Zadanie 2.12.
a)
64^{\frac{1}{3}}=\sqrt[3]{64}=4
lub
64^{\frac{1}{3}}=(4^3)^{\frac{1}{3}}=4^1=4
b)
(\frac{9}{16})^{\frac{1}{2}}=\sqrt{\frac{9}{16}}=\frac{3}{4}
c)
(\frac{4}{9})^{-\frac{1}{2}}=(\frac{9}{4})^{\frac{1}{2}}=\sqrt{\frac{9}{4}}=\frac{3}{2}
lub
(\frac{4}{9})^{-\frac{1}{2}}=[(\frac{2}{3})^2]^{-\frac{1}{2}}=(\frac{2}{3})^{-1}=\frac{3}{2}
d)
(\frac{125}{8})^{-\frac{1}{3}}=[(\frac{5}{2})^3]^{-\frac{1}{3}}=(\frac{5}{2})^{-1}=\frac{2}{5}
lub
(\frac{125}{8})^{-\frac{1}{3}}=(\frac{8}{125})^{\frac{1}{3}}=\sqrt[3]{\frac{8}{125}}=\frac{2}{5}
e)
(0,125)^{-\frac{1}{3}}=(0,5^3)^{-\frac{1}{3}}=(\frac{5}{10})^{-1}=\frac{10}{5}=2
lub
(0,125)^{-\frac{1}{3}}=(\frac{125}{1000})^{-\frac{1}{3}}=(\frac{1000}{125})^{\frac{1}{3}}=8^{\frac{1}{3}}=(2^3)^{\frac{1}{3}}=2^1=2
f)
25^{\frac{1}{2}}=\sqrt{25}=5
g)
(0,027)^{\frac{2}{3}}=(0,3^3)^{\frac{2}{3}}=0,3^2=0,09
h)
1000^{-\frac{1}{3}}=(10^3)^{-\frac{1}{3}}=10^{-1}=10^{-1}=\frac{1}{10}
i)
2*27^{\frac{2}{3}}=2*(3^3)^{\frac{2}{3}}=2^1*2^2=2^3
j)
(\frac{49}{25})^{-\frac{1}{2}}*(\frac{2}{7})^{-1}
k)
256^{0,25}-3^{-1}*21^0=(2^8)^{\frac{1}{4}}-\frac{1}{3}*1=2^2-\frac{1}{3}=4-\frac{1}{3}=3\frac{3}{3}-\frac{1}{3}=3\frac{2}{3}
l)
5^{-3}*125^{\frac{2}{3}}*25^{\frac{3}{2}}=5^{-3}*(5^3)^{\frac{2}{3}}*(5^2)^{\frac{3}{2}}=5^{-3}*5^2*5^3
m)
27^{0,333...}=(3^3)^{\frac{1}{3}}=3^1=3
n)
0,25^{-2}*\frac{1}{25}=(\frac{1}{4})^{-2}*(\frac{1}{5})^2=4^2*(\frac{1}{5})^2=(4*\frac{1}{5})^2=(\frac{4}{5})^2=\frac{16}{25}
o)
[-0,0(21)]^{-1}
zamiana ułamka okresowego na zwykły:
a=0,0(21)=0,021021…
1000a=21,0(21)
1000a-a=21,0(21)-0,0(21)
999a=21
a=\frac{21}{999}=\frac{7}{333}
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[-0,0(21)]^{-1}=(\frac{7}{333})^{-1}=\frac{333}{7}=47\frac{4}{7}