Tìm min ( max ) : ` \frac{x^2-12x+2012}{x^2}`

Tìm min ( max ) :
` \frac{x^2-12x+2012}{x^2}`

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  1. 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:

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

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