faktorisieren (1+x)^{5/2}-(1-x)^{5/2}

Lösung

faktorisieren

(1 + x)^{\frac{5}{2}} - (1 - x)^{\frac{5}{2}}

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

Schritte zur Lösung

(1 + x)^{\frac{5}{2}} - (1 - x)^{\frac{5}{2}}

Schreibe

(1 + x)^{\frac{5}{2}}

um:

(1 + x)^{5}

= (1 + x)^{5} - (1 - x)^{\frac{5}{2}}

Schreibe

(1 - x)^{\frac{5}{2}}

um:

(1 - x)^{5}

= (1 + x)^{5} - (1 - x)^{5}

Wende Regel zum Faktorisieren an:

x^{n} - y^{n} = (x - y) (x^{n - 1} + x^{n - 2} y + \dots + x y^{n - 2} y^{n - 1})

(1 + x)^{5} - (1 - x)^{5} = (1 + x - 1 - x) ((1 + x)^{4} + (1 + x)^{3} 1 - x + (1 + x)^{2} (1 - x)^{2} + 1 + x (1 - x)^{3} + (1 - x)^{4})

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

Vereinfache

(1 + x - 1 - x) ((1 + x)^{4} + (1 + x)^{3} 1 - x + (1 + x)^{2} (1 - x)^{2} + 1 + x (1 - x)^{3} + (1 - x)^{4}) : (1 + x - 1 - x) ((1 + x)^{2} + (1 + x)^{3} 1 - x + (1 + x) (1 - x) + (1 - x)^{3} 1 + x + (1 - x)^{2})

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

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

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

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

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

Multipliziere aus

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

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

\frac{2 - x}{4} - 1 \geq 0

f a c t or 8 x^{2} - 10 x - 3

n - 9 \leq - 5 and - 10 n < 40

f a c t or b^{2} + 18 b + 80

5 x + 3 x + 1 = 0