(3x^2+2x-8)/(x^2-2x)<0

Lösung

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

- 2 < x < 0 or \frac{4}{3} < x < 2

Intervall-Notation

(- 2, 0) \cup (\frac{4}{3}, 2)

Dezimale

- 2 < x < 0 or 1.33333 \dots < x < 2

Schritte zur Lösung

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

Faktorisiere

\frac{3 x ^{2} + 2 x - 8}{x ^{2} - 2 x} : \frac{( 3 x - 4 ) ( x + 2 )}{x ( x - 2 )}

\frac{( 3 x - 4 ) ( x + 2 )}{x ( x - 2 )} < 0

Identifiziere die Intervalle

- 2 < x < 0 or \frac{4}{3} < x < 2

x^{4} - 5 x^{2} + 4 < 0

\frac{( x - 1 )}{( x + 1 )} < x

\frac{4}{( x - 1 )} \geq \frac{1}{( 2 x - 1 )}

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

\frac{( ( x - 1 ) ^{2} \cdot ( x + 1 ) ^{3} )}{( x ^{4} \cdot ( x - 2 ) )} \leq 0