1/((1+x))+1/((x-1))+2/((x+2))<0

Lösung

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

x < - 2 or - \frac{1 + 3}{2} < x < - 1 or \frac{3 - 1}{2} < x < 1

Intervall-Notation

(- \infty, - 2) \cup (- \frac{1 + 3}{2}, - 1) \cup (\frac{3 - 1}{2}, 1)

Dezimale

x < - 2 or - 1.36602 \dots < x < - 1 or 0.36602 \dots < x < 1

Schritte zur Lösung

\frac{1}{1 + x} + \frac{1}{x - 1} + \frac{2}{x + 2} < 0

Rewrite in standard form

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

Faktor

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

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

Teile beide Seiten durch

2

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

Vereinfache

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

Identifiziere die Intervalle

x < - 2 or - \frac{1 + 3}{2} < x < - 1 or \frac{3 - 1}{2} < x < 1

(x - 2)^{3} (x - 4)^{2} (x - 10) 17 > 0

x (x + 3) < 0

\frac{3 x ( x - 1 )}{5 x} \geq 0

x^{4} + 6 x \leq 6 x^{3} + x^{2}

(x - 1) (x - 5) > 4