(x+9)(x-3)>0

Lösung

(x + 9) (x - 3) > 0

x < - 9 or x > 3

Intervall-Notation

(- \infty, - 9) \cup (3, \infty)

Schritte zur Lösung

(x + 9) (x - 3) > 0

Identifiziere die Intervalle

x < - 9 or x > 3

(x - 1)^{2} - 11 < (x - 2)^{2}

4 x^{2} + 2 > 0

x^{2} \geq \frac{1}{2}

x (x + 5)^{2} (x - 3) < 0

\frac{( [ x ] - 1 )}{( [ x ] + 3 )} > 1 - [x]