|x-2|>= 5

Lösung

∣ x - 2 ∣ \geq 5

x \leq - 3 or x \geq 7

Intervall-Notation

(- \infty, - 3] \cup [7, \infty)

Schritte zur Lösung

∣ x - 2 ∣ \geq 5

Wende Absoluteregel an: Wenn

∣ u ∣ \geq a, a > 0

dann

u \leq - a

u \geq a

x - 2 \leq - 5 or x - 2 \geq 5

x - 2 \leq - 5 or x - 2 \geq 5

x \leq - 3 or x \geq 7

- 4 m \geq - 4

2 x^{2} - 4 x - 70 = 0

f a c t or x^{2 - x + 1}

(x - 1) (x + 4) (4 - x) < 0

7 x^{2} = 112