(x^2-4x+4)(1-2x)(x^2+3x-1)x>0

Lösung

(x^{2} - 4 x + 4) (1 - 2 x) (x^{2} + 3 x - 1) x > 0

\frac{- 3 - 13}{2} < x < 0 or \frac{13 - 3}{2} < x < \frac{1}{2}

Intervall-Notation

(\frac{- 3 - 13}{2}, 0) \cup (\frac{13 - 3}{2}, \frac{1}{2})

Dezimale

- 3.30277 \dots < x < 0 or 0.30277 \dots < x < 0.5

Schritte zur Lösung

(x^{2} - 4 x + 4) (1 - 2 x) (x^{2} + 3 x - 1) x > 0

Faktor

(x^{2} - 4 x + 4) (1 - 2 x) (x^{2} + 3 x - 1) x : - x (x - 2)^{2} (2 x - 1) (x - \frac{- 3 + 13}{2}) (x - \frac{- 3 - 13}{2})

- x (x - 2)^{2} (2 x - 1) (x - \frac{- 3 + 13}{2}) (x - \frac{- 3 - 13}{2}) > 0

Multipliziere beide Seiten mit

- 1

(drehe die Ungleichung um)

(- x (x - 2)^{2} (2 x - 1) (x - \frac{- 3 + 13}{2}) (x - \frac{- 3 - 13}{2})) (- 1) < 0 \cdot (- 1)

Vereinfache

x (x - 2)^{2} (2 x - 1) (x - \frac{- 3 + 13}{2}) (x - \frac{- 3 - 13}{2}) < 0

Identifiziere die Intervalle

\frac{- 3 - 13}{2} < x < 0 or \frac{13 - 3}{2} < x < \frac{1}{2}

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

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

\frac{( x ^{2} - 5 ) ( x ^{2} - 4 )}{( x - 1 )} < 0

3 x^{2} - 8 x + 2 < 0

x (x - 1) (x - 2) (x - 3) > 0