A = x2 - x + 1
= x2 - 2\(\dfrac{1}{2}\)x + \(\left(\dfrac{1}{2}\right)^2\)- \(\dfrac{1}{4}\) +1
= [ x2 - 2\(\dfrac{1}{2}\)x +\(\left(\dfrac{1}{2}\right)^2\)]+ ( \(-\dfrac{1}{4}\)+ 1 )
= (x - \(\dfrac{1}{2}\) )2 + \(\dfrac{3}{4}\)
Ta có :
(x - \(\dfrac{1}{2}\) )2≥ 0 ∀ x
(x - \(\dfrac{1}{2}\) )2 + \(\dfrac{3}{4}\) ≥ \(\dfrac{3}{4}\)
=> GTNN của (x - \(\dfrac{1}{2}\) )2 + \(\dfrac{3}{4}\) là \(\dfrac{3}{4}\) khi (x - \(\dfrac{1}{2}\) )2 = 0
x - \(\dfrac{1}{2}\) = 0
x = \(\dfrac{1}{2}\)
Vậy GTNN của đa thức trên là \(\dfrac{3}{4}\)khi x = \(\dfrac{1}{2}\)
