|u-2|>= 5

Lösung

∣ u - 2 ∣ \geq 5

u \leq - 3 or u \geq 7

Intervall-Notation

(- \infty, - 3] \cup [7, \infty)

Schritte zur Lösung

∣ u - 2 ∣ \geq 5

Wende Absoluteregel an: Wenn

∣ u ∣ \geq a, a > 0

dann

u \leq - a

u \geq a

u - 2 \leq - 5 or u - 2 \geq 5

u - 2 \leq - 5 or u - 2 \geq 5

u \leq - 3 or u \geq 7

- 32.8 \cdot 12 + 8.5 \cdot 9 \cdot 4.5 + x = 0

2 x^{3} + 6 \leq 5 x^{2} + x

4 - \frac{r}{3} = 5

f a c t or 3 x^{2} y - 2 x y

2 x + 3 x = 4 x - 3