Question

tìm min của \(A=\frac{x^2-2x+2018}{x^2}\)

Answer 1

\(A=\frac{x^2-2x+2018}{x^2}=1-\frac{2}{x}+\frac{2018}{x^2}\)

\(=2018t^2-2t+1\left(\frac{1}{x}=t\right)\)

\(=2018\left(t^2-\frac{2t}{2018}+\frac{1}{2018}\right)\)

\(=2018\left(t-\frac{1}{2018}\right)^2+\frac{2017}{2018}\ge\frac{2017}{2018}\)