\(A=\dfrac{x^2-2x+2025}{x^2}=1-\dfrac{2}{x}+\dfrac{2025}{x^2}\left(x\ne0\right)\)
\(\Rightarrow A=2025\left(\dfrac{1}{x^2}-2.\dfrac{1}{x}.\dfrac{1}{2025}+\dfrac{1}{2025^2}\right)+1-\dfrac{1}{2025}\)
\(\Rightarrow A=2025\left(\dfrac{1}{x}-\dfrac{1}{2025}\right)^2+\dfrac{2024}{2025}\)
mà \(2025\left(\dfrac{1}{x}-\dfrac{1}{2025}\right)^2\ge0;\forall x\ne0\)
\(\Rightarrow A\ge\dfrac{2024}{2025}\)
Dấu "=" xảy ra khi \(\dfrac{1}{x}-\dfrac{1}{2025}=0\Leftrightarrow x=\dfrac{1}{2025}\)
Vậy \(A\left(min\right)=\dfrac{2024}{2025}\left(tại.x=\dfrac{1}{2025}\right)\)
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