erweitern ((2x-1)^3)/9

Lösung

erweitern

\frac{( 2 x - 1 ) ^{3}}{9}

\frac{8 x ^{3}}{9} - \frac{4 x ^{2}}{3} + \frac{2 x}{3} - \frac{1}{9}

Schritte zur Lösung

\frac{( 2 x - 1 ) ^{3}}{9}

(2 x - 1)^{3} = 8 x^{3} - 12 x^{2} + 6 x - 1

= \frac{8 x ^{3} - 12 x ^{2} + 6 x - 1}{9}

Wende Bruchregel an:

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

\frac{8 x ^{3} - 12 x ^{2} + 6 x - 1}{9} = \frac{8 x ^{3}}{9} - \frac{12 x ^{2}}{9} + \frac{6 x}{9} - \frac{1}{9}

= \frac{8 x ^{3}}{9} - \frac{12 x ^{2}}{9} + \frac{6 x}{9} - \frac{1}{9}

Streiche

\frac{12 x ^{2}}{9} : \frac{4 x ^{2}}{3}

= \frac{8 x ^{3}}{9} - \frac{4 x ^{2}}{3} + \frac{6 x}{9} - \frac{1}{9}

Streiche

\frac{6 x}{9} : \frac{2 x}{3}

= \frac{8 x ^{3}}{9} - \frac{4 x ^{2}}{3} + \frac{2 x}{3} - \frac{1}{9}

e x p an d 5 x + \frac{- 8 e ^{5 (- t + c)} + 7}{160}

e x p an d cot (x) (sin (x) + sec (x))

e x p an d (x - \frac{13}{6}) \cdot (x + \frac{1}{6})

e x p an d sin ((2 n + 1) x)

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