|x-1|>0

Lösung

∣ x - 1 ∣ > 0

x < 1 or x > 1

Intervall-Notation

(- \infty, 1) \cup (1, \infty)

Schritte zur Lösung

∣ x - 1 ∣ > 0

Wende Absoluteregel an: Wenn

∣ u ∣ > 0

dann

u < 0

u > 0

x - 1 < 0 or x - 1 > 0

x - 1 < 0 or x - 1 > 0

x < 1 or x > 1

3 \geq ∣ q - 4 ∣

420 = 2 (r + 10)

f a c t or [(a - b)^{2} - c^{2}]

x < \frac{1}{3} or x > - \frac{1}{3}

6^{2} + 7^{2} = c^{2}