Nowa MATeMAtyka 1. Podręcznik. Zakres podstawowy | Strona 44
Zadanie 1. Zapisz liczbę w postaci potęgi o podstawie 2.
a)
\[ \sqrt{2} = 2^{\frac{1}{2}}\]
b)
\[ \sqrt[3]{2} = 2^{\frac{1}{3}}\]
c)
\[ \sqrt{0,5} = \sqrt{\frac{1}{2}} = \sqrt{2^{- 1}} = 2^{- \frac{1}{2}}\]
d)
\[ \sqrt[3]{4} = 4^{\frac{1}{3}} = (2^2)^{\frac{1}{3}} = 2^{\frac{2}{3}}\]
e)
\[ \sqrt{512} = 512^{\frac{1}{2}} = (2^9)^{\frac{1}{2}} = 2^{\frac{9}{2}}\]
f)
\[ \sqrt[3]{1024} = 1024^{\frac{1}{3}} = (2^{10})^{\frac{1}{3}} = 2^{\frac{10}{3}}\]
g)
\[ 2\sqrt{2} = 2 \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2}} = 2^{\frac{3}{2}}\]
h)
\[ 2\sqrt[3]{2} = 2 \cdot 2^{\frac{1}{3}} = 2^{1 + \frac{1}{3}} = 2^{\frac{4}{3}}\]
Zadanie 2. Zapisz liczbę w postaci potęgi o podstawie 7.
a)
\[\sqrt{7^3} = (7^3)^{\frac{1}{2}} = 7^{\frac{3}{2}}\]
b)
\[\sqrt[3]{7^2} = (7^2)^{\frac{1}{3}} = 7^{\frac{2}{3}}\]
c)
\[\frac{1}{\sqrt{7}} = (\sqrt{7})^{- 1} = 7^{- \frac{1}{2}}\]
d)
\[\frac{1}{\sqrt[3]{7}} = (\sqrt[3]{7})^{- 1} = 7^{- \frac{1}{3}}\]
e)
\[\sqrt{\frac{1}{7^3}} = (\frac{1}{7^3})^{\frac{1}{2}} = (7^{- 3})^{\frac{1}{2}} = 7^{- \frac{3}{2}}\]
f)
\[\frac{1}{\sqrt[5]{7^3}} = (\sqrt[5]{7^3})^{- 1} = (7^{\frac{3}{5}})^{- 1} = 7^{- \frac{3}{5}}\]
g)
\[49\sqrt{7} = 7^2 \cdot 7^{\frac{1}{2}} = 7^{2\frac{1}{2}} = 7^{\frac{5}{2}}\]
h)
\[7\sqrt[5]{7} = 7^1 \cdot 7^{\frac{1}{5}} = 7^{1\frac{1}{5}} = 7^{\frac{6}{5}}\]
Zadanie 3. Oblicz.
a)
\[9^{\frac{3}{2}} = (3^2)^{\frac{3}{2}} = 3^{\frac{6}{2}} = 3^3 = 27\]
b)
\[125^{\frac{2}{3}} = (5^2)^{\frac{2}{3}} = 5^{\frac{6}{3}} = 5^2 = 25\]
c)
\[8^{- \frac{4}{3}} = (2^3)^{- \frac{4}{3}} = 2^{- \frac{12}{3}} = 2^{- 4} = (\frac{1}{2})^4 = \frac{1}{16}\]
d)
\[27^{- \frac{2}{3}} = (3^3)^{- \frac{2}{3}} = 3^{- \frac{6}{3}} = 3^{- 2} = (\frac{1}{3})^2 = \frac{1}{9}\]
e)
\[144^{- 0,5} = (12^2)^{- \frac{1}{2}} = 12^{- \frac{2}{2}} = 12^{- 1} = \frac{1}{12}\]
f)
\[16^{0,75} = (2^4)^{\frac{3}{4}} = 2^{\frac{12}{4}} = 2^3 = 8\]
g)
\[10000^{0,25} = (10^4)^{\frac{1}{4}} = 10^{\frac{4}{4}} = 10\]
h)
\[81^{- 0,125} = (3^4)^{- \frac{1}{8}} = 3^{- \frac{4}{8}} = 3^{- \frac{1}{2}} = (\frac{1}{3})^{\frac{1}{2}} = \frac{1}{\sqrt{3}}\]
\[\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}\]
Zadanie 4. Oblicz.
a)
\[0,04^{\frac{3}{2}} = (\frac{4}{100})^{\frac{3}{2}} = (\frac{100}{4})^{- \frac{3}{2}} = 25^{- \frac{3}{2}} = (5^2)^{- \frac{3}{2}} = 5^{- \frac{6}{2}} = 5^{- 3} = (\frac{1}{5})^3 = \frac{1}{125}\]
b)
\[0,16^{- \frac{1}{2}} = (\frac{16}{100})^{- \frac{1}{2}} = (\frac{4}{25})^{\frac{1}{2}} = (\frac{25}{4})^{\frac{1}{2}} = \sqrt{\frac{25}{4}} = \frac{5}{2}\]
c)
\[0,16^{- \frac{1}{2}} = (\frac{16}{100})^{- \frac{1}{2}} = (\frac{4}{25})^{\frac{1}{2}} = (\frac{25}{4})^{\frac{1}{2}} = \sqrt{\frac{25}{4}} = \frac{5}{2}\]
d)
\[0,027^{\frac{2}{3}} = (\frac{27}{1000})^{\frac{2}{3}} = (\sqrt[3]{\frac{27}{10}})^2 = (\frac{3}{10})^2 = \frac{9}{100}\]
e)
\[0,0625^{- \frac{5}{4}} = (\frac{625}{10000})^{- \frac{5}{4}} = (\frac{10000}{625})^{\frac{5}{4}} = (\sqrt[4]{\frac{10000}{625}})^5 = (\frac{10}{5})^5 = \frac{100000}{3125} = 32\]
f)
\[0,00032^{\frac{3}{5}} = (\frac{32}{100000})^{\frac{3}{5}} = (\sqrt[5]{\frac{32}{100000}})^3 = (\frac{2}{10})^3 = \frac{8}{1000}\]
g)
\[0,0081^{- 1,25} = 0,0081^{- \frac{5}{4}} = (\frac{81}{10000})^{- \frac{5}{4}} = (\frac{10000}{81})^{\frac{5}{4}} = (\sqrt[4]{\frac{10000}{81}})^5 = (\frac{10}{3})^5 = \frac{10000}{243} = 411\frac{127}{243}\]
h)
\[0,00000256^{0,375} = 0,00000256^{\frac{3}{8}} = (\frac{256}{10000000})^{\frac{3}{8}} = (\frac{1}{390625})^{\frac{3}{8}} = (\sqrt[8]{\frac{1}{390625}})^3 = (\frac{1}{5})^3 = \frac{1}{125}\]
Zadanie 5. Oblicz.
a)
\[2^2 \cdot (2^3)^{\frac{2}{3}} = 2^2 \cdot 2^{\frac{6}{3}} = 2^2 \cdot 2^2 = 4 \cdot 4 = 16\]
b)
\[2^4:32^{\frac{1}{5}} = 2^4:(2^5)^{\frac{1}{5}} = 2^4:2^1 = 2^3 = 8\]
c)
\[3^3 \cdot 27^{- \frac{4}{3}} = 3^3 \cdot (3^3)^{- \frac{4}{3}} = 3^3 \cdot 3^{- \frac{12}{3}} = 3^3 \cdot 3^{- 4} = 3^{- 1} = \frac{1}{3}\]
d)
\[27^{\frac{2}{3}}:9^{- \frac{3}{2}} = (3^3)^{\frac{2}{3}}:(3^3)^{- \frac{3}{2}} = 3^{\frac{6}{3}}:3^{- \frac{9}{3}} = 3^2:3^{- 3} = 3^{2 – ( – 3)} = 3^5 = 243\]
e)
\[0,008^{\frac{1}{3}} \cdot \sqrt[3]{125} = \sqrt[3]{0,008} \cdot \sqrt[3]{125} = \sqrt[3]{1} = 1\]
f)
\[0,04^{\frac{1}{2}}:64^{\frac{1}{3}} = \sqrt{0,04}:\sqrt[3]{64} = \sqrt{\frac{4}{100}}:4 = \frac{2}{10}:4 = 0,05\]
g)
\[0,0256^{\frac{3}{4}} \cdot (\sqrt[3]{10})^9 = (\frac{256}{10000})^{\frac{3}{4}} \cdot (10^{\frac{1}{3}})^9 = (\frac{16}{625})^{\frac{3}{4}} \cdot 10^{\frac{9}{3}} = (\sqrt[4]{\frac{16}{625}})^3 \cdot 10^3 = (\frac{2}{5})^3 \cdot 1000 = \frac{8}{125} \cdot 1000 = 64\]
h)
\[0,027^{\frac{2}{3}}:\sqrt[6]{27^2} = (0,3^3)^{\frac{2}{3}}:27^{\frac{2}{6}} = 0,3^{\frac{6}{3}}:(3^3)^{\frac{2}{6}} = 0,3^2:3^{\frac{6}{6}} = 0,3^2:3^1 = 0,03\]
Zadanie 6. Zapisz liczbę w postaci potęgi o podstawie 3.
a)
\[\sqrt{3} = 3^{\frac{1}{2}}\]
b)
\[\sqrt[4]{3} = 3^{\frac{1}{4}}\]
c)
\[\sqrt[3]{9} = \sqrt[3]{3^2} = 3^{\frac{2}{3}}\]
d)
\[\sqrt{\frac{1}{3}} = \sqrt{3^{- 1}} = 3^{- \frac{1}{2}}\]
e)
\[\sqrt[5]{81} = \sqrt[5]{3^4} = 3^{\frac{4}{5}}\]
f)
\[\sqrt[3]{243} = \sqrt[3]{3^5} = 3^{\frac{5}{3}}\]
g)
\[3\sqrt{3} = 3 \cdot 3^{\frac{1}{2}} = 3^{1 + \frac{1}{2}} = 3^{\frac{3}{2}}\]
h)
\[9\sqrt[3]{3} = 3^2 \cdot 3^{\frac{1}{3}} = 3^{2 + \frac{1}{3}} = 3^{\frac{7}{3}}\]
Zadanie 7. Oblicz.
a)
\[25^{\frac{1}{2}} + 8^{\frac{1}{3}} = \sqrt{25} + \sqrt[3]{8} = 5 + 2 = 7\]
b)
\[49^{\frac{1}{2}} – 27^{\frac{1}{3}} = \sqrt{49} – \sqrt[3]{27} = 7 – 3 = 4\]
c)
\[32^{\frac{1}{5}} + 81^{\frac{1}{4}} = \sqrt[5]{32} + \sqrt[4]{81} = 2 + 3 = 5\]
d)
\[64^{\frac{1}{6}} – 16^{\frac{1}{4}} = \sqrt[6]{64} – \sqrt[4]{16} = 2 – 2 = 0\]
e)
\[4^{- \frac{1}{2}} + 8^{- \frac{1}{3}} = \sqrt{\frac{1}{4}} + \sqrt[3]{\frac{1}{8}} = \frac{1}{2} + \frac{1}{2} = 1\]
f)
\[9^{- \frac{1}{2}} – 27^{- \frac{1}{3}} = \sqrt{\frac{1}{9}} – \sqrt[3]{\frac{1}{27}} = \frac{1}{3} – \frac{1}{3} = 0\]
g)
\[(\frac{9}{16})^{- \frac{1}{2}} + (\frac{27}{64})^{- \frac{1}{3}} = \sqrt{\frac{16}{9}} + \sqrt[3]{\frac{64}{27}} = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}\]
h)
\[0,008^{- \frac{1}{3}} – 0,25^{- \frac{1}{2}} = \sqrt[3]{\frac{1000}{8}} – \sqrt{\frac{100}{25}} = \frac{10}{2} – \frac{10}{5} = 5 – 2 = 3\]
Zadanie 8. Oblicz.
a)
\[16^{\frac{5}{4}} = (2^4)^{\frac{5}{4}} = 2^{\frac{20}{4}} = 2^5 = 32\]
b)
\[64^{\frac{2}{3}} = (2^6)^{\frac{2}{3}} = 2^{\frac{12}{3}} = 2^4 = 16\]
c)
\[8^{- \frac{5}{3}} = (2^3)^{- \frac{5}{3}} = 2^{- \frac{15}{3}} = 2^{- 5} = (\frac{1}{2})^5 = \frac{1}{32}\]
d)
\[25^{- \frac{3}{2}} = (5^2)^{- \frac{3}{2}} = 5^{- \frac{6}{2}} = 5^{- 3} = (\frac{1}{5})^3 = \frac{1}{125}\]
e)
\[81^{- \frac{3}{4}} = (3^4)^{- \frac{3}{4}} = 3^{- \frac{12}{4}} = 3^{- 3} = (\frac{1}{3})^3 = \frac{1}{27}\]
Zadanie 9. Oblicz.
a)
\[2^{\frac{5}{2}} \cdot \sqrt{2} = 2^{\frac{5}{2}} \cdot 2^{\frac{1}{2}} = 2^{\frac{6}{2}} = 2^3 = 8\]
b)
\[49^{\frac{1}{2}}:49^{\frac{1}{4}} = 7:(7^2)^{\frac{1}{4}} = 7:(7)^{\frac{1}{2}} = 7^{1 – \frac{1}{2}} = 7^{\frac{1}{2}} = \sqrt{7}\]
c)
\[\sqrt{7^5} \cdot 7^{- \frac{3}{2}} = 7^{\frac{5}{2}} \cdot 7^{- \frac{3}{2}} = 7^{\frac{5}{2} – \frac{3}{2}} = 7^{\frac{2}{2}} = 7\]
d)
\[9^{- \frac{3}{4}}:27^{- \frac{3}{2}} = (3^2)^{- \frac{3}{4}}:(3^3)^{- \frac{3}{2}} = 3^{- \frac{6}{4}}:3^{- \frac{9}{2}} = 3^{- \frac{6}{4} – ( – \frac{9}{2})} = 3^{- \frac{6}{4} + \frac{18}{4}} = 3^{\frac{12}{4}} = 3^3 = 27\]