((5x+3)(3x-2))/((x^3-7)^2)>= 0

Lösung

\frac{( 5 x + 3 ) ( 3 x - 2 )}{( x ^{3} - 7 ) ^{2}} \geq 0

x \leq - \frac{3}{5} or \frac{2}{3} \leq x < 37 or x > 37

Intervall-Notation

(- \infty, - \frac{3}{5}] \cup [\frac{2}{3}, 37) \cup (37, \infty)

Dezimale

x \leq - 0.6 or 0.66666 \dots \leq x < 1.91293 \dots or x > 1.91293 \dots

Schritte zur Lösung

\frac{( 5 x + 3 ) ( 3 x - 2 )}{( x ^{3} - 7 ) ^{2}} \geq 0

Faktor

\frac{( 5 x + 3 ) ( 3 x - 2 )}{( x ^{3} - 7 ) ^{2}} : \frac{( 5 x + 3 ) ( 3 x - 2 )}{( x - 3 7 ) ^{2} ( x ^{2} + 3 7 x + 7 ^{\frac{2}{3}} ) ^{2}}

\frac{( 5 x + 3 ) ( 3 x - 2 )}{( x - 3 7 ) ^{2} ( x ^{2} + 3 7 x + 7 ^{\frac{2}{3}} ) ^{2}} \geq 0

Identifiziere die Intervalle

x \leq - \frac{3}{5} or \frac{2}{3} \leq x < 37 or x > 37

3 x^{2} - 15 x + 12 = 0

(5 x - 3) \cdot (4 x + 7) = 0

8 x \frac{2}{3} - 43 x \frac{1}{3} + 15 = 0

2 g > - 8

f a c t or 15 q^{3} + 40 q^{2} + 3 q + 8