|2x-4|>8

Lösung

∣ 2 x - 4 ∣ > 8

x < - 2 or x > 6

Intervall-Notation

(- \infty, - 2) \cup (6, \infty)

Schritte zur Lösung

∣ 2 x - 4 ∣ > 8

Wende Absoluteregel an: Wenn

∣ u ∣ > a, a > 0

dann

u < - a

u > a

2 x - 4 < - 8 or 2 x - 4 > 8

2 x - 4 < - 8 or 2 x - 4 > 8

x < - 2 or x > 6

6 x^{2} - 17 x - 28 = 0

x^{3} + 7 \geq 0

x^{2} + 7 x + 12 = 0

\frac{2}{3} (6 - 9 x) = 16

f a c t or 8 x^{3} - 64 x^{2} + x - 8