|x+5|>= 5

Lösung

∣ x + 5 ∣ \geq 5

x \leq - 10 or x \geq 0

Intervall-Notation

(- \infty, - 10] \cup [0, \infty)

Schritte zur Lösung

∣ x + 5 ∣ \geq 5

Wende Absoluteregel an: Wenn

∣ u ∣ \geq a, a > 0

dann

u \leq - a

u \geq a

x + 5 \leq - 5 or x + 5 \geq 5

x + 5 \leq - 5 or x + 5 \geq 5

x \leq - 10 or x \geq 0

- ∣ u - 420 ∣ - 6 \geq - 21

0 = - 16 t^{2} + 6 t + 5

f a c t or b^{2} + 8 b + 7

3 (x - 1) = 6 x

\frac{2 x + 2}{3} \leq 2