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[tex]a) \ 2^{82}:4^{40}= 2^{82}:(2^{2})^{40} = 2^{82}:2^{2\cdot40} = 2^{82}:2^{80} = 2^{82-80} = 2^{2} = 4\\\\\\b) \ \frac{18^{4}}{9^{4}} = (\frac{18}{9})^{4} = 2^{4} = 16\\\\\\c) \ \frac{2^{15}\cdot27^{4}}{6^{15}} = \frac{2^{15}\cdot(3^{3})^{4}}{(2\cdot3)^{15}} = \frac{2^{15}\cdot3^{12}}{2^{15}\cdot3^{15}} = \frac{3^{12}}{3^{15}} = 3^{12-15} = 3^{-3} = (\frac{1}{3})^{3} = \frac{1}{27}[/tex]

[tex]d) \ (5^{2})^{3}\cdot2^{6} = 5^{5}\cdot2^{6} = (5\cdot2)^{6} = 10^{6} = 1 \ 000 \ 000[/tex]

[tex]e) \ 0,2^{8}\cdot25^{4} = (\frac{2}{10})^{8}\cdot(5^{2})^{4} = (\frac{1}{5})^{8}\cdot5^{8} =5^{-8}\cdot5^{8} = 5^{-5+5} = 5^{0} = 1[/tex]

[tex]f) \ 0,01^{12}\cdot100^{9} =(\frac{1}{100})^{12}\cdot100^{9} =100^{-12}\cdot100^{9} = 100^{-12+9} = 100^{-3} = (\frac{1}{100})^{3} =\\\\=0,000001[/tex]

[tex]g) \ 25^{3}\cdot0,5^{-6} = (5^{2})^{3}\cdot(\frac{5}{10})^{-6} = 5^{6}\cdot(\frac{1}{2})^{-6} = 5^{6}\cdot2^{6} = (5\cdot2)^{6} = 10^{6} = 1 \ 000 \ 000[/tex]

[tex]h) \ 5^{8}:0,2^{-7} = 5^{8}:(\frac{2}{10})^{-7} = 5^{8}:(\frac{1}{5})^{-7} = 5^{8}:5^{7} = 5[/tex]

[tex]i) \ \frac{125^{3}\cdot(5^{-2})^{4}}{25^{4}\cdot25^{-6}} = \frac{(5^{3})^{3}\cdot5^{-8}}{(5^{2})^{4}\cdot(5^{2})^{-5}}=\frac{5^{9}\cdot5^{-8}}{5^{8}\cdot5^{-10}}=\frac{5^{1}}{5^{-2}} = 5^{1-(-2)} = 5^{3} = 125[/tex]

[tex]j) \ \frac{6^{-3}\cdot(6^{2})^{-5}}{(\frac{1}{36})^{6}} = \frac{6^{-3}\cdot6^{-10}}{36^{-6}} = \frac{6^{-13}}{(6^{2})^{-6}} = \frac{6^{-13}}{6^{-12}} = 6^{-13-(-12)} = 6^{-1} = \frac{1}{6}[/tex]

[tex]k) \ \frac{(121^{2})^{3}}{(\frac{1}{11})^{-8}}\cdot11^{-2} = \frac{((11^{2})^{2})^{3}}{11^{8}}\cdot11^{-2} = \frac{11^{12}}{11^{8}}\cdot11^{-2} = 11^{4}\cdot11^{-2} = 11^{2} =121[/tex]

[tex]l) \ \frac{7^{5}:49}{7}\cdot(\frac{1}{7})^{-2} = \frac{7^{5}:7^{2}}{7^{1}}\cdot7^{2} = \frac{7^{3}}{7^{1}}\cdot7^{2} = 7^{2}\cdot7^{2} = 7^{4} = 2401[/tex]

Korzystamy z twierdzeń dotyczących potęgi:

[tex]a^{0} = 1 \ \ \ dla \ a\neq 0\\\\a^{1} = a\\\\a^{m}\cdot a^{n} = a^{m+n}\\\\a^{m}:a^{n} = a^{m-n}\\\\a^{n}\cdot b^{n}=(a\cdot b)^{n}\\\\a^{n}:b^{n} = (a:b)^{n}\\\\(a^{m})^{n} = a^{m\cdot n}\\\\(\frac{a}{b})^{-n} = (\frac{b}{a})^{n}[/tex]

MertBviz MertBviz · Answer 2

Cześć!

Szczegółowe wyjaśnienie:

15. Oblicz.

a)

[tex]2 {}^{82} : 4 {}^{40} = 2 {}^{82} : 2 {}^{80} = 2 {}^{82 - 80} = 2 {}^{2} = 4[/tex]

b)

[tex] \frac{18 {}^{4} }{9 {}^{4} } = \frac{2 {}^{4} \cdot9 {}^{4} }{9 {}^{4} } = 2 {}^{4} = 16[/tex]

c)

[tex] \frac{2 {}^{15} \cdot27 {}^{4} }{6 {}^{15} } = \frac{2 {}^{15} \cdot3 {}^{12} }{3 {}^{15} \cdot2 {}^{15} } = \frac{1}{3 {}^{3} } = \frac{1}{27} [/tex]

d)

[tex](5 {}^{2} ) {}^{3} \cdot2 {}^{6} = 5 {}^{2 \cdot3} = 5 {}^{6} \cdot2 {}^{6} = 10 {}^{6} [/tex]

e)

[tex]0.2 {}^{8} \cdot25 {}^{4} = 5 {}^{ - 8} \cdot5 {}^{8} = 1[/tex]

f)

[tex]0.01 {}^{12} \cdot100 {}^{9} = 10 {}^{ - 24} \cdot10 {}^{18} = 10 {}^{ - 6} = \frac{1}{10 {}^{6} } [/tex]

g)

[tex]25 {}^{3} \cdot0.5 {}^{ - 6} = 25 {}^{3} \cdot (\frac{1}{2} ) {}^{ - 6} = 25 {}^{3} \cdot2 {}^{6} = 50 {}^{3} \cdot2 {}^{3} = (50 \cdot2) {}^{3} = 100 {}^{3} [/tex]

h)

[tex]5 {}^{8} : 0.2 {}^{ - 7} = 5 {}^{8} : 5 {}^{7} = 5 {}^{8 - 7} = 5 {}^{1} = 5[/tex]

i)

[tex] \frac{125 {}^{3} \cdot(5 {}^{ - 2} ) {}^{4} }{25 {}^{4} \cdot25 {}^{ - 5} } = \frac{5 {}^{9} \cdot5 {}^{ - 8} }{5 {}^{ - 2} } = 5 {}^{11} \cdot5 {}^{ - 8} = 5 {}^{11 - 8} = 5 {}^{3} [/tex]

j)

[tex] \frac{6 {}^{ - 3} \cdot(6 {}^{2}) {}^{ - 5} }{( \frac{1}{36}) {}^{6} } = \frac{6 {}^{ - 3} \cdot6 {}^{ - 10} }{6 {}^{ - 12} } = \frac{6 {}^{ - 13} }{6 {}^{ - 12} } = \frac{1}{6} [/tex]

k)

[tex] \frac{(121 {}^{2} ) {}^{3} }{( \frac{1}{11} ) {}^{ - 8} } \cdot11 {}^{ - 2} = \frac{121 {}^{6} }{11 {}^{8} } \cdot11 {}^{ - 2} = \frac{11 {}^{12} }{11 {}^{ - 8} } \cdot11 {}^{ - 2} = 11 {}^{12 - 8} = 11 {}^{4} \cdot11 {}^{ - 2} = 11 {}^{4 - 2} = 11 {}^{2} = 121[/tex]

l)

[tex] \frac{7 {}^{5} : 49}{7} \cdot( \frac{1}{7} ) {}^{ - 2} = \frac{7 {}^{5} : 7 {}^{2} }{7} \cdot7 {}^{2} = \frac{7 {}^{3} }{7} \cdot7 {}^{2} = 7 {}^{2} \cdot7 {}^{2} = 7 {}^{2 + 2} = 7 {}^{4} [/tex]