| x/((x^2-x-1))|<2

Lösung

\frac{x}{( x ^{2} - x - 1 )} < 2

x < \frac{1 - 17}{4} or - \frac{1}{2} < x < \frac{1 + 17}{4} or x > 2

Intervall-Notation

(- \infty, \frac{1 - 17}{4}) \cup (- \frac{1}{2}, \frac{1 + 17}{4}) \cup (2, \infty)

Dezimale

x < - 0.78077 \dots or - 0.5 < x < 1.28077 \dots or x > 2

Schritte zur Lösung

\frac{x}{x ^{2} - x - 1} < 2

Wende Absoluteregel an: Wenn

∣ u ∣ < a, a > 0

dann

- a < u < a

- 2 < \frac{x}{x ^{2} - x - 1} < 2

\frac{x}{x ^{2} - x - 1} > - 2 and \frac{x}{x ^{2} - x - 1} < 2

(x < \frac{1 - 17}{4} or \frac{1 - 5}{2} < x < \frac{1 + 17}{4} or x > \frac{1 + 5}{2}) and (x < \frac{1 - 5}{2} or - \frac{1}{2} < x < \frac{1 + 5}{2} or x > 2)

Füge die sich überschneidenden Intervalle zusammen

x < \frac{1 - 17}{4} or - \frac{1}{2} < x < \frac{1 + 17}{4} or x > 2

∣ 4 x - 3 ∣ + 2 < 10

\frac{2}{∣ x - 4 ∣} > 1

lo g_{2} (∣ x - 2 ∣) > lo g_{2} (∣ x + 4 ∣)

∣ x + 1 ∣^{2} + 2 ∣ x + 2 ∣ \geq 2

x^{2} - 2 x + ∣ x - 4 ∣ < x^{2} - 3 x + 4