$\text{4. A = |2x - $\frac{1}{3}$| - $\frac{7}{4}$ }$
Ta có: $\text{|2x - $\frac{1}{3}$| ≥ 0 ∀x }$
$\text{⇒ A = |2x - $\frac{1}{3}$| - $\frac{7}{4}$ ≥ $\frac{-7}{4}$ ∀x}$
$\text{A = $\frac{-7}{4}$ ⇔ |2x - $\frac{1}{3}$| = 0 }$
$\text{⇔ 2x - $\frac{1}{3}$ = 0 }$
$\text{⇔ 2x = $\frac{1}{3}$ }$
$\text{⇔ x = $\frac{1}{3}$ : 2 }$
$\text{⇔ x = $\frac{1}{3}$ . $\frac{1}{2}$ }$
$\text{⇔ x = $\frac{1}{6}$ }$
Vậy $\text{Amin = $\frac{-7}{4}$ ⇔ x = $\frac{1}{6}$ }$
- minimum <min>: tối thiểu, ít nhất
$\text{5. A = $\frac{9}{4}$ - $\frac{1}{4}$ |1 + 2x|}$
Ta có: $\text{|1 + 2x| ≥ 0 ∀x }$
$\text{⇒ $\frac{1}{4}$ |1 + 2x| ≥ 0 ∀x }$
$\text{⇒ -$\frac{1}{4}$ |1 + 2x| ≤ 0 ∀x }$
$\text{⇒ A = $\frac{9}{4}$ - $\frac{1}{4}$ |1 + 2x| ≤ $\frac{9}{4}$ ∀x}$
$\text{A = $\frac{9}{4}$ }$
$\text{⇔ |1 + 2x| = 0 }$
$\text{⇔ 1 + 2x = 0 }$
$\text{⇔ 2x = -1 }$
$\text{⇔ x = $\frac{-1}{2}$ }$
Vậy $\text{Amax = $\frac{9}{4}$ ⇔ x = $\frac{-1}{2}$ }$
- maximum <max>: tối đa, nhiều nhất
