Tìm min ( max ) : ` \frac{x-2x+2000}{x^2}`

$\begin{array}{l}\underline{\text{Đáp án:}}\\Min_A=\dfrac{1999}{2000}↔x=2000\\\underline{\text{Giải thích các bước giải:}}\\Đặt \,\, A=\dfrac{x^2-2x+2000}{x^2}\\ĐK:x \neq 0\\Xét \,\, A-\dfrac{1999}{2000}\\=\dfrac{2000x^2-4000x+2000^2-1999x^2}{2000x^2}\\=\dfrac{x^2-2.x.2000+2000^2}{2000x^2}\\=\dfrac{(x-2000)^2}{2000x^2} \geq 0\\→A-\dfrac{1999}{2000} \geq 0\\→A \geq \dfrac{1999}{2000}\\\text{Dấu =xảy ra khi}\\x-2000=0\\→x=2000\\Vậy \,\, Min_A=\dfrac{1999}{2000}↔x=2000\\\underline{\text{CHÚC BẠN HỌC TỐT}}\\\end{array}$