|2x+1|>0

Lösung

∣ 2 x + 1 ∣ > 0

x < - \frac{1}{2} or x > - \frac{1}{2}

Intervall-Notation

(- \infty, - \frac{1}{2}) \cup (- \frac{1}{2}, \infty)

Dezimale

x < - 0.5 or x > - 0.5

Schritte zur Lösung

∣ 2 x + 1 ∣ > 0

Wende Absoluteregel an: Wenn

∣ u ∣ > 0

dann

u < 0

u > 0

2 x + 1 < 0 or 2 x + 1 > 0

2 x + 1 < 0 or 2 x + 1 > 0

x < - \frac{1}{2} or x > - \frac{1}{2}

2 x \geq 16

∣ x - 2 ∣ \geq 3

\frac{3 x}{5} - \frac{x}{3} = - 3

- 5 w + 6 = - 12 w + 62

f a c t or 3 (x + 1) - 12