vereinfachen 8^{3/2}*2/3

Lösung

vereinfachen

8^{\frac{3}{2}} \cdot \frac{2}{3}

\frac{32}{3} \cdot 2^{\frac{1}{2}}

Dezimale

15.08494 \dots

Schritte zur Lösung

8^{\frac{3}{2}} \cdot \frac{2}{3}

Primfaktorzerlegung von

8 : 2^{3}

= (2^{3})^{\frac{3}{2}} \frac{2}{3}

Wende Exponentenregel an:

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

2^{3} = 2^{2} \cdot 2

= (2^{2} \cdot 2)^{\frac{3}{2}} \frac{2}{3}

Wende Exponentenregel an:

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

(2^{2} \cdot 2)^{\frac{3}{2}} = (2^{2})^{\frac{3}{2}} \cdot 2^{\frac{3}{2}}

= (2^{2})^{\frac{3}{2}} \cdot 2^{\frac{3}{2}} \cdot \frac{2}{3}

(2^{2})^{\frac{3}{2}} = 8

= 8 \cdot 2^{\frac{3}{2}} \cdot \frac{2}{3}

8 \cdot 2^{\frac{3}{2}} = 16 \cdot 2^{\frac{1}{2}}

= 16 \cdot 2^{\frac{1}{2}} \cdot \frac{2}{3}

16 \cdot 2^{\frac{1}{2}} \cdot \frac{2}{3} = \frac{32}{3} \cdot 2^{\frac{1}{2}}

= \frac{32}{3} \cdot 2^{\frac{1}{2}}

∣ x + 5 ∣ \geq 9

s im pl i f y (x - 3 - i) (x - 3 + i)

(x + 7) (x + 3) (x - 3) \geq 0

f a c t or 2 c^{2} + 11 c + 5

5^{2} + b^{2} = 1 3^{2}