(((3-1)/(3x))}{((3+1)/2)}>= ((\frac{3x-5)/2))/(((1-2)/3))

Lösung

\frac{( \frac{3 - 1}{3 x} )}{( \frac{3 + 1}{2} )} \geq \frac{( \frac{3 x - 5}{2} )}{( \frac{1 - 2}{3} )}

0 < x \leq \frac{15 - 201}{18} or x \geq \frac{15 + 201}{18}

Intervall-Notation

(0, \frac{15 - 201}{18}] \cup [\frac{15 + 201}{18}, \infty)

Dezimale

0 < x \leq 0.04569 \dots or x \geq 1.62096 \dots

Schritte zur Lösung

\frac{\frac{3 - 1}{3 x}}{\frac{3 + 1}{2}} \geq \frac{\frac{3 x - 5}{2}}{\frac{1 - 2}{3}}

Rewrite in standard form

\frac{4 + 18 x ( 3 x - 5 )}{x} \geq 0

Faktor

\frac{4 + 18 x ( 3 x - 5 )}{x} : \frac{2 ( x - \frac{15 + 201}{18} ) ( x - \frac{15 - 201}{18} )}{x}

\frac{2 ( x - \frac{15 + 201}{18} ) ( x - \frac{15 - 201}{18} )}{x} \geq 0

Teile beide Seiten durch

2

\frac{\frac{2 ( x - \frac{15 + 201}{18} ) ( x - \frac{15 - 201}{18} )}{x}}{2} \geq \frac{0}{2}

Vereinfache

\frac{( x - \frac{15 + 201}{18} ) ( x - \frac{15 - 201}{18} )}{x} \geq 0

Identifiziere die Intervalle

0 < x \leq \frac{15 - 201}{18} or x \geq \frac{15 + 201}{18}

3 x + 4 = 9 x - 8

2 x^{2} + 8 x = - 1

f a c t or 3 x^{2} - 19 x y + 20 y^{2}

s im pl i f y \frac{5 - 4 2}{4 - 3 2}

f a c t or 7 y^{2} - 5 y - 2