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Bereken m.b.v. de rekenregels (zonder ZRM)

1. \(\sqrt[6]{ (\frac{196}{225})^{3} }\)
2. \(\sqrt[12]{ (\frac{27}{64})^{4} }\)
3. \(\sqrt[3]{ (\frac{7}{12})^{-6} }\)
4. \(\sqrt[6]{ (\frac{1}{8})^{2} }\)
5. \( \sqrt{ (2)^{6} } \)
6. \(\sqrt[12]{ (8)^{4} }\)
7. \(\sqrt[4]{ (\frac{5}{4})^{-8} }\)
8. \(\sqrt[6]{ (\frac{64}{169})^{3} }\)
9. \(\sqrt[8]{ (\frac{81}{16})^{2} }\)
10. \(\sqrt[4]{ (\frac{361}{400})^{2} }\)
11. \(\sqrt[4]{ (\frac{2}{3})^{-16} }\)
12. \(\sqrt[4]{ (\frac{1}{2})^{-12} }\)

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Bereken m.b.v. de rekenregels (zonder ZRM)

Verbetersleutel

1. \(\sqrt[6]{ (\frac{196}{225})^{3} }\\= (\frac{196}{225})^{\frac{-3}{6}}\\= (\frac{196}{225})^{\frac{-1}{2}}\\= \sqrt{ \frac{225}{196} } =\frac{15}{14}\)
2. \(\sqrt[12]{ (\frac{27}{64})^{4} }\\= (\frac{27}{64})^{\frac{-4}{12}}\\= (\frac{27}{64})^{\frac{-1}{3}}\\=\sqrt[3]{ \frac{64}{27} }=\frac{4}{3}\)
3. \(\sqrt[3]{ (\frac{7}{12})^{-6} }\\= (\frac{7}{12})^{\frac{-6}{3}}\\= (\frac{7}{12})^{-2}\\= (\frac{12}{7})^{2}= \frac{144}{49}\)
4. \(\sqrt[6]{ (\frac{1}{8})^{2} }\\= (\frac{1}{8})^{\frac{2}{6}}\\= (\frac{1}{8})^{\frac{1}{3}}\\=\sqrt[3]{ \frac{1}{8} }=\frac{1}{2}\)
5. \( \sqrt{ (2)^{6} } \\= (2)^{\frac{6}{2}}\\= (2)^{3}=8\)
6. \(\sqrt[12]{ (8)^{4} }\\= (8)^{\frac{4}{12}}\\= (8)^{\frac{1}{3}}\\=\sqrt[3]{ 8 }=2\)
7. \(\sqrt[4]{ (\frac{5}{4})^{-8} }\\= (\frac{5}{4})^{\frac{-8}{4}}\\= (\frac{5}{4})^{-2}\\= (\frac{4}{5})^{2}= \frac{16}{25}\)
8. \(\sqrt[6]{ (\frac{64}{169})^{3} }\\= (\frac{64}{169})^{\frac{3}{6}}\\= (\frac{64}{169})^{\frac{1}{2}}\\= \sqrt{ \frac{64}{169} } =\frac{8}{13}\)
9. \(\sqrt[8]{ (\frac{81}{16})^{2} }\\= (\frac{81}{16})^{\frac{-2}{8}}\\= (\frac{81}{16})^{\frac{-1}{4}}\\=\sqrt[4]{ \frac{16}{81} }=\frac{2}{3}\)
10. \(\sqrt[4]{ (\frac{361}{400})^{2} }\\= (\frac{361}{400})^{\frac{-2}{4}}\\= (\frac{361}{400})^{\frac{-1}{2}}\\= \sqrt{ \frac{400}{361} } =\frac{20}{19}\)
11. \(\sqrt[4]{ (\frac{2}{3})^{-16} }\\= (\frac{2}{3})^{\frac{-16}{4}}\\= (\frac{2}{3})^{-4}\\= (\frac{3}{2})^{4}= \frac{81}{16}\)
12. \(\sqrt[4]{ (\frac{1}{2})^{-12} }\\= (\frac{1}{2})^{\frac{-12}{4}}\\= (\frac{1}{2})^{-3}\\= (2)^{3}= 8\)