1<|((x+1))/((x+2))|<3

Lösung

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

x < - \frac{5}{2} or - \frac{7}{4} < x < - \frac{3}{2}

Intervall-Notation

(- \infty, - \frac{5}{2}) \cup (- \frac{7}{4}, - \frac{3}{2})

Dezimale

x < - 2.5 or - 1.75 < x < - 1.5

Schritte zur Lösung

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

Wenn

a < u < b

dann

a < u and u < b

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

1 < \frac{x + 1}{x + 2} : x < - 2 or - 2 < x < - \frac{3}{2}

\frac{x + 1}{x + 2} < 3 : x < - \frac{5}{2} or x > - \frac{7}{4}

Kombiniere die Bereiche

(x < - 2 or - 2 < x < - \frac{3}{2}) and (x < - \frac{5}{2} or x > - \frac{7}{4})

Füge die sich überschneidenden Intervalle zusammen

x < - \frac{5}{2} or - \frac{7}{4} < x < - \frac{3}{2}

- 1 \leq \frac{( x - 2 )}{( x + 7 )} \leq 1

\frac{- 25}{6} < \frac{1}{2} - 2 \frac{x}{3} < 2

\frac{- 1}{2} < \frac{x}{( x ^{2} - x - 1 )} < 2

- 1 < \frac{( x + 2 )}{( x - 1 )} \leq 1

x + 1 < 3 x - 5 < 4 x - 7