A = x2 + x + 1 = ( x2 + x + 1/4 ) + 3/4 = ( x + 1/2 )2 + 3/4 ≥ 3/4 ∀ x
Dấu "=" xảy ra khi x = -1/2
=> MinA = 3/4 <=> x = -1/2
B = -x2 - 4x + 12 = -( x2 + 4x + 4 ) + 16 = -( x + 2 )2 + 16 ≤ 16 ∀ x
Dấu "=" xảy ra khi x = -2
=> MaxB = 16 <=> x = -2
C = \(\frac{5}{x^2+6}\)
Ta có : x2 + 6 ≥ 6 ∀ x
<=> \(\frac{1}{x^2+6}\le\frac{1}{6}\forall x\)
<=> \(\frac{5}{x^2+6}\le\frac{5}{6}\forall x\)
Dấu "=" xảy ra khi x = 0
=> MaxC = 5/6 <=> x = 0