Question

10 БАЛЛОВ хелп, алгебра​

Answer 1

$1)\; \; \frac{a-125}{a^{2/3}-25}=\frac{(a^{1/3}-5)(a^{2/3}+5a^{1/3}+25)}{(a^{1/3}-5)(a^{1/3}+5)}=\frac{a^{2/3}+5a^{1/3}+25}{a^{1/3}+5}$

$2)\; \; \frac{24^{1/4}-8^{1/4}}{6^{1/4}-2^{1/4}}=\frac{(6\cdot 4)^{1/4}-(4\cdot 2)^{1/4}}{6^{1/4}-2^{1/4}}=\frac{4^{1/4}\cdot (6^{1/4}-2^{1/4})}{6^{1/4}-2^{1/4}}=4^{\frac{1}{4}}$

$3)\; \; \frac{5^{3/2}\, \cdot \, 8^{1/12}\, \cdot \, 8^{1/4}}{9^{1/6}\, \cdot \, 5^{5/2}\, \cdot \, 9^{1/3}}=\frac{8^{1/3}}{9^{1/2}\, \cdot \, 5}=\frac{2}{3\, \cdot \, 5}=\frac{2}{15}$

$4)\; \; \Big(\frac{3^{-5/6}\, \cdot \, 7^{-5/6}}{21^{-1}\, \cdot \, 5^{1/3}}\Big)^{-6}=\Big(\frac{21^{-5/6}\, \cdot \, 21}{5^{1/3}}\Big)^{-6}=\frac{21^{5}\, \cdot \, 21^{-6}}{5^{-2}}=\frac{21^{-1}}{5^{-2}}=\frac{5^2}{21}=\\\\=\frac{25}{21}=1\frac{4}{21}$