x(3-x)(x-7)<0

Lösung

x (3 - x) (x - 7) < 0

0 < x < 3 or x > 7

Intervall-Notation

(0, 3) \cup (7, \infty)

Schritte zur Lösung

x (3 - x) (x - 7) < 0

Identifiziere die Intervalle

0 < x < 3 or x > 7

x^{2} - 1 + x^{2} - 16 = 2

- 3 x^{2} + 6 x - 3 = 0

4 x - 22 = - 2

\frac{j}{3} > 3

x^{2} + 2 x + 4 > 0