m-2<-8\lor m/8 >1

Lösung

m - 2 < - 8 or \frac{m}{8} > 1

m < - 6 or m > 8

Intervall-Notation

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

Schritte zur Lösung

m - 2 < - 8 or \frac{m}{8} > 1

m - 2 < - 8 : m < - 6

\frac{m}{8} > 1 : m > 8

Kombiniere die Bereiche

m < - 6 or m > 8

(x - 3) (x + 1) > 0

- 6 x > - 12

4 a < 12

x^{2} - x - 56 < 0

f a c t or \frac{- 2 x + 6}{x ^{2} - 9}