Nowa MATeMAtyka 1. Podręcznik. Zakres podstawowy | Strona 43
Ćwiczenie 5. Oblicz.
a)
\[ 2^{\frac{2}{3}} \cdot 2^{\frac{4}{3}} = 2^{\frac{2}{3} + \frac{4}{3}} = 2^{\frac{6}{3}} = 2^2 = 4\]
b)
\[ 5^{\frac{4}{3}} \cdot 5^{\frac{5}{3}} = 5^{\frac{4}{3} + \frac{5}{3}} = 5^{\frac{9}{3}} = 5^3 = 125\]
c)
\[ 4^{\frac{9}{2}}:4^{\frac{5}{2}} = 4^{\frac{9}{2} – \frac{5}{2}} = 4^{\frac{4}{2}} = 4^2 = 16\]
d)
\[ 2^{7,5}:2^{2,5} = 2^{7,5 – 2,5} = 2^5 = 32\]
e)
\[ 3^{\frac{1}{2}} \cdot 3^{- \frac{1}{3}} \cdot 3^{\frac{1}{6}} = 3^{\frac{1}{2} – \frac{1}{3} + \frac{1}{6}} = 3^{\frac{3}{6} – \frac{2}{6} + \frac{1}{6}} = 3^{\frac{2}{6}} = \sqrt[3]{3}\]
Ćwiczenie 6. Oblicz.
a)
\[ 12^{\frac{3}{2}} \cdot 3^{\frac{3}{2}} = 36^{\frac{3}{2}} = (\sqrt{36})^3 = 6^3 = 216\]
b)
\[ 12^{\frac{5}{2}}:3^{\frac{5}{2}} = 4^{\frac{5}{2}} = (\sqrt{4})^5 = 2^5 = 32\]
c)
\[ 24^{\frac{2}{3}} \cdot 3^{\frac{1}{3}} = 8^{\frac{2}{3}} \cdot 3^{\frac{2}{3}} \cdot 3^{\frac{1}{3}} = (\sqrt[3]{8})^2 \cdot 3^{\frac{2}{3} + \frac{1}{3}} = 2^2 \cdot 3^1 = 4 \cdot 3 = 12\]
d)
\[ 9^{\frac{4}{3}}:24^{\frac{2}{3}} = (3^2)^{\frac{4}{3}} \cdot 3^{- \frac{2}{3}} \cdot 8^{- \frac{2}{3}} = 3^{\frac{8}{3}} \cdot 3^{- \frac{2}{3}} \cdot 8^{- \frac{2}{3}} = 3^{\frac{8}{3} – \frac{2}{3}} \cdot 8^{- \frac{2}{3}} = = 3^{\frac{6}{3}} \cdot 8^{- \frac{2}{3}} = 3^2 \cdot (\sqrt[3]{8})^{- 2} = 9 \cdot \frac{1}{4} = 2\frac{1}{4}\]
e)
\[ 375^{\frac{1}{3}} \cdot 3^{\frac{2}{3}} = 125^{\frac{1}{3}} \cdot 3^{\frac{1}{3}} \cdot 3^{\frac{2}{3}} = \sqrt[3]{125} \cdot 3^{\frac{1}{3} + \frac{2}{3}} = 5 \cdot 3 = 15\]
Ćwiczenie 7. Oblicz.
a)
\[ (9^{\frac{3}{5}})^{\frac{10}{3}} = 9^{\frac{30}{15}} = 9^2 = 81\]
b)
\[(8^{\frac{3}{2}})^{\frac{4}{9}} = 8^{\frac{12}{18}} = 8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 2^2 = 4\]
c)
\[(125^{\frac{2}{3}})^{0,75} = (125^{\frac{2}{3}})^{\frac{3}{4}} = 125^{\frac{2}{4}} = 125^{\frac{1}{2}} = \sqrt{125} = 5\sqrt{5}\]
d)
\[((\frac{4}{9})^{- 2,5})^{- 0,6} = (\frac{4}{9})^{1,5} = (\frac{4}{9})^{\frac{3}{2}} = \sqrt{(\frac{4}{9})^3} = (\sqrt{\frac{4}{9}})^3 = (\frac{2}{3})^3 = \frac{8}{27}\]
e)
\[(32^{\frac{28}{25}})^{- \frac{5}{7}} = (32)^{- \frac{4}{5}} = \frac{1}{32^{\frac{4}{5}}} = \frac{1}{(2^5)^{\frac{4}{5}}} = \frac{1}{2^4} = \frac{1}{16}\]
Ćwiczenie 8. Korzystając z powyższych przybliżeń, podaj z dokładnością do czterech miejsc po przecinku przybliżoną wartość potęgi:
a) 102,3 = 102 · 100,3 ≈ 199,5262
b) 103,45 = 103 · 100,45 ≈ 2818,3829
c)
\[10^{\frac{13}{8}} = 10^1 \cdot 10^{0,625} \approx 42,1697\]
d)
\[10^{- 1,7} = 10^{0,3}:10^2 \approx 0,0200\]