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Beliebt Algebra >

erweitern (\sqrt[4]{-3125n^7})^6

Lösung

erweitern

(4 - 3125 n^{7})^{6}

Lösung

- 781255 i n^{\frac{21}{2}}

Schritte zur Lösung

(4 - 3125 n^{7})^{6}

4 - 3125 n^{7} = 5 4 - 1 45 n 4 n^{3}

= (5 4 - 1 45 n 4 n^{3})^{6}

Wende Exponentenregel an:

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

= 5^{6} (45)^{6} (4 - 1)^{6} n^{6} (4 n^{3})^{6}

(4 - 1)^{6} : (- 1)^{\frac{3}{2}}

= (- 1)^{\frac{3}{2}} \cdot 5^{6} (45)^{6} n^{6} (4 n^{3})^{6}

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

= i^{3} \cdot 5^{6} (45)^{6} n^{6} (4 n^{3})^{6}

(45)^{6} : 5^{\frac{3}{2}}

= i^{3} \cdot 5^{6} \cdot 5^{\frac{3}{2}} n^{6} (4 n^{3})^{6}

5^{\frac{3}{2}} = 55

= 55

(4 n^{3})^{6} : (n^{3})^{\frac{3}{2}}

= i^{3} \cdot 5^{6} \cdot 55 n^{6} (n^{3})^{\frac{3}{2}}

(n^{3})^{\frac{3}{2}} : n^{\frac{9}{2}}

= i^{3} \cdot 5^{6} \cdot 55 n^{6} n^{\frac{9}{2}}

Wende Exponentenregel an:

a^{b} \cdot a^{c} = a^{b + c}

5^{6} \cdot 5 = 5^{6 + 1}

= i^{3} \cdot 5^{6 + 1} 5 n^{6} n^{\frac{9}{2}}

Addiere die Zahlen:

6 + 1 = 7

= i^{3} \cdot 5^{7} 5 n^{6} n^{\frac{9}{2}}

Wende Exponentenregel an:

a^{b} \cdot a^{c} = a^{b + c}

n^{6} n^{\frac{9}{2}} = n^{6 + \frac{9}{2}}

= i^{3} \cdot 5^{7} 5 n^{6 + \frac{9}{2}}

n^{6 + \frac{9}{2}} = n^{\frac{21}{2}}

= 5^{7} 5 i^{3} n^{\frac{21}{2}}

i^{3} = - i

= - 5^{7} 5 i n^{\frac{21}{2}}

5^{7} = 78125

= - 781255 i n^{\frac{21}{2}}

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