(4x)/(x^2-9)<=-1

Lösung

\frac{4 x}{x ^{2} - 9} \leq - 1

- 2 - 13 \leq x < - 3 or - 2 + 13 \leq x < 3

Intervall-Notation

[- 2 - 13, - 3) \cup [- 2 + 13, 3)

Dezimale

- 5.60555 \dots \leq x < - 3 or 1.60555 \dots \leq x < 3

Schritte zur Lösung

\frac{4 x}{x ^{2} - 9} \leq - 1

Rewrite in standard form

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

Faktor

\frac{4 x + x ^{2} - 9}{x ^{2} - 9} : \frac{( x + 2 - 13 ) ( x + 2 + 13 )}{( x + 3 ) ( x - 3 )}

\frac{( x + 2 - 13 ) ( x + 2 + 13 )}{( x + 3 ) ( x - 3 )} \leq 0

Identifiziere die Intervalle

- 2 - 13 \leq x < - 3 or - 2 + 13 \leq x < 3

\frac{2 x - 3}{4} - \frac{3 + 2 x}{9} = 2 - \frac{x + 1}{6}

x + \frac{5}{3} \leq 2 x + \frac{1}{3} \leq 3 x + \frac{4}{3}

f a c t or 648 a + 1029 a^{2}

54 + x = 180

- 5.2 + 57.9 = 15.9 + 1.8 x