faktorisieren 1000y^9-m^{15}

Lösung

faktorisieren

1000 y^{9} - m^{15}

(10 y^{3} - m^{5}) (100 y^{6} + 10 y^{3} m^{5} + m^{10})

Schritte zur Lösung

1000 y^{9} - m^{15}

Wende Exponentenregel an

= (10 y^{3})^{3} - (m^{5})^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(10 y^{3})^{3} - (m^{5})^{3} = (10 y^{3} - m^{5}) ((10 y^{3})^{2} + 10 y^{3} m^{5} + (m^{5})^{2})

= (10 y^{3} - m^{5}) ((10 y^{3})^{2} + 10 y^{3} m^{5} + (m^{5})^{2})

Vereinfache

(10 y^{3})^{2} + 10 y^{3} m^{5} + (m^{5})^{2} : 100 y^{6} + 10 y^{3} m^{5} + m^{10}

= (10 y^{3} - m^{5}) (100 y^{6} + 10 y^{3} m^{5} + m^{10})

f a c t or 25 n^{3} + 15 n^{2} + 20 n + 12

- 5 x > 20

- 2 ∣ 7 d ∣ + 9 \geq - 7

3 (3 x - 8) = 4 x + 6

5 x + 6 = 3 (- x - 1)