Đáp án:
`a,`
`A = 2x^2 - 8x + 1`
`⇔ A = 2x^2 - 8x + 8 - 7`
`⇔ A = 2 (x^2 - 4x + 4) - 7`
`⇔ A = 2 (x^2 - 2 . 2x + 2^2) - 7`
`⇔ A = 2 (x - 2)^2 - 7`
Vì $(x-2)^2 \geqslant 0 ∀ x$
$→ 2 (x-2)^2 \geqslant 0 x$
$→ 2 (x - 2)^2 - 7 \geqslant -7$
$→ A \geqslant -7$
`-> min A = -7`
Dấy "`=`" xảy ra khi :
`⇔ x-2=0⇔x=2`
Vậy `min A = -7 ⇔x=2`
$\\$
`b,`
`B = -5x^2 - 4x + 1`
`⇔ B = -5x^2 - 4x - 20/25 + 45/25`
`⇔ B = -5x^2 - 4x - 20/25 + 9/5`
`⇔ B = -5 (x^2 + 4/5x + 4/25) + 9/5`
`⇔ B = -5 [x^2 + 2 . 2/5x + (2/5)^2] + 9/5`
`⇔ B = -5 (x + 2/5)^2 + 9/5`
Vì `(x + 2/5)^2` $\geqslant 0 ∀ x$
`-> -5 (x + 2/5)^2` $\leqslant 0 ∀ x$
`-> -5 (x + 2/5)^2 + 9/5` $\leqslant$ `9/5`
`-> B` $\leqslant$ `9/5`
`-> max B = 9/5`
Dấu "`=`" xảy ra khi :
`⇔ x + 2/5 =0 ⇔ x = (-2)/5`
Vậy `max B = 9/5 ⇔ x = (-2)/5`
