1/(3(x-1))< 1/4 (x+2)< 1/(6(x+4))

Lösung

\frac{1}{3 ( x - 1 )} < \frac{1}{4} (x + 2) < \frac{1}{6 ( x + 4 )}

\frac{- 3 - 129}{6} < x < \frac{15 - 9}{3}

Intervall-Notation

(\frac{- 3 - 129}{6}, \frac{15 - 9}{3})

Dezimale

- 2.39296 \dots < x < - 1.70900 \dots

Schritte zur Lösung

\frac{1}{3 ( x - 1 )} < \frac{1}{4} (x + 2) < \frac{1}{6 ( x + 4 )}

Wenn

a < u < b

dann

a < u and u < b

\frac{1}{3 ( x - 1 )} < \frac{1}{4} (x + 2) and \frac{1}{4} (x + 2) < \frac{1}{6 ( x + 4 )}

\frac{1}{3 ( x - 1 )} < \frac{1}{4} (x + 2) : \frac{- 3 - 129}{6} < x < 1 or x > \frac{129 - 3}{6}

\frac{1}{4} (x + 2) < \frac{1}{6 ( x + 4 )} : x < - \frac{9 + 15}{3} or - 4 < x < \frac{15 - 9}{3}

Kombiniere die Bereiche

(\frac{- 3 - 129}{6} < x < 1 or x > \frac{129 - 3}{6}) and (x < - \frac{9 + 15}{3} or - 4 < x < \frac{15 - 9}{3})

Füge die sich überschneidenden Intervalle zusammen

\frac{- 3 - 129}{6} < x < \frac{15 - 9}{3}

x + 1 \leq x^{2} - 3 x + 4 < 14 x - 56

- 2 < 2 x + 2 \leq 8

2 \leq x^{4} \leq 16

- 3 < 11 - 2 x < 1

- 12 < 3 x + 6 < 18