Hallo,
\( \displaystyle a^{\frac{m}{n}}=\sqrt[n]{a^{m}} \)
e) \( \displaystyle \sqrt{\sqrt{2}}=\sqrt{2^{\frac{1}{2}}}=\left(2^{\frac{1}{2}}\right)^{\frac{1}{2}}=2^{\frac{1}{4}}=\sqrt[4]{2} \)
f) \(\displaystyle \sqrt[4]{9} \cdot \sqrt[4]{3}=9^{\frac{1}{4}} \cdot 3^{\frac{1}{4}}=27^{\frac{1}{4}}=\sqrt[4]{27} \)
9) \(\displaystyle \sqrt[3]{7} \cdot \sqrt[5]{7}=7^{\frac{1}{3}} \cdot 7^{\frac{1}{5}}=7^{\frac{1}{3}+\frac{1}{5}}=7^{\frac{8}{15}}=\sqrt[15]{7^{8}} \)
h) \(\displaystyle \sqrt{\sqrt[3]{5}}=\sqrt[3]{5^{\frac{1}{2}}}=\left(5^{\frac{1}{3}}\right)^{\frac{1}{2}}=5^{\frac{1}{6}}=\sqrt[6]{5} \)
i) \(\displaystyle \sqrt[4]{2} \cdot \sqrt[4]{32}=2^{\frac{1}{4}} \cdot 32^{\frac{1}{4}} = 64^{\frac{1}{4}}=\sqrt[4]{64}=\sqrt[4]{2^{6}}=\sqrt{2^{3}}=2\sqrt{2} \)
j) \(\displaystyle \sqrt[n]{3}: \sqrt[2 n]{3}=3^{\frac{1}{n}}: 3^{\frac{1}{2 n}}=3^{\frac{1}{n}-\frac{1}{2 n}}=3^{\frac{1}{2 n}}=\sqrt[2 n]{3} \)
k) \(\displaystyle \sqrt[4]{10 y} :\sqrt[4]{2 y} =10 y^\frac{1}{4}: 2 y^{\frac{1}{4}}=5^{\frac{1}{4}}=\sqrt[4]{5} \)
l) \( \displaystyle \sqrt[n]{\sqrt[3 ]{a}}=\sqrt[3]{a}^{\frac{1}{n}}=\left(a^{\frac{1}{3}}\right)^{\frac{1}{n}}=a^{\frac{1}{3 n}}=\sqrt[3 n]{a} \)
Bei Fragen bitte melden.
Gruß, Silvia
