vereinfachen 3(1/3)^3-10(1/3)^2-27(1/3)+10

Lösung

vereinfachen

3 (\frac{1}{3})^{3} - 10 (\frac{1}{3})^{2} - 27 (\frac{1}{3}) + 10

0

Schritte zur Lösung

3 (\frac{1}{3})^{3} - 10 (\frac{1}{3})^{2} - 27 (\frac{1}{3}) + 10

Apply rule:

(a) = a

(\frac{1}{3}) = \frac{1}{3}

= 3 (\frac{1}{3})^{3} - 10 (\frac{1}{3})^{2} - 27 \cdot \frac{1}{3} + 10

(\frac{1}{3})^{3} = \frac{1}{27}

(\frac{1}{3})^{2} = \frac{1}{9}

27 \cdot \frac{1}{3} = 9

= 3 \cdot \frac{1}{27} - 10 \cdot \frac{1}{9} - 9 + 10

3 \cdot \frac{1}{27} - 10 \cdot \frac{1}{9} - 9 + 10 = \frac{\frac{1}{27} \cdot 81 - \frac{1}{9} \cdot 270}{27} + \frac{1}{1}

= \frac{\frac{1}{27} \cdot 81 - \frac{1}{9} \cdot 270}{27} + \frac{1}{1}

\frac{\frac{1}{27} \cdot 81 - \frac{1}{9} \cdot 270}{27} = - 1

\frac{1}{1} = 1

= - 1 + 1

Addiere die Zahlen:

- 1 + 1 = 0

= 0

s im pl i f y (\frac{1}{125 x ^{3}})^{- \frac{1}{3}}

x + 5 \geq 3

s im pl i f y 78

f a c t or 12 - 10 x

∣ 4 x - 3 ∣ = 8