Oblicz (doprowadź do jednej potęgi)

[asiarst]

a) 8^2/3
b) 9^1/2
c) 81^-2/4
d) 9^-2 * 3^-1
e)2^-1/2 * 4^-2 * (1/16)^2
f) (1/5)^-2 * 25^1/3 * 125^3
g) 3^-2 * (1/9)^-2 * 27^-1
h) 2^5 * (1/4)^2 * 8
------------------------------- =
2^-3 * 16^-1

i) 3^-2 * 2^5 * 9^-3
--------------------------- =
4^-2 * (1/3)^2

j) 5^-2 * (1/2)^2 * 4^-1
-------------------------------- =
25^-3 * 8^-3

[domino21]

a. \(8^{\frac{2}{3}}=\sqrt[3]{8^2}=\sqrt[3]{64}=4\)

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

c. \(81^{-\frac{2}{4}}=(\frac{1}{81})^{\frac{1}{2}}=\sqrt{\frac{1}{81}}=\frac{1}{9}\)

d. \(9^{-2}\cdot 3^{-1}=(\frac{1}{9})^2 \cdot (\frac{1}{3})^1=\frac{1}{81} \cdot \frac{1}{3}=\frac{1}{243}\)

e. \(2^{-\frac{1}{2}} \cdot 4^{-2} \cdot (\frac{1}{16})^2=(\frac{1}{2})^{\frac{1}{2}}\cdot (\frac{1}{4})^2 \cdot (\frac{1}{16})^2=\sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{4}}\cdot \frac{1}{256}=\sqrt{\frac{1}{8}}\cdot \frac{1}{256}=\frac{\sqrt{2}}{4} \cdot \frac{1}{256}=\frac{\sqrt{2}}{1024}\)

[Lbubsazob]

h) \(\frac{2^5 \cdot \left( \frac{1}{4} \right)^2 \cdot 8 }{2^{-3} \cdot 16^{-1}}= \frac{2^5 \cdot 2^{-4} \cdot 2^3}{2^{-3} \cdot 2^{-4}}= \frac{2^4}{2^{-7}}=2^{11}\)

i) \(\frac{3^{-2} \cdot 2^5 \cdot 9^{-3}}{4^{-2} \cdot \left( \frac{1}{3} \right)^2 }= \frac{3^{-2} \cdot 2^5 \cdot 3^{-6}}{3^{-2} \cdot 2^{-4}} = \frac{3^{-8}}{3^{-2}} \cdot \frac{2^5}{2^{-4}}=3^{-6} \cdot 2^9\)

j) \(\frac{5^{-2} \cdot \left( \frac{1}{2} \right)^2 \cdot 4^{-1} }{25^{-3} \cdot 8^{-3}}= \frac{5^{-2} \cdot 2^{-2} \cdot 2^{-2}}{5^{-6} \cdot 2^{-9}}=5^4 \cdot 2^{5}\)

[Lbubsazob]

f) \(\left( \frac{1}{5} \right)^{-2} \cdot 25 ^{ \frac{1}{3} } \cdot 125^3= \left(5^{-1} \right)^{-2} \cdot \left(5^2 \right) ^{ \frac{1}{3} } \cdot \left(5^3 \right)^3=5^2 \cdot 5^{ \frac{2}{3}} \cdot 5^9=5^{11 \frac{2}{3} }\)

g) \(3^{-2} \cdot \left( \frac{1}{9} \right)^2 \cdot 27^{-1}=3^{-2} \cdot 3^{-4} \cdot 3^{-3}=3^{-9}\)

To chyba wszystko.