\(A=2019x^2-2x+1\)
\(A=2019\left(x^2-\frac{2}{2019}x+\frac{1}{2019}\right)\)
\(A=2019\left(x^2-2\cdot x\cdot\frac{1}{2019}+\frac{1}{2019^2}+\frac{2018}{2019^2}\right)\)
\(A=2019\left[\left(x-\frac{1}{2019}\right)^2+\frac{2018}{2019^2}\right]\)
\(A=2019\left(x-\frac{1}{2019}\right)^2+\frac{2018}{2019}\ge\frac{2018}{2019}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2019}\)