erweitern (x_{0}+αx^{1/2})^3

Lösung

erweitern

(x_{0} + α x^{\frac{1}{2}})^{3}

x_{0}^{3} + 3 x_{0}^{2} α x^{\frac{1}{2}} + 3 x_{0} α^{2} x + α^{3} x^{\frac{3}{2}}

Schritte zur Lösung

(x_{0} + α x^{\frac{1}{2}})^{3}

Wende Formel für perfekte dritte Potenzen an:

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

(x_{0} + α x^{\frac{1}{2}})^{3} = x_{0}^{3} + 3 x_{0}^{2} α x^{\frac{1}{2}} + 3 x_{0} (α x^{\frac{1}{2}})^{2} + (α x^{\frac{1}{2}})^{3}

= x_{0}^{3} + 3 x_{0}^{2} α x^{\frac{1}{2}} + 3 x_{0} (α x^{\frac{1}{2}})^{2} + (α x^{\frac{1}{2}})^{3}

(α x^{\frac{1}{2}})^{2} = α^{2} x

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

= x_{0}^{3} + 3 x_{0}^{2} α x^{\frac{1}{2}} + 3 x_{0} α^{2} x + α^{3} x^{\frac{3}{2}}

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

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

e x p an d (2 - \frac{x}{2})^{7}

e x p an d (2 - \frac{x}{2})^{5}

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