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Trả lời:
$A=\dfrac{x^2-2x+2005}{x^2}$ $(ĐK: x \neq 0)$
$=\dfrac{2005}{x^2}-\dfrac{2}{x}+1$
$=\dfrac{2005}{x^2}-2.\dfrac{\sqrt{2005}}{x}.\dfrac{1}{\sqrt{2005}}+\dfrac{1}{2005}+\dfrac{2004}{2005}$
$=\bigg{(}\dfrac{\sqrt{2005}}{x}-\dfrac{1}{\sqrt{2005}}\bigg{)}^2+\dfrac{2004}{2005}$
Ta có: $\bigg{(}\dfrac{\sqrt{2005}}{x}-\dfrac{1}{\sqrt{2005}}\bigg{)}^2 \geq 0 ⇒ A \geq \dfrac{2004}{2005}$
Dấu "=" xảy ra $⇔\dfrac{\sqrt{2005}}{x}=\dfrac{1}{\sqrt{2005}}⇔x=2005$.
Vậy $A_{min}=\dfrac{2004}{2005}$ khi $x=2005$.
