4)A = 25x² + 10x + 1
= ( 5x ) ² + 2 . 5x . 1 + 1
= ( 5x + 1 )²
Vì ( 5x + 1 )² ≥ 0 ∀x
Dấu "= " xảy ra khi :
5x + 1 = 0
⇔ 5x = -1
⇔ x = $\frac{-1}{5}$
Vậy A min = 0 khi x = $\frac{-1}{5}$
B = x² - x + 1
= x² - 2 .x . $\frac{1}{2}$ + ($\frac{1}{2}$ )² - $\frac{1}{4}$ + 1
= ( x - $\frac{1}{2}$ )² + $\frac{3}{4}$
Vì ( x - $\frac{1}{2}$ )² ≥ 0 ∀x nên ( x - $\frac{1}{2}$ )² + $\frac{3}{4}$ ≥ $\frac{3}{4}$ ∀x
Dấu "= " xảy ra khi :
( x - $\frac{1}{2}$ )² = 0
⇔ x = $\frac{1}{2}$
Vậy A min = $\frac{3}{4}$ khi x = $\frac{1}{2}$
C = 8x - x² + 5
= - x² + 8x + 5
= - ( x² - 8x - 5 )
= - ( x² - 2. x . 4 + 4² - 16 -5 )
= - [ ( x - 4 )² - 21 ]
= - ( x - 4 )² + 21
Vì ( x - 4 )² ≥ 0 ∀x ⇒ - ( x - 4 )² ≤ 0 ∀x nên - ( x - 4 )² + 21 ≤ 21 ∀x
Dấu "= " xảy ra khi :
x - 4 = 0
⇔ x = 4
Vậy C max = 0 khi x = 4
