((2x^2+3x+5))/((2x^2+8x+1))<= 0

Lösung

\frac{( 2 x ^{2} + 3 x + 5 )}{( 2 x ^{2} + 8 x + 1 )} \leq 0

- \frac{4 + 14}{2} < x < \frac{14 - 4}{2}

Intervall-Notation

(- \frac{4 + 14}{2}, \frac{14 - 4}{2})

Dezimale

- 3.87082 \dots < x < - 0.12917 \dots

Schritte zur Lösung

\frac{2 x ^{2} + 3 x + 5}{2 x ^{2} + 8 x + 1} \leq 0

Faktor

2 x^{2} + 8 x + 1 : (x - \frac{- 4 + 14}{2}) (x + \frac{4 + 14}{2})

\frac{2 x ^{2} + 3 x + 5}{( x - \frac{- 4 + 14}{2} ) ( x + \frac{4 + 14}{2} )} \leq 0

Identifiziere die Intervalle

- \frac{4 + 14}{2} < x < \frac{14 - 4}{2}

\frac{( x ^{2} - 3 x + 7 )}{( x ^{2} + 1 )} < 1

\frac{13}{( x - 3 )} < 0

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

(t + 2)^{2} \cdot t (t + 1) > t^{2} \cdot (t + 1)^{2}

\frac{lo g _{10} ( 1 )}{2 ( 5 - 2 x )} < 0