|x+6|>= 8

Lösung

∣ x + 6 ∣ \geq 8

x \leq - 14 or x \geq 2

Intervall-Notation

(- \infty, - 14] \cup [2, \infty)

Schritte zur Lösung

∣ x + 6 ∣ \geq 8

Wende Absoluteregel an: Wenn

∣ u ∣ \geq a, a > 0

dann

u \leq - a

u \geq a

x + 6 \leq - 8 or x + 6 \geq 8

x + 6 \leq - 8 or x + 6 \geq 8

x \leq - 14 or x \geq 2

176 = 8 (1 + 3 x)

3 - 5 = 2 - 1

2 c - 3 = 2 (6 - c) + 7 c

s im pl i f y (1 + \frac{1}{3})^{3}

\frac{( x + 1 )}{( x + 5 )} < 2