|4x+40|<= 0

Lösung

∣ 4 x + 40 ∣ \leq 0

x = - 10

Intervall-Notation

x = - 10

Schritte zur Lösung

∣ 4 x + 40 ∣ \leq 0

Wende Absoluteregel an: Wenn

∣ u ∣ \leq 0

dann

u = 0

4 x + 40 = 0

4 x + 40 = 0 : x = - 10

x = - 10

s im pl i f y 6 (\frac{1}{3})^{0}

co m pl e t es q u a re x^{2} + 12 x

x^{2} > 25 x

6 < x + 3 < 8

9 = 12 - 3 x