|x+4|>= 2

Lösung

∣ x + 4 ∣ \geq 2

x \leq - 6 or x \geq - 2

Intervall-Notation

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

Schritte zur Lösung

∣ x + 4 ∣ \geq 2

Wende Absoluteregel an: Wenn

∣ u ∣ \geq a, a > 0

dann

u \leq - a

u \geq a

x + 4 \leq - 2 or x + 4 \geq 2

x + 4 \leq - 2 or x + 4 \geq 2

x \leq - 6 or x \geq - 2

2 a^{2} + 5 a - 5 = 0

0.0014 > 0.000012

2 - x = - 3 + x

\frac{4}{x - 1} \geq 0

f a c t or 9 x^{2} - 100