|2x-3|>= 7

Lösung

∣ 2 x - 3 ∣ \geq 7

x \leq - 2 or x \geq 5

Intervall-Notation

(- \infty, - 2] \cup [5, \infty)

Schritte zur Lösung

∣ 2 x - 3 ∣ \geq 7

Wende Absoluteregel an: Wenn

∣ u ∣ \geq a, a > 0

dann

u \leq - a

u \geq a

2 x - 3 \leq - 7 or 2 x - 3 \geq 7

2 x - 3 \leq - 7 or 2 x - 3 \geq 7

x \leq - 2 or x \geq 5

x^{2} - 11 x + 24 > 0

f a c t or n^{2} + 4 n - 45

x^{2} = 11 x - 28

\frac{2 - 3 x}{1 + x} > 1

4 x^{2} - 6 x + 5 = 0