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Beliebt Algebra >

faktorisieren (2mn)^2+(m^2-n^2)^2

Lösung

faktorisieren

(2 mn)^{2} + (m^{2} - n^{2})^{2}

Lösung

(m^{2} + n^{2})^{2}

Schritte zur Lösung

(2 mn)^{2} + (m^{2} - n^{2})^{2}

Wende Exponentenregel an:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} m^{2} n^{2} + (m^{2} - n^{2})^{2}

Faktorisiere

m^{2} - n^{2} : (m + n) (m - n)

= 2^{2} m^{2} n^{2} + ((m + n) (m - n))^{2}

Wende Exponentenregel an:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} m^{2} n^{2} + (m + n)^{2} (m - n)^{2}

2^{2} = 4

= 4 m^{2} n^{2} + (m + n)^{2} (m - n)^{2}

(m + n)^{2} : m^{2} + 2 mn + n^{2}

= 4 m^{2} n^{2} + (m^{2} + 2 mn + n^{2}) (m - n)^{2}

(m - n)^{2} : m^{2} - 2 mn + n^{2}

= 4 m^{2} n^{2} + (m^{2} + 2 mn + n^{2}) (m^{2} - 2 mn + n^{2})

Multipliziere aus

(m^{2} + 2 mn + n^{2}) (m^{2} - 2 mn + n^{2}) : m^{4} - 2 m^{2} n^{2} + n^{4}

= 4 m^{2} n^{2} + m^{4} - 2 m^{2} n^{2} + n^{4}

Vereinfache

4 m^{2} n^{2} + m^{4} - 2 m^{2} n^{2} + n^{4} : m^{4} + 2 m^{2} n^{2} + n^{4}

= m^{4} + 2 m^{2} n^{2} + n^{4}

Zerlege die Ausdrücke in Gruppen

= (m^{4} + m^{2} n^{2}) + (m^{2} n^{2} + n^{4})

Klammere

m^{2}

aus

m^{4} + m^{2} n^{2}

aus

: m^{2} (m^{2} + n^{2})

Klammere

n^{2}

aus

m^{2} n^{2} + n^{4}

aus

: n^{2} (m^{2} + n^{2})

= m^{2} (m^{2} + n^{2}) + n^{2} (m^{2} + n^{2})

Klammere gleiche Terme aus

m^{2} + n^{2}

= (m^{2} + n^{2}) (m^{2} + n^{2})

Fasse zusammen

= (m^{2} + n^{2})^{2}

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