Đáp án:
$\min\left(\dfrac{x^2 - 12x + 2012}{x^2}\right) = \dfrac{494}{503}\Leftrightarrow x = \dfrac{1006}{3}$
Giải thích các bước giải:
$\begin{array}{l}\quad \dfrac{x^2 - 12x + 2012}{x^2}\qquad (x \ne 0)\\ = 1 - \dfrac{12}{x} + \dfrac{2012}{x^2}\\ = 2012\left(\dfrac{1}{x^2} - 2\cdot \dfrac{1}{x} \cdot \dfrac{3}{1006} + \dfrac{9}{1006^2}\right)+\dfrac{494}{503}\\ = 2012\left(\dfrac{1}{x} - \dfrac{3}{1006}\right)^2 + \dfrac{494}{503}\\ \text{Ta có:}\\ \quad \left(\dfrac{1}{x} - \dfrac{3}{1006}\right)^2 \geq 0\quad \forall x \ne 0\\ \to 2012\left(\dfrac{1}{x} - \dfrac{3}{1006}\right)^2 + \dfrac{494}{503}\geq \dfrac{494}{503}\\ Hay\,\, \dfrac{x^2 - 12x + 2012}{x^2}\geq \dfrac{494}{503}\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow \dfrac{1}{x} - \dfrac{3}{1006} = 0\Leftrightarrow x = \dfrac{1006}{3}\\ Vậy\,\,\min\left(\dfrac{x^2 - 12x + 2012}{x^2}\right) = \dfrac{494}{503}\Leftrightarrow x = \dfrac{1006}{3}\end{array}$
