2x-5<-11\lor 5x+1>= 6

Lösung

2 x - 5 < - 11 or 5 x + 1 \geq 6

x < - 3 or x \geq 1

Intervall-Notation

(- \infty, - 3) \cup [1, \infty)

Schritte zur Lösung

2 x - 5 < - 11 or 5 x + 1 \geq 6

2 x - 5 < - 11 : x < - 3

5 x + 1 \geq 6 : x \geq 1

Kombiniere die Bereiche

x < - 3 or x \geq 1

5 x^{2} = 14 x + 3

- 10 x < 80

\frac{13}{9} x + x = 88

x (x - 15) = 1 0^{2}

x^{3} + 2 x^{2} - x - 2 < 0