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Beliebt Algebra >

|x+1||x+4|=1

Lösung

∣ x + 1 ∣ ∣ x + 4 ∣ = 1

Lösung

x = \frac{- 5 - 13}{2} or x = - \frac{5 + 5}{2} or x = - \frac{5 - 5}{2} or x = \frac{- 5 + 13}{2}

+1

Dezimale

x = - 4.30277 \dots, x = - 3.61803 \dots, x = - 1.38196 \dots, x = - 0.69722 \dots

Schritte zur Lösung

∣ x + 1 ∣ ∣ x + 4 ∣ = 1

Finde positive und negative Intervalle

x < - 4, - 4 \leq x < - 1, x \geq - 1

Löse die Ungleichung für jedes Intervall

x = \frac{- 5 - 13}{2} or (x = - \frac{5 + 5}{2} or x = - \frac{5 - 5}{2}) or x = \frac{- 5 + 13}{2}

x = \frac{- 5 - 13}{2} or x = - \frac{5 + 5}{2} or x = - \frac{5 - 5}{2} or x = \frac{- 5 + 13}{2}

Graph

Beliebte Beispiele

x^{2} - 7 ∣ x ∣ - 8 = 0

|x^2-x+1|=|x^2-x-1|

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

x^{2} - 5 x + 2 = 4

∣ x ∣ + ∣ x - 2 ∣ = 2

x^{2} - ∣ x ∣ - 2 = 0