\(A=\frac{5}{x^2-4x+2019}=\frac{5}{\left(x-2\right)^2+2015}\)
Có: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2+2015\ge2015\forall x\\ \Rightarrow\frac{5}{\left(x-2\right)^2+2015}\le\frac{5}{2015}=\frac{1}{403}\forall x\\ \Leftrightarrow A\le\frac{1}{403}\forall x\)
\(\Rightarrow\max\limits_A=\frac{1}{403}\)
Dấu "=" xảy ra: \(\Leftrightarrow\left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\)