Đáp án đúng: A
Giải chi tiết:Điều kiện : \(x \ne 0.\)
Ta có: \(A = \frac{{{x^2} - 3x + 2019}}{{{x^2}}} = 1 - \frac{3}{x} + \frac{{2019}}{{{x^2}}}\)
Đặt \(t = \frac{1}{x}\,\,\left( {t \ne 0} \right)\), khi đó ta có:
\(\begin{array}{l}A = 1 - 3t + 2019{t^2} = 2019\left( {{t^2} - \frac{1}{{673}}t} \right) + 1\\\,\,\,\,\, = 2019\left[ {{t^2} - 2t.\frac{1}{{1346}} + {{\left( {\frac{1}{{1346}}} \right)}^2}} \right] - 2019.{\left( {\frac{1}{{1346}}} \right)^2} + 1\\\,\,\,\,\, = 2019{\left( {t - \frac{1}{{1346}}} \right)^2} + \frac{{2689}}{{2692}}\end{array}\)
Ta có
\(\begin{array}{l}{\left( {t - \frac{3}{{4038}}} \right)^2} \ge 0\,\,\,\,\,\forall t \Leftrightarrow 2019{\left( {t - \frac{3}{{4038}}} \right)^2} \ge 0\,\,\,\,\forall t\\ \Leftrightarrow {\left( {t - \frac{3}{{4038}}} \right)^2} + \frac{{2689}}{{2692}} \ge \frac{{2689}}{{2692}}\,\,\,\,\forall t\\ \Rightarrow A \ge \frac{{2689}}{{2692}}\,\,\,\forall t\end{array}\)
Dấu "=" xảy ra \( \Leftrightarrow t = \frac{1}{{1346}}\,\,\,\left( {tm} \right)\).
Vậy \(\min A = \frac{{2689}}{{2692}}\) đạt được khi \(t = \frac{1}{{1346}} \Leftrightarrow x = 1346\,\,\,\left( {tm} \right).\,\,\)
Chọn A.
