`A=x^2+4x+7=x^2+4x+4+3=(x+2)^2+3`
$\text{Vì $(x+2)^2$ ≥ 0 ∀ x}$
`⇒ (x+2)^2+3≥ 3 ∀ x`
$\text{Dấu "=" xảy ra}$
`⇔(x+2)^2=0⇔x=-2`
$\text{Vậy Min A = 3 khi x = -2}$
`B=2x^2-8x+3=x^2-4x+4+x^2-4x+4-5`
`=(x-2)^2+(x-2)^2-5=2(x-2)^2-5`
$\text{Vì $(x-2)^2$ ≥ 0 ∀ x}$
`⇒2(x-2)^2-5≥-5∀x`
$\text{Dấu "=" xảy ra}$
`⇔(x-2)^2=0⇔x=2`
$\text{Vậy Min B = -5 khi x = 2}$
