de

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Beliebt Algebra >

(x^2-3x+2)(x^3-3x^2)(4-x^2)>= 0

Lösung

(x^{2} - 3 x + 2) (x^{3} - 3 x^{2}) (4 - x^{2}) \geq 0

Lösung

x \leq - 2 or x = 0 or 1 \leq x \leq 3

+1

Intervall-Notation

(- \infty, - 2] \cup x = 0 \cup [1, 3]

Schritte zur Lösung

(x^{2} - 3 x + 2) (x^{3} - 3 x^{2}) (4 - x^{2}) \geq 0

Faktorisiere

(x^{2} - 3 x + 2) (x^{3} - 3 x^{2}) (4 - x^{2}) : - x^{2} (x - 2)^{2} (x - 1) (x - 3) (x + 2)

- x^{2} (x - 2)^{2} (x - 1) (x - 3) (x + 2) \geq 0

Multipliziere beide Seiten mit

- 1

(drehe die Ungleichung um)

(- x^{2} (x - 2)^{2} (x - 1) (x - 3) (x + 2)) (- 1) \leq 0 \cdot (- 1)

Vereinfache

x^{2} (x - 2)^{2} (x - 1) (x - 3) (x + 2) \leq 0

Identifiziere die Intervalle

x \leq - 2 or x = 0 or 1 \leq x \leq 3

Graph

Beliebte Beispiele

((x^2-5x+4))/((x-4))>= 0

\frac{( x ^{2} - 5 x + 4 )}{( x - 4 )} \geq 0

x^{3} < 7 x + 6

((x+2))/((x-3))<0

\frac{( x + 2 )}{( x - 3 )} < 0

(1+12)/(x^2)< 7/x

\frac{1 + 12}{x ^{2}} < \frac{7}{x}

x^2*(x+2)*(x-1)^3*(x^2+1)>0

x^{2} \cdot (x + 2) \cdot (x - 1)^{3} \cdot (x^{2} + 1) > 0