faktorisieren a^5b^{10}-c^{15}d^5

Lösung

faktorisieren

a^{5} b^{10} - c^{15} d^{5}

(a b^{2} - c^{3} d) (a^{4} b^{8} + a^{3} b^{6} c^{3} d + a^{2} b^{4} c^{6} d^{2} + a b^{2} c^{9} d^{3} + c^{12} d^{4})

Schritte zur Lösung

a^{5} b^{10} - c^{15} d^{5}

Schreibe

a^{5} b^{10}

um:

(a b^{2})^{5}

Schreibe

c^{15} d^{5}

um:

(c^{3} d)^{5}

= (a b^{2})^{5} - (c^{3} d)^{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})

(a b^{2})^{5} - (c^{3} d)^{5} = (a b^{2} - c^{3} d) ((a b^{2})^{4} + c^{3} d (a b^{2})^{3} + (c^{3} d)^{2} (a b^{2})^{2} + (c^{3} d)^{3} a b^{2} + (c^{3} d)^{4})

= (a b^{2} - c^{3} d) ((a b^{2})^{4} + c^{3} d (a b^{2})^{3} + (c^{3} d)^{2} (a b^{2})^{2} + (c^{3} d)^{3} a b^{2} + (c^{3} d)^{4})

Vereinfache

(a b^{2} - c^{3} d) ((a b^{2})^{4} + c^{3} d (a b^{2})^{3} + (c^{3} d)^{2} (a b^{2})^{2} + (c^{3} d)^{3} a b^{2} + (c^{3} d)^{4}) : (a b^{2} - c^{3} d) (a^{4} b^{8} + a^{3} b^{6} c^{3} d + a^{2} b^{4} c^{6} d^{2} + a b^{2} c^{9} d^{3} + c^{12} d^{4})

= (a b^{2} - c^{3} d) (a^{4} b^{8} + a^{3} b^{6} c^{3} d + a^{2} b^{4} c^{6} d^{2} + a b^{2} c^{9} d^{3} + c^{12} d^{4})

\frac{d}{d y} ((x (y)) (y^{2}))

s im pl i f y [- 4.4]

s im pl i f y (1 + cos (2 θ)) (1 - cos (2 θ))

roo t s f (t)^{2}

f a c t or 3 18 + 22