|x+5|>0

Lösung

∣ x + 5 ∣ > 0

x < - 5 or x > - 5

Intervall-Notation

(- \infty, - 5) \cup (- 5, \infty)

Schritte zur Lösung

∣ x + 5 ∣ > 0

Wende Absoluteregel an: Wenn

∣ u ∣ > 0

dann

u < 0

u > 0

x + 5 < 0 or x + 5 > 0

x + 5 < 0 or x + 5 > 0

x < - 5 or x > - 5

5 [3 (x + 2) + 4] + 2 x = 67

4 (y + 3)^{2} = 0

(2 x - 1)^{2} = 26

8 (h - 1) = 6 h + 4 + 2 h

\frac{x}{4} - \frac{x - 2}{8} = \frac{x - 9}{12}