erweitern (x^2-1/(2x))^3

Lösung

erweitern

(x^{2} - \frac{1}{2 x})^{3}

\frac{8 x ^{9} - 12 x ^{6} - 1 + 6 x ^{3}}{8 x ^{3}}

Schritte zur Lösung

(x^{2} - \frac{1}{2 x})^{3}

Wende Formel für perfekte dritte Potenzen an:

(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}

(x^{2} - \frac{1}{2 x})^{3} = (x^{2})^{3} - 3 (x^{2})^{2} \frac{1}{2 x} + 3 x^{2} (\frac{1}{2 x})^{2} - (\frac{1}{2 x})^{3}

= (x^{2})^{3} - 3 (x^{2})^{2} \frac{1}{2 x} + 3 x^{2} (\frac{1}{2 x})^{2} - (\frac{1}{2 x})^{3}

Vereinfache

(x^{2})^{3} - 3 (x^{2})^{2} \frac{1}{2 x} + 3 x^{2} (\frac{1}{2 x})^{2} - (\frac{1}{2 x})^{3} : \frac{8 x ^{9} - 12 x ^{6} - 1 + 6 x ^{3}}{8 x ^{3}}

= \frac{8 x ^{9} - 12 x ^{6} - 1 + 6 x ^{3}}{8 x ^{3}}

e x p an d \frac{3 ( - x ^{2} - 1 )}{( x ^{2} - 1 ) ^{2}}

e x p an d (2 x y^{2} + 3 x^{2}) d x + (2 x^{2} y + 4 y^{3}) d y

e x p an d 2 (x + h)^{3} - 6 (x + h) - 2 x^{3} - 6 x

s im pl i f y (x^{2} - 2 x + 1) (x^{2} + 2 x + 1)

e x p an d (2 a - \frac{1}{b})^{4}