\(C=\frac{3}{4x^2-4x+5}\)
\(C=\frac{3}{4x^2-4x+1+4}\)
\(C=\frac{3}{\left(4x^2-4x+1\right)+4}\)
\(C=\frac{3}{\left(4x-1\right)^2+4}\)
Ta thấy: \(\left(4x-1\right)^2\ge0\Rightarrow\left(4x-1\right)^2+4\ge4\)
\(\Rightarrow\frac{3}{\left(4x-1\right)^2+4}\le\frac{3}{4}\)
\(Max_A=\frac{3}{4}\)
Dấu " = " xảy ra \(\Leftrightarrow x=\frac{1}{2}\)