\(Max_A=\frac{2}{3}\)
Dấu \("="\) xảy ra\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)