|x^2+x-1|=1

Lösung

x^{2} + x - 1 = 1

x = - 2 or x = - 1 or x = 0 or x = 1

Schritte zur Lösung

x^{2} + x - 1 = 1

Wende Absoluteregel an: Wenn

∣ u ∣ = a, a > 0

dann

u = a

u = - a

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

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

(x = - 1 or x = 0) or (x = - 2 or x = 1)

x = - 2 or x = - 1 or x = 0 or x = 1

\frac{2}{3} x + \frac{3}{2} (x - 5) \leq x

f a c t or x^{4} - 2 x^{3} + 27 x - 54

x (2 x + 4) = 0

8 - 2 x \leq 2

\frac{x ^{2} - 16}{x ^{2} - 36} \leq 0