a,
(x-1)²+(x+2)²+2(x+2)(x-1)
=(x-1)²+2(x+2)(x-1)+(x+2)²
=(x-1+x+2)²
=(2x+1)²
b,
A=16x²-8x+2020
A=(16x²-8x+1)+2019
A=(4x-1)²+2019
Ta có: (4x-1)² ≥0; ∀x
⇒ (4x-1)²+2019 ≥2019
A ≥2019
Dấu bằng xảy ra ⇔(4x-1)²=0
⇔ 4x-1=0
⇔ 4x=1
⇔ x=$\frac{1}{4}$
Vậy GTNN A=2019 khi x=$\frac{1}{4}$.