a, A=(x-3)2+7/4
ta có: (x-3)2 ≥ 0 ∀x
⇒(x-3)2+7/4 ≥ 7/4 ∀x
Dấu "=" xảy ra ⇔ (x-3)2 ⇔x-3=0 ⇔x=3
Vậy Amin=7/4 ⇔ x=3
b,B= \(\frac{7}{4}-\left|\frac{x+2}{3}\right|\)
Ta có: \(\left|\frac{x+2}{3}\right|\) ≥ 0 ∀x
⇒-\(\left|\frac{x+2}{3}\right|\) ≤ 0 ∀x
⇒\(\frac{7}{4}-\left|\frac{x+2}{3}\right|\) ≤ \(\frac{7}{4}\) ∀x
Dấu "=" xảy ra ⇔\(\left|\frac{x+2}{3}\right|\) =0
⇔ \(\frac{x+2}{3}\) =0 ⇔x=-2
Vậy Bmax = 7/4 ⇔ x=-2