Question

Tìm GTNN
4x²-10x+2022
(mong mn giải thích đầy đủ các bước làm giúp e ạ, e cảm ơn)

Answer 1

\(=4\left(x^2-\dfrac{5}{2}x+\dfrac{2022}{4}\right)\)

\(=4\left(x^2-2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{8063}{16}\right)\)

\(=4\left(x-\dfrac{5}{4}\right)^2-\dfrac{8063}{4}>=-\dfrac{8063}{4}\)

Dấu '=' xảy rakhi x=5/4

Answer 2

Đặt `A=4x^2 -10x +2022`

`=(2x)^2 - 2*2x*5/2 + (5/2)^2 +2022 -(5/2)^2`

`= (2x - 5/2)^2 + 2015,75`

Vì ` (2x - 5/2)^2 + 2015,75 >= 2015,75 ∀ x `

'=' khi  `(2x - 5/2)^2=0`

`<=> 2x -5/2 =0`

`<=> x = 5/4`

Vậy GTNN :  `A   =  2015,75 [khi : x=5/4]`