Toán Tìm min(max): ` \frac{x^2-12x+2012}{x^2}` 22/11/2021 By Amara Tìm min(max): ` \frac{x^2-12x+2012}{x^2}`
Đặt `A=(x^2-12x+2012)/(x^2)` $(Đk:x\neq0)$ `⇒A=(503x^2-503.12x+503.2012)/(503x^2)` `⇒A=(9x^2-2.1006.3x+1006^2+494x^2)/(503x^2)` `⇒A=(3x-1006)^2/(x^2)+(494)/(503)` Vì `(3x-1006)^2/(x^2)≥0∀x,x\ne0` `⇒(3x-1006)^2/(x^2)+(494)/(503)≥(494)/(503)∀x,x\ne0` Dấu ”=” xảy ra khi `3x-1006=0⇔x=1006/3` (tm đk) Vậy $A_{min}=$ `(494)/(503)⇔x=1006/3`. Trả lời
Đặt `A=(x^2-12x+2012)/(x^2)` $(Đk:x\neq0)$
`⇒A=(503x^2-503.12x+503.2012)/(503x^2)`
`⇒A=(9x^2-2.1006.3x+1006^2+494x^2)/(503x^2)`
`⇒A=(3x-1006)^2/(x^2)+(494)/(503)`
Vì `(3x-1006)^2/(x^2)≥0∀x,x\ne0`
`⇒(3x-1006)^2/(x^2)+(494)/(503)≥(494)/(503)∀x,x\ne0`
Dấu ”=” xảy ra khi `3x-1006=0⇔x=1006/3` (tm đk)
Vậy $A_{min}=$ `(494)/(503)⇔x=1006/3`.