oblicz: potegowanie

czubek1 · Post autor: **czubek1** » 16 sie 2011, o 18:54

\(\displaystyle{ \frac{ \sqrt[3]{90} \cdot \sqrt[8]{4}} { \sqrt[4]{20 \cdot \sqrt[3]{ 3^{2} \cdot 10 } } }}\)

aalmond · Post autor: **aalmond** » 16 sie 2011, o 19:00

\(\displaystyle{ \sqrt[8]{4} = \sqrt[4]{2}}\)-- 16 sierpnia 2011, 19:02 --\(\displaystyle{ \sqrt[4]{20 \cdot \sqrt[3]{ 3^{2} \cdot 10 } } = \sqrt[4]{20} \cdot \sqrt[12]{90}}\)

irena_1 · Post autor: **irena_1** » 16 sie 2011, o 19:21

\(\displaystyle{ = \frac{3^{ \frac{2}{3} }\cdot2^{ \frac{1}{3}} \cdot5^{ \frac{1}{3} }\cdot2^ {\frac{1}{4} }}{2^ {\frac{1}{2} }\cdot5^ {\frac{1}{4} }\cdot3^ {\frac{1}{6} }\cdot2^{ \frac{1}{12} }\cdot5^ {\frac{1}{12}} } =\\=3^{ \frac{2}{3} - \frac{1}{6} }\cdot2^{ \frac{1}{3} + \frac{1}{4} - \frac{1}{2} - \frac{1}{12} }\cdot5^{ \frac{1}{3} - \frac{1}{4} - \frac{1}{12} }=\\=3^ \frac{1}{2} \cdot2^0\cdot5^0=\sqrt{3}\cdot1\cdot1=\sqrt{3}}\)