Odpowiedź i szczegółowe wyjaśnienie:
[tex]a)\\\\sqrt{1:4}=\sqrt1:\sqrt4=1:2=\dfrac12\\\\\\\sqrt{4:25}=\sqrt4:\sqrt{25}=2:5=\dfrac25\\\\\\\sqrt{\dfrac{16}{81}}=\dfrac{\sqrt{16}}{\sqrt{81}}=\dfrac49\\\\\\\sqrt{\dfrac{49}{100}}=\dfrac{\sqrt{49}}{\sqrt{100}}=\dfrac{7}{10}[/tex]
[tex]b)\\\sqrt[3]{64:(-125)}}=\sqrt[3]{64}:\sqrt[3]{-125}=\sqrt[3]{4^3}:\sqrt[3]{(-5)^3}=4:(-5)=-\dfrac45\\\\\\\sqrt[3]{(-8)^3:27}=\sqrt[3]{-8}:\sqrt[3]{27}=\sqrt[3]{(-2)^3}:\sqrt[3]{3^3}=-\dfrac23\\\\\\\sqrt[3]{\dfrac{1}{729}}=\dfrac{\sqrt[3]1}{\sqrt[3]{729}}=\dfrac{1}{\sqrt[3]{9^3}}=\dfrac19\\\\\\\sqrt[3]{\dfrac{343}{8000}}=\dfrac{\sqrt[3]{343}}{\sqrt[3]{8000}}=\dfrac{\sqrt[3]{7^3}}{20^3}=\dfrac{7}{20}[/tex]
[tex]c)\\\sqrt[4]{(-1):(-16)}=\sqrt[4]{1:16}=\sqrt[4]1:\sqrt[4]{16}=\sqrt[4]{1}:\sqrt[4]{2^4}=1:2=\dfrac12\\\\\\\sqrt[4]{81:16}=\sqrt[4]{81}:\sqrt[4]{16}=\sqrt[4]{3^4}:\sqrt[4]{2^4}=3:4=\dfrac34\\\\\\\sqrt[5]{\dfrac{32}{3125}}=\dfrac{\sqrt[5]{32}}{\sqrt[5]{3125}}=\dfrac{\sqrt[5]{2^5}}{\sqrt[5]{5^5}}=\dfrac25\\\\\\\sqrt[5]{\dfrac{243}{100000}}=\dfrac{\sqrt[5]{243}}{\sqrt[5]{100000}}=\dfrac{\sqrt[5]{3^5}}{10^5}=\dfrac{3}{10}[/tex]
Wykorzystano podstawowe własności pierwiastkowania:
[tex]\sqrt[n]{a^n}=a\\\\\sqrt[n]{a\cdot b}=\sqrt[n]a\cdot \sqrt[n]b\\\\\sqrt[n]{a:b}=\sqrt[n]a:\sqrt[n]b[/tex]
