Cho ​\(\begin{array}{l} A = \frac{{2019{x^2} - 2 \cdot x \cdot 2019 + {{2019}^2}}}{{2019{x^2}}} \end{array}\) (x>0). Tìm giá trị nhỏ nhất của A

* Hướng dẫn giải

 \(\begin{array}{l} A = \frac{{2019{x^2} - 2 \cdot x \cdot 2019 + {{2019}^2}}}{{2019{x^2}}} = \frac{{{{(x - 2019)}^2}}}{{2019{x^2}}} + \frac{{2018{x^2}}}{{2019{x^2}}} = \frac{{{{(x - 2019)}^2}}}{{2019{x^2}}} + \frac{{2018}}{{2019}}\\ Ta\,có\,\frac{{{{(x - 2019)}^2}}}{{2019{x^2}}} > 0,\forall x > 0 \Rightarrow \frac{{{{(x - 2019)}^2}}}{{2019{x^2}}} + \frac{{2018}}{{2019}} \ge \frac{{2018}}{{2019}} \end{array}\)

Vậy \(MinA = \frac{{2018}}{{2019}}\)