Rozwiązane

[tex]a) \ 81^{\frac{1}{4} } =\\b) \ (\frac{8}{27})^{-1\frac{1}{3}} =\\c) \ (\frac{1}{16})^{-\frac{1}{4} } =[/tex]



Odpowiedź :

AstraySpirit2308

1)

[tex]81^{\frac{1}{4}}=\sqrt[4]{81}=\sqrt[4]{3^4}=3\\[/tex]

2)

[tex](\frac{8}{27})^{-1\frac{1}{3}}=(\frac{8}{27})^{-\frac{4}{3}}=(\frac{27}{8})^{\frac{4}{3}}=((\frac{3}{2})^3)^{\frac{4}{3}}=(\frac{3}{2})^4=\frac{3^4}{2^4}=\frac{81}{16}=5\frac{1}{16}\\[/tex]

3)

[tex](\frac{1}{16})^{-\frac{1}{4}}=16^{\frac{1}{4}}=\sqrt[4]{16}=\sqrt[4]{2^4}=2[/tex]

MatmFizyk

Przypomnijmy sobie:

[tex]a^{\frac{m}{n}} = (\sqrt[n]{a})^m[/tex]

Przykłady:

[tex]a) \ 81^{\frac{1}{4}}= (\sqrt[4]{81})^1 = \boxed{3}[/tex]

[tex]b) \ (\frac{8}{27})^{-1\frac{1}{3}}= (\sqrt{\frac{27}{8}})^4 = (\frac{3}{2})^4= \frac{81}{16} = \boxed{5\frac{1}{16}}[/tex]

[tex]c) \ (\frac{1}{16})^{-\frac{1}{4}}= (\sqrt[4]{16})^1 = \boxed{2}[/tex]

Inne Pytanie