Ta có:
a) A = x2 + 6x + 10 = (x2 + 6x + 9) + 1 = (x + 3)2 + 1 \(\ge\)1 \(\forall\)x
Dấu "=" xảy ra <=> x + 3 = 0 <=> x = -3
Vậy MinA = 1 <=> x = -3
b) B = 4x2 - 12x + 13 = 4(x2 - 3x + 9/4) + 4 = 4(x - 3/2)2 + 4 \(\ge\)4 \(\forall\)x
Dấu "=" xảy ra <=> x - 3/2 = 0 <=> x = 3/2
Vậy MinB = 4 <=> x = 3/2