|x-7|>6

Lösung

∣ x - 7 ∣ > 6

x < 1 or x > 13

Intervall-Notation

(- \infty, 1) \cup (13, \infty)

Schritte zur Lösung

∣ x - 7 ∣ > 6

Wende Absoluteregel an: Wenn

∣ u ∣ > a, a > 0

dann

u < - a

u > a

x - 7 < - 6 or x - 7 > 6

x - 7 < - 6 or x - 7 > 6

x < 1 or x > 13

3 (x - 5) - 3 = 6

lo g_{8} (\frac{1}{64}) = - 2

x - 4 \geq - 1

∣ 5 x - 2 ∣ = 8

\frac{3}{x + 5} \geq 2