\(\frac{5}{4x^2-4x+21}=\frac{5}{4x^2-2x-2x+1+20}=\frac{5}{\left(2x-1\right)^2+20}\)
\(\left(2x-1\right)^2\ge0\Rightarrow\left(2x-1\right)^2+20\ge20\)
dấu = xảy ra khi (2x-1)2=0
=> \(x=\frac{1}{2}\)
Vậy max \(\frac{5}{4x^2-4x+21}=\frac{5}{20}=\frac{1}{4}\)