|(x-3)/2 |>= 5

Lösung

\frac{x - 3}{2} \geq 5

x \leq - 7 or x \geq 13

Intervall-Notation

(- \infty, - 7] \cup [13, \infty)

Schritte zur Lösung

\frac{x - 3}{2} \geq 5

Wende Absoluteregel an: Wenn

∣ u ∣ \geq a, a > 0

dann

u \leq - a

u \geq a

\frac{x - 3}{2} \leq - 5 or \frac{x - 3}{2} \geq 5

\frac{x - 3}{2} \leq - 5 or \frac{x - 3}{2} \geq 5

x \leq - 7 or x \geq 13

x^{3} - 3 x^{2} = 0

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

\frac{1}{4} m + 9 = - 16

1 \leq x \leq 2

x^{3} > 9 x