1/(x+1)+1/(x+8)<= 1/(x+9)

Lösung

\frac{1}{x + 1} + \frac{1}{x + 8} \leq \frac{1}{x + 9}

x \leq - 9 - 22 or - 9 < x < - 8 or - 9 + 22 \leq x < - 1

Intervall-Notation

(- \infty, - 9 - 22] \cup (- 9, - 8) \cup [- 9 + 22, - 1)

Dezimale

x \leq - 11.82842 \dots or - 9 < x < - 8 or - 6.17157 \dots \leq x < - 1

Schritte zur Lösung

\frac{1}{x + 1} + \frac{1}{x + 8} \leq \frac{1}{x + 9}

Rewrite in standard form

\frac{x ^{2} + 18 x + 73}{( x + 1 ) ( x + 8 ) ( x + 9 )} \leq 0

Faktor

x^{2} + 18 x + 73 : (x + 9 - 22) (x + 9 + 22)

\frac{( x + 9 - 2 2 ) ( x + 9 + 2 2 )}{( x + 1 ) ( x + 8 ) ( x + 9 )} \leq 0

Identifiziere die Intervalle

x \leq - 9 - 22 or - 9 < x < - 8 or - 9 + 22 \leq x < - 1

14 r - 11 > - 227

s im pl i f y 5108

(x - 5) (x + 2)^{2} > 0

s im pl i f y \frac{10 !}{0 ! ( 10 - 0 ) !}

s im pl i f y \frac{5}{6} \cdot 2