Przedstaw podaną liczbę...- Zadanie 5: NOWA MATeMAtyka 2. Maturalne karty pracy. Zakres podstawowy i rozszerzony

Rozwiązanie

Zapiszmy dane liczby tak, aby w podstawie miały liczby całkowite.

$9^{\frac{2}{3} 3} \cdot 9^{\frac{1}{3}} = 9^{\frac{2 3}{3}} \cdot 9^{\frac{3}{3}} = 9^{\frac{2 3}{3} + \frac{3}{3}} = 9^{\frac{3 3}{3}} = 9^{3}$

$1 0^{28} : 1 0^{82} = 1 0^{42} : 1 0^{82} = 1 0^{42 - 82} = 1 0^{- 42}$

$\frac{( \frac{1}{2} ) ^{\frac{1}{2} 2} : ( \frac{1}{2} ) ^{\frac{1}{2} 8}}{( \frac{1}{2} ) ^{\frac{1}{2} 32}} = \frac{( \frac{1}{2} ) ^{\frac{2}{2}} : ( \frac{1}{2} ) ^{2}}{( \frac{1}{2} ) ^{22}} = \frac{( \frac{1}{2} ) ^{- \frac{2}{2}}}{( \frac{1}{2} ) ^{22}} = (\frac{1}{2})^{- \frac{5 2}{2}} = 2^{\frac{5 2}{2}}$

$1 2^{\frac{2}{2}} \cdot 0, 5^{0, 52} = 1 2^{\frac{2}{2}} \cdot (\frac{1}{2})^{\frac{2}{2}} = (12 \cdot \frac{1}{2})^{\frac{2}{2}} = 6^{\frac{2}{2}}$

$2 0^{5} \cdot (25)^{25} = 2 0^{5} \cdot (4 \cdot 5)^{5} = 2 0^{5} \cdot 2 0^{5} = (20 \cdot 20)^{5} = 40 0^{5}$

$1 8^{3 - 1} : (\frac{1}{3})^{1 - 3} = 1 8^{3 - 1} : 3^{3 - 1} = (18 : 3)^{3 - 1} = 6^{3 - 1}$

$2 4^{6} : (3^{6} \cdot 8^{6}) = 2 4^{6} : 2 4^{6} = 2 4^{0} = 1$

$(\frac{1}{4})^{- \frac{2}{2}} \cdot (\frac{1}{8})^{22} = (2^{- 2})^{- \frac{2}{2}} \cdot (2^{- 3})^{22} = 2^{2} \cdot 2^{- 62} = 2^{- 52}$

$(5^{\frac{1}{5 - 1}})^{5 + 1} = 5^{\frac{5 + 1}{5 - 1}} = 5^{\frac{( 5 + 1 ) ^{2}}{5 - 1}} = 5^{\frac{6 + 2 5}{4}} = 5^{\frac{3 + 5}{2}} = 5^{\frac{3 + 5}{4}}$

$0, 12 5^{1 - 2} : 0, 2 5^{1 + 2} = (2^{- 3})^{1 - 2} : (2^{- 2})^{1 + 2} = 2^{- 3 + 32} : 2^{- 2 - 22} = 2^{- 1 + 52}$

$\frac{0 , 1 ^{10 - 3} \cdot 2 ^{10}}{8} = \frac{1 0 ^{3 - 10} \cdot 2 ^{10}}{2 ^{3}} = \frac{2 ^{3 - 10} \cdot 5 ^{3 - 10} \cdot 2 ^{10}}{2 ^{3}} = \frac{2 ^{3} \cdot 5 ^{3 - 10}}{2 ^{3}} = 5^{3 - 10}$

$\frac{6 ^{6 - 2}}{( 3 ^{2} ) ^{3 - 1}} = \frac{( 6 ^{\frac{1}{2}} ) ^{6 - 2}}{( 3 ^{\frac{1}{2}} ) ^{2 \cdot (3 - 1)}} = \frac{6 ^{\frac{6 - 2}{2}}}{3 ^{\frac{6 - 2}{2}}} = (6 : 3)^{\frac{6 - 2}{2}} = 2^{\frac{6 - 2}{2}}$