Đặt \(A=\frac{x^2-2x+2020}{x^2}\)
\(=1-\frac{2}{x}+\frac{2020}{x^2}\)
\(=2020\left(\frac{1}{x^2}-\frac{1}{1010\cdot x}+\frac{1}{2020}\right)\)
\(=2020\left(\frac{1}{x^2}-2\cdot\frac{1}{x}\cdot\frac{1}{2020}+\frac{1}{2020^2}+\frac{1}{2020}-\frac{1}{2020^2}\right)=2020\left(\frac{1}{x}-\frac{1}{2020}\right)^2+1-\frac{1}{2020}=2020\left(\frac{1}{x}-\frac{1}{2020}\right)^2+\frac{2019}{2020}\ge\frac{2019}{2020}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\frac{1}{x}-\frac{1}{2020}=0\)
=>x=2020
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