|x-1|>3

Lösung

∣ x - 1 ∣ > 3

x < - 2 or x > 4

Intervall-Notation

(- \infty, - 2) \cup (4, \infty)

Schritte zur Lösung

∣ x - 1 ∣ > 3

Wende Absoluteregel an: Wenn

∣ u ∣ > a, a > 0

dann

u < - a

u > a

x - 1 < - 3 or x - 1 > 3

x - 1 < - 3 or x - 1 > 3

x < - 2 or x > 4

3 t - \frac{1}{4} = 0.75

x - \frac{3}{3} + 9 x = 7 x - 9 + (- 7 + 7)

∣ 1 - x ∣ \leq 2

(q - 1)^{2} = 0.75

3 (x + 1) + 4 = 2 (x - 1) + x