Упр.431 ГДЗ Колмогоров 10-11 класс (Алгебра)

а) $$8^{\frac12} : \left(8^{\frac16}\cdot 9^{\frac32}\right)=8^{\frac12}\cdot 8^{-\frac16}\cdot 9^{-\frac32}=8^{\frac13}\cdot 9^{-\frac32}=2\cdot \frac{1}{27}=\frac{2}{27}.$$

б) $$\sqrt[3]{100}\cdot (\sqrt2)^{\frac83}\cdot \left(\frac15\right)^{\frac53}=\left(2^2\cdot 5^2\right)^{\frac13}\cdot \left(2^{\frac12}\right)^{\frac83}\cdot \frac{1}{5^{\frac53}}$$
$$=2^{\frac23}\cdot 5^{\frac23}\cdot 2^{\frac43}\cdot 5^{-\frac53}=2^2\cdot 5^{-\!1}=\frac45.$$

в) $$8^{2\frac13}:81^{0{,}75}=8^{\frac73}:81^{\frac34}=2^7\cdot 3^{-3}=\frac{128}{27}=4\frac{20}{27}.$$

г) $$\left(1\frac{11}{25}\right)^{-0{,}5}\cdot \left(4\frac{17}{25}\right)^{-\frac13}=\left(\frac{36}{25}\right)^{-\frac12}\cdot \left(\frac{117}{25}\right)^{-\frac13}$$
$$=\left(\frac65\right)^{-1}\cdot \left(\frac53\right)^{-1}=\frac56\cdot \frac35=\frac12.$$

Ответ: а) $$\frac{2}{27}$$; б) $$\frac45$$; в) $$4\frac{20}{27}$$; г) $$\frac12$$.