Odpowiedzi #
1.4.3 Zapisz w postaci jednej potęgi #
a) $ \frac{64 \cdot 4^2}{\sqrt{256}} = \frac{2^6 \cdot 16 }{16}= 2^6$
b) $ (\frac{1}{2})^{-2} \cdot \sqrt[3]{64} = \frac{1}{(\frac{1}{2})^2} \cdot 4 = \frac{1}{\frac{1}{2^2}} \cdot 4 = \frac{1}{\frac{1}{4}} \cdot 4 = 4 \cdot 4 = 16 = 2^{4}$
c) $ \sqrt{\sqrt{\sqrt{256}}} = \sqrt{\sqrt{16}} = \sqrt{4} = 2 = 2^1$
d) $ 15 \cdot 3^{17} - 6 \cdot 3^{17} = 9 \cdot 3^{17} = 3^2 \cdot 3^{17} = 3^{2+17} = 3^{19}$
e) $ 0.5 \cdot 3^{123} + 8.5 \cdot 3^{123} = {\frac{1}{2}} \cdot 3^{123}+ 8 \frac {1}{2} \cdot 3^{123} = 9 \cdot 3^{123} = 3^2 \cdot 3^{123} = 3^{125} $
f) $ 5^8 \cdot 16^{-2} = \frac{5^8}{16^{2}} = \frac{5^8}{(2^4)^{2}} = \frac{5^8}{2^8} = (\frac{5}{2})^8 $
g) $ \frac{27^{\frac{1}{2}} \cdot 9^{-2}}{3^{-0.5}} = \frac {(3^{3})^{\frac{1}{2}} \cdot ({3^2})^{-2}}{3^{- \frac{1}{2}}} = \frac{{3^{ \frac{3}{2}} \cdot 3^{-4}}}{3^{- \frac{1}{2}}} = \frac {3^{ \frac{3}{2} - \frac{8}{2}}}{3^{- \frac{1}{2}}} = \frac {3^{-\frac{5}{2}}}{3^{- \frac{1}{2}}} = 3^{-\frac{5}{2} + \frac{1}{2} } = 3^{-\frac{4}{2} }= 3^{-2} $
h) $ (9^{\frac{1}{4}} \cdot 27^{\frac{1}{5}}): (\sqrt{3})^{\frac{4}{5}} = [(3^2)^{\frac{1}{4}} \cdot (3^3)^{\frac{1}{5}}]:(3^{\frac{1}{2}})^{\frac{4}{5}} = (3^{\frac{1}{2}} \cdot 3^{\frac{3}{5}}):3^{\frac{2}{5}}= $ $ = 3^{\frac{1}{2} + \frac{3}{5}}: 3^{\frac{2}{5}}= 3^{\frac{1}{2} + \frac{3}{5} - \frac{2}{5} } = 3^{\frac{1}{2} + \frac{1}{5}}=3^{\frac{5}{10} + \frac{2}{10}} = 3^{\frac{7}{10}}$