|5-x|>= 3

Lösung

∣ 5 - x ∣ \geq 3

x \leq 2 or x \geq 8

Intervall-Notation

(- \infty, 2] \cup [8, \infty)

Schritte zur Lösung

∣ 5 - x ∣ \geq 3

Wende Absoluteregel an: Wenn

∣ u ∣ \geq a, a > 0

dann

u \leq - a

u \geq a

5 - x \leq - 3 or 5 - x \geq 3

5 - x \leq - 3 or 5 - x \geq 3

x \leq 2 or x \geq 8

324 k^{2} + 36 k + 1 = 0

x + 10 = 2 x + 3

(x - 4) (x - 2) (x + 7) (x + 6) \leq 0

7 v + 23 = 4 v + 8 v - 22

x^{2} - 48 x + 576 = 0