(x+1)/(2x^2-x-1)>= 0

Lösung

\frac{x + 1}{2 x ^{2} - x - 1} \geq 0

- 1 \leq x < - \frac{1}{2} or x > 1

Intervall-Notation

[- 1, - \frac{1}{2}) \cup (1, \infty)

Dezimale

- 1 \leq x < - 0.5 or x > 1

Schritte zur Lösung

\frac{x + 1}{2 x ^{2} - x - 1} \geq 0

Faktorisiere

\frac{x + 1}{2 x ^{2} - x - 1} : \frac{x + 1}{( 2 x + 1 ) ( x - 1 )}

\frac{x + 1}{( 2 x + 1 ) ( x - 1 )} \geq 0

Identifiziere die Intervalle

- 1 \leq x < - \frac{1}{2} or x > 1

∣ x + 5 ∣ \geq 2 x

36 = 25 + u

\frac{2 x - 8}{4} > 9

\frac{n}{2} - 1 \geq 1

8 n^{2} + n - 3 = - 3 n