|3x|>6

Lösung

∣ 3 x ∣ > 6

x < - 2 or x > 2

Intervall-Notation

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

Schritte zur Lösung

∣ 3 x ∣ > 6

Wende Absoluteregel an: Wenn

∣ u ∣ > a, a > 0

dann

u < - a

u > a

3 x < - 6 or 3 x > 6

3 x < - 6 or 3 x > 6

x < - 2 or x > 2

\frac{1}{x - 10} \leq \frac{x}{x + 2}

7 b - 15 = 5 b - 3

2 (4 x + 4) < 5

\frac{x}{7} - 4 = \frac{2}{3}

3 x^{2} - 10 x + 8 = 0