Bereken m.b.v. de rekenregels (zonder ZRM)
- \(\sqrt[8]{ (\frac{81}{16})^{2} }\)
- \(\sqrt[6]{ (8)^{2} }\)
- \(\sqrt[12]{ (\frac{81}{16})^{3} }\)
- \(\sqrt[6]{ (\frac{361}{400})^{3} }\)
- \(\sqrt[4]{ (\frac{1}{2})^{-12} }\)
- \(\sqrt[12]{ (\frac{64}{27})^{4} }\)
- \(\sqrt[9]{ (\frac{27}{64})^{3} }\)
- \( \sqrt{ (\frac{3}{2})^{8} } \)
- \(\sqrt[3]{ (\frac{17}{18})^{-6} }\)
- \(\sqrt[16]{ (\frac{16}{81})^{4} }\)
- \(\sqrt[4]{ (\frac{1}{2})^{12} }\)
- \(\sqrt[6]{ (\frac{16}{25})^{3} }\)
Bereken m.b.v. de rekenregels (zonder ZRM)
Verbetersleutel
- \(\sqrt[8]{ (\frac{81}{16})^{2} }\\= (\frac{81}{16})^{\frac{2}{8}}\\= (\frac{81}{16})^{\frac{1}{4}}\\=\sqrt[4]{ \frac{81}{16} }=\frac{3}{2}\)
- \(\sqrt[6]{ (8)^{2} }\\= (8)^{\frac{-2}{6}}\\= (8)^{\frac{-1}{3}}\\=\sqrt[3]{ \frac{1}{8} }=\frac{1}{2}\)
- \(\sqrt[12]{ (\frac{81}{16})^{3} }\\= (\frac{81}{16})^{\frac{3}{12}}\\= (\frac{81}{16})^{\frac{1}{4}}\\=\sqrt[4]{ \frac{81}{16} }=\frac{3}{2}\)
- \(\sqrt[6]{ (\frac{361}{400})^{3} }\\= (\frac{361}{400})^{\frac{3}{6}}\\= (\frac{361}{400})^{\frac{1}{2}}\\= \sqrt{ \frac{361}{400} } =\frac{19}{20}\)
- \(\sqrt[4]{ (\frac{1}{2})^{-12} }\\= (\frac{1}{2})^{\frac{-12}{4}}\\= (\frac{1}{2})^{-3}\\= (2)^{3}= 8\)
- \(\sqrt[12]{ (\frac{64}{27})^{4} }\\= (\frac{64}{27})^{\frac{-4}{12}}\\= (\frac{64}{27})^{\frac{-1}{3}}\\=\sqrt[3]{ \frac{27}{64} }=\frac{3}{4}\)
- \(\sqrt[9]{ (\frac{27}{64})^{3} }\\= (\frac{27}{64})^{\frac{3}{9}}\\= (\frac{27}{64})^{\frac{1}{3}}\\=\sqrt[3]{ \frac{27}{64} }=\frac{3}{4}\)
- \( \sqrt{ (\frac{3}{2})^{8} } \\= (\frac{3}{2})^{\frac{8}{2}}\\= (\frac{3}{2})^{4}=\frac{81}{16}\)
- \(\sqrt[3]{ (\frac{17}{18})^{-6} }\\= (\frac{17}{18})^{\frac{-6}{3}}\\= (\frac{17}{18})^{-2}\\= (\frac{18}{17})^{2}= \frac{324}{289}\)
- \(\sqrt[16]{ (\frac{16}{81})^{4} }\\= (\frac{16}{81})^{\frac{-4}{16}}\\= (\frac{16}{81})^{\frac{-1}{4}}\\=\sqrt[4]{ \frac{81}{16} }=\frac{3}{2}\)
- \(\sqrt[4]{ (\frac{1}{2})^{12} }\\= (\frac{1}{2})^{\frac{12}{4}}\\= (\frac{1}{2})^{3}=\frac{1}{8}\)
- \(\sqrt[6]{ (\frac{16}{25})^{3} }\\= (\frac{16}{25})^{\frac{3}{6}}\\= (\frac{16}{25})^{\frac{1}{2}}\\= \sqrt{ \frac{16}{25} } =\frac{4}{5}\)