Đáp án:
Giải thích các bước giải:
$A^{}$ = $\frac{x²-2x+2020}{x²}$ ( với x $\neq$ 0 )
⇒ $\frac{2020}{2019}$$A^{}$ = $\frac{ 2020x²-2.2020x+2020²}{2019x²}$
⇔ $\frac{2020}{2019}$$A^{}$ = $\frac{ 2019x²+ (x² -2.2020x+2020²)}{2019x²}$
⇔ $\frac{2020}{2019}$$A^{}$ = $\frac{2019x²}{2019x²}$ + $\frac{(x-2020)^{2}}{2019x²}$
⇔ $\frac{2020}{2019}$$A^{}$ = 1 + $\frac{(x-2020)^{2}}{2019x²}$
Có $\frac{(x-2020)^{2}}{2019x²}$ $\geq$ 0
⇒ $\frac{2020}{2019}$$A^{}$ $\geq$ 1.
⇒ $A^{}$$\geq$ 1 : $\frac{2020}{2019}$ = 1 . $\frac{2019}{2020}$ = $\frac{2019}{2020}$
Dấu ''='' xảy ra khi $\frac{(x-2020)^{2}}{2019x²}$ = 0
⇔ $(x-2020)^{2}$ = 0 ⇔ x- 2020 = 0
⇔ x = 0 + 2020 = 2020.
Vậy Min$A^{}$ $\geq$ $\frac{2019}{2020}$. Dấu ''='' ⇔ x = 2020.
