erweitern (5+4/(x^3))(2x+3/(x^2))^{15}

Lösung

erweitern

(5 + \frac{4}{x ^{3}}) (2 x + \frac{3}{x ^{2}})^{15}

\frac{5 ( 2 x ^{3} + 3 ) ^{15}}{x ^{30}} + \frac{4 ( 2 x ^{3} + 3 ) ^{15}}{x ^{33}}

Schritte zur Lösung

(5 + \frac{4}{x ^{3}}) (2 x + \frac{3}{x ^{2}})^{15}

= (2 x + \frac{3}{x ^{2}})^{15} (5 + \frac{4}{x ^{3}})

Wende das Distributivgesetz an:

a (b + c) = ab + a c

(2 x + \frac{3}{x ^{2}})^{15} (5 + \frac{4}{x ^{3}}) = (2 x + \frac{3}{x ^{2}})^{15} \cdot 5 + (2 x + \frac{3}{x ^{2}})^{15} \frac{4}{x ^{3}}

= (2 x + \frac{3}{x ^{2}})^{15} \cdot 5 + (2 x + \frac{3}{x ^{2}})^{15} \frac{4}{x ^{3}}

Vereinfache

(2 x + \frac{3}{x ^{2}})^{15} \cdot 5 + (2 x + \frac{3}{x ^{2}})^{15} \frac{4}{x ^{3}} : \frac{5 ( 2 x ^{3} + 3 ) ^{15}}{x ^{30}} + \frac{4 ( 2 x ^{3} + 3 ) ^{15}}{x ^{33}}

= \frac{5 ( 2 x ^{3} + 3 ) ^{15}}{x ^{30}} + \frac{4 ( 2 x ^{3} + 3 ) ^{15}}{x ^{33}}

e x p an d (- 3, - 11) v er t e x (0, - 2)

e x p an d e^{\frac{y ^{3}}{3}} [3 x^{2} y^{2} d x + (2 x^{3} y + x^{3} y^{4}) d y]

e x p an d \frac{( 16 m ^{3} + n ^{2} ) ^{2}}{16}

e x p an d (4 - \frac{1}{n})^{6}

e x p an d (4 - \frac{1}{n})^{7}