(x^2-4x-12)/(x-2)<= 0

Lösung

\frac{x ^{2} - 4 x - 12}{x - 2} \leq 0

x \leq - 2 or 2 < x \leq 6

Intervall-Notation

(- \infty, - 2] \cup (2, 6]

Schritte zur Lösung

\frac{x ^{2} - 4 x - 12}{x - 2} \leq 0

Faktorisiere

\frac{x ^{2} - 4 x - 12}{x - 2} : \frac{( x + 2 ) ( x - 6 )}{x - 2}

\frac{( x + 2 ) ( x - 6 )}{x - 2} \leq 0

Identifiziere die Intervalle

x \leq - 2 or 2 < x \leq 6

3 y^{2} = 21 y

4 x^{\frac{3}{2}}

- x^{3} + 7 x^{2} + 4 x > 28

w^{2} + 14 w - 1800 = 0

(2 x - 5)^{2} = 49