((x^2+6x+4))/((3x^2-9x+4))<= 0

Lösung

\frac{( x ^{2} + 6 x + 4 )}{( 3 x ^{2} - 9 x + 4 )} \leq 0

- 3 - 5 \leq x \leq - 3 + 5 or \frac{9 - 33}{6} < x < \frac{9 + 33}{6}

Intervall-Notation

[- 3 - 5, - 3 + 5] \cup (\frac{9 - 33}{6}, \frac{9 + 33}{6})

Dezimale

- 5.23606 \dots \leq x \leq - 0.76393 \dots or 0.54257 \dots < x < 2.45742 \dots

Schritte zur Lösung

\frac{x ^{2} + 6 x + 4}{3 x ^{2} - 9 x + 4} \leq 0

Faktor

x^{2} + 6 x + 4 : (x + 3 - 5) (x + 3 + 5)

\frac{( x + 3 - 5 ) ( x + 3 + 5 )}{3 x ^{2} - 9 x + 4} \leq 0

Faktor

3 x^{2} - 9 x + 4 : (x - \frac{9 + 33}{6}) (x - \frac{9 - 33}{6})

\frac{( x + 3 - 5 ) ( x + 3 + 5 )}{( x - \frac{9 + 33}{6} ) ( x - \frac{9 - 33}{6} )} \leq 0

Identifiziere die Intervalle

- 3 - 5 \leq x \leq - 3 + 5 or \frac{9 - 33}{6} < x < \frac{9 + 33}{6}

(x - 2) (x + 7) < 0

(x - 1) - 7 > (x - 2)^{2}

6 x \cdot (- 5) x - 3 x^{2} - 2 x + 6 < 4

\frac{( x + 1 )}{( x ^{2} - x - 12 )} > 2

\frac{( 4 x - 1 )}{( 3 x + 1 )} > 0