| 1/(x-10)|<= | x/(x+2)|

Lösung

\frac{1}{x - 10} \leq \frac{x}{x + 2}

x < - 2 or - 2 < x \leq \frac{11 - 129}{2} or \frac{9 - 73}{2} \leq x \leq \frac{9 + 73}{2} or x \geq \frac{11 + 129}{2}

Intervall-Notation

(- \infty, - 2) \cup (- 2, \frac{11 - 129}{2}] \cup [\frac{9 - 73}{2}, \frac{9 + 73}{2}] \cup [\frac{11 + 129}{2}, \infty)

Dezimale

x < - 2 or - 2 < x \leq - 0.17890 \dots or 0.22799 \dots \leq x \leq 8.77200 \dots or x \geq 11.17890 \dots

Schritte zur Lösung

\frac{1}{x - 10} \leq \frac{x}{x + 2}

Finde positive und negative Intervalle

x < - 2, - 2 \leq x < 0, 0 \leq x < 10, x \geq 10

Löse die Ungleichung für jedes Intervall

x < - 2 or - 2 < x \leq \frac{11 - 129}{2} or \frac{9 - 73}{2} \leq x \leq \frac{9 + 73}{2} or x \geq \frac{11 + 129}{2}

7 b - 15 = 5 b - 3

2 (4 x + 4) < 5

\frac{x}{7} - 4 = \frac{2}{3}

3 x^{2} - 10 x + 8 = 0

s im pl i f y \frac{16 a ^{3} b ^{6} c ^{- 2}}{10 a b ^{- 2} c ^{3}}