faktorisieren 1/64 x^3-8/1000 y^3

Lösung

faktorisieren

\frac{1}{64} x^{3} - \frac{8}{1000} y^{3}

(\frac{x}{4} - \frac{y}{5}) (\frac{x ^{2}}{16} + \frac{x y}{20} + \frac{y ^{2}}{25})

Schritte zur Lösung

\frac{1}{64} x^{3} - \frac{8}{1000} y^{3}

Faktorisiere

\frac{x ^{3}}{64} - \frac{8}{1000} y^{3} : (\frac{x}{4} - \frac{1}{5} y) ((\frac{1}{4})^{2} x^{2} + \frac{1}{5} \cdot \frac{1}{4} x y + (\frac{1}{5})^{2} y^{2})

= (\frac{x}{4} - \frac{1}{5} y) ((\frac{1}{4})^{2} x^{2} + \frac{1}{5} \cdot \frac{1}{4} x y + (\frac{1}{5})^{2} y^{2})

Vereinfache

(\frac{x}{4} - \frac{1}{5} y) ((\frac{1}{4})^{2} x^{2} + \frac{1}{5} \cdot \frac{1}{4} x y + (\frac{1}{5})^{2} y^{2}) : (\frac{x}{4} - \frac{y}{5}) (\frac{x ^{2}}{16} + \frac{x y}{20} + \frac{y ^{2}}{25})

= (\frac{x}{4} - \frac{y}{5}) (\frac{x ^{2}}{16} + \frac{x y}{20} + \frac{y ^{2}}{25})

s im pl i f y - (\frac{1}{3})^{- 1}

4 < 3 x - 2 \leq 10

p^{2} + 2 p + 1 = 0

7 y \geq - 21

24.93 = 0.375 x^{2} + 7.235