erweitern ((4x-1)^4)/(16)

Lösung

erweitern

\frac{( 4 x - 1 ) ^{4}}{16}

16 x^{4} - 16 x^{3} + 6 x^{2} - x + \frac{1}{16}

Schritte zur Lösung

\frac{( 4 x - 1 ) ^{4}}{16}

(4 x - 1)^{4} = 256 x^{4} - 256 x^{3} + 96 x^{2} - 16 x + 1

= \frac{256 x ^{4} - 256 x ^{3} + 96 x ^{2} - 16 x + 1}{16}

Wende Bruchregel an:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{256 x ^{4} - 256 x ^{3} + 96 x ^{2} - 16 x + 1}{16} = \frac{256 x ^{4}}{16} - \frac{256 x ^{3}}{16} + \frac{96 x ^{2}}{16} - \frac{16 x}{16} + \frac{1}{16}

= \frac{256 x ^{4}}{16} - \frac{256 x ^{3}}{16} + \frac{96 x ^{2}}{16} - \frac{16 x}{16} + \frac{1}{16}

\frac{256 x ^{4}}{16} = 16 x^{4}

\frac{256 x ^{3}}{16} = 16 x^{3}

\frac{96 x ^{2}}{16} = 6 x^{2}

\frac{16 x}{16} = x

= 16 x^{4} - 16 x^{3} + 6 x^{2} - x + \frac{1}{16}

s im pl i f y \frac{1}{2} (u - \frac{1}{2} sin (2 u))

e x p an d (1 - sin (x))^{2} + (cos (x))^{2}

e x p an d \exists x (x^{3} = - 1)

e x p an d x^{\frac{1}{3}} (x^{2} - 49)

e x p an d - \frac{48 x}{( x ^{2} + 12 ) ^{2}}