vereinfachen (2x-1)^3-(2x-1)^2-2(2x-1)-1

Lösung

vereinfachen

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

8 x^{3} - 16 x^{2} + 6 x - 1

Schritte zur Lösung

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

Schreibe

(2 x - 1)^{3}

um:

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

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

Schreibe

- (2 x - 1)^{2}

um:

- 4 x^{2} + 4 x - 1

= 8 x^{3} - 12 x^{2} + 6 x - 1 - 4 x^{2} + 4 x - 1 - 2 (2 x - 1) - 1

Schreibe

- 2 (2 x - 1)

um:

- 4 x + 2

= 8 x^{3} - 12 x^{2} + 6 x - 1 - 4 x^{2} + 4 x - 1 - 4 x + 2 - 1

Fasse gleiche Terme zusammen

= 8 x^{3} - 12 x^{2} - 4 x^{2} + 6 x + 4 x - 4 x - 1 - 1 - 1 + 2

Addiere gleiche Elemente:

- 12 x^{2} - 4 x^{2} = - 16 x^{2}

= 8 x^{3} - 16 x^{2} + 6 x + 4 x - 4 x - 1 - 1 - 1 + 2

Addiere gleiche Elemente:

6 x + 4 x - 4 x = 6 x

= 8 x^{3} - 16 x^{2} + 6 x - 1 - 1 - 1 + 2

Addiere die Zahlen:

- 1 - 1 - 1 + 2 = - 1

= 8 x^{3} - 16 x^{2} + 6 x - 1

s im pl i f y ((((a^{2})^{2})^{3})^{2})^{2}

f a c t or x^{3} - x^{2} - x \cdot y^{2} + y^{2}

s im pl i f y \frac{5 ( 2 x - 3 ) - 3 ( x + 4 )}{9}

s im pl i f y \frac{2 cos ( x )}{sin ^{2} ( x )}

s im pl i f y a^{3} + 9 b^{3} + (a + b)^{3}