|x-1|>= 7

Lösung

∣ x - 1 ∣ \geq 7

x \leq - 6 or x \geq 8

Intervall-Notation

(- \infty, - 6] \cup [8, \infty)

Schritte zur Lösung

∣ x - 1 ∣ \geq 7

Wende Absoluteregel an: Wenn

∣ u ∣ \geq a, a > 0

dann

u \leq - a

u \geq a

x - 1 \leq - 7 or x - 1 \geq 7

x - 1 \leq - 7 or x - 1 \geq 7

x \leq - 6 or x \geq 8

s im pl i f y \frac{( x - y ) ^{2} + x ^{2}}{x y}

∣ 2 - 3 x ∣ \geq - 6

2 a^{2} + 2 a = 4

x^{2} = 3 6^{2}

13 x^{2} + 1 = 6 x