Упр.512 ГДЗ Колягин Ткачёва 10 класс (Алгебра)

2) 65^0 : 8^-2, 16^1/4* 32^1/5, (1/15)^-1 : 9^1/2, 8^1/3 * (1/2)4 : 16^-1;
3)

1. $$\sqrt[3]{6^3\cdot 5^3}=\sqrt[3]{(6\cdot 5)^3}=6\cdot 5=30.$$

   $$\sqrt[4]{324}\cdot \sqrt[4]{4}=\sqrt[4]{324\cdot 4}=\sqrt[4]{1296}=\sqrt[4]{6^4}=6.$$

   $$\sqrt[4]{15\frac{5}{8}}:\sqrt[4]{\frac{2}{5}}=\sqrt[4]{\frac{125}{8}}:\sqrt[4]{\frac{2}{5}}=\sqrt[4]{\frac{625}{16}}=\frac{5}{2}.$$

2. $$65^0:8^{-2}=1:\left(\frac{1}{8}\right)^2=1\cdot 8^2=64.$$

   $$16^{\frac14}\cdot 32^{\frac15}=(2^4)^{\frac14}\cdot (2^5)^{\frac15}=2\cdot 2=4.$$

   $$\left(\frac{1}{15}\right)^{-1}:9^{\frac12}=15:3=5.$$

   $$8^{\frac13}\cdot \left(\frac12\right)^4:16^{-1}=(2^3)^{\frac13}\cdot \frac{1}{2^4}\cdot 16=2\cdot \frac{1}{16}\cdot 16=2.$$

3. $$\frac{6^{\frac14}\cdot 6^{-\frac14}}{6^2}=\frac{6^0}{6^2}=\frac{1}{36}.$$

   $$\frac{9^{\frac73}\cdot 9^{-\frac43}}{9^2}=\frac{9^{1}}{9^2}=\frac{1}{9}.$$

   $$\frac{(0{,}5)^{0{,}3}\cdot (0{,}5)^{-1}}{(0{,}5)^{1{,}3}}=(0{,}5)^{0{,}3-1-1{,}3}=(0{,}5)^{-2}=2^2=4.$$

Ответ

1) $$30;\ 6;\ \frac{5}{2}.$$
2) $$64;\ 4;\ 5;\ 2.$$
3) $$\frac{1}{36};\ \frac{1}{9};\ 4.$$