5x^3-17x^2+17x-5<= 0

Lösung

5 x^{3} - 17 x^{2} + 17 x - 5 \leq 0

x \leq \frac{6 - 11}{5} or 1 \leq x \leq \frac{6 + 11}{5}

Intervall-Notation

(- \infty, \frac{6 - 11}{5}] \cup [1, \frac{6 + 11}{5}]

Dezimale

x \leq 0.53667 \dots or 1 \leq x \leq 1.86332 \dots

Schritte zur Lösung

5 x^{3} - 17 x^{2} + 17 x - 5 \leq 0

Faktor

5 x^{3} - 17 x^{2} + 17 x - 5 : (x - 1) (x - \frac{6 + 11}{5}) (x - \frac{6 - 11}{5})

(x - 1) (x - \frac{6 + 11}{5}) (x - \frac{6 - 11}{5}) \leq 0

Identifiziere die Intervalle

x \leq \frac{6 - 11}{5} or 1 \leq x \leq \frac{6 + 11}{5}

(x + 2)^{2} - (x - 2)^{2} > 8 x - 2

\frac{( x ^{3} + 1 )}{( 7 x ^{2} - 1 )} \leq 0

\frac{1}{( x ^{2} )} \leq 4

\frac{( 3 x - x ^{4} )}{( ( x + 4 ) ^{2} )} \geq 0

25 x^{2} - 40 x + 16 < 0