A=2/3x-x^2-1 Tìm giá trị lớn nhất

tickerbell2006

2 năm trước

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nhilehoyen

`#Sad`

`A= (2)/(3)x-x^2-1`

`= -x^2+(2)/(3)x-1`

`= -(x^2-(2)/(3)x+1)`

`= -(x^2-(2)/(3)x+1/9)-8/9`

`= -(x-1/3)^2-8/9`

`\text{Vì}` `-(x-1/3)^2 <= 0` `AA` `x`

`⇒ -(x-1/3)^2-8/9 <= -8/9`

`\text{Dấu "=" xảy ra:}`

`⇔ -(x-1/3)^2 = 0`

`⇔ x-1/3 = 0`

`⇔ x = (1)/(3)`

`\text{Vậy}` `Max_A` `= -(8)/9` `⇔` `x=(1)/(3)`

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trieuquocvuong

Đáp án:

`\text{Max}_A = -8/9 <=>x =1/3`

Giải thích các bước giải:

`A = 2/3x - x^2 - 1`

`= - (x^2 - 2/3x + 1)`

`= - (x^2 - 2/3x+ 1/9) -8/9`

` = - (x - 1/3)^2 - 8/9`

`\forall x` ta có :

`(x-1/3)^2 \ge 0`

`=> - (x-1/3)^2 \le 0`

`=> - (x-1/3)^2 - 8/9 \le -8/9`

`=> A \le -8/9`

Dấu `=` xảy ra `<=> x -1/3=0`

`<=>x=1/3`

Vậy `\text{Max}_A = -8/9 <=>x =1/3`

Vote (0) Phản hồi (0) 2 năm trước

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