Упр.10.15 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) 12^(1/3)·6^(2/3)·(0,5)^(1/3);
2) 25^1,5+(0,25)^(-0,5)-81^0,75;
3) (1/16)^(-3/4)+(1/8)^(-2/3)·(0,81)^(-0,5);
4) 16^(1/8)·8^(-5/6)·4^1,5;
5) (10 000^0,4·10^0,5)/(100^0,3·1 000^(1/6));
6) (5^(3/2)·8^(1/12))/9^(1/6)·8^(1/4)/(5^(5/2)·9^(1/3));
7) (72^(2/3))^(1/2)·2^(-4/3):36^(-1/6);
8) (((3^(-5/6)·7^(-5/6))/(21^(-1)·5^(1/3)))^(-6).

1. $$12^{\frac13}\cdot 6^{\frac23}\cdot (0{,}5)^{\frac13}=(6\cdot 2)^{\frac13}\cdot 6^{\frac23}\cdot \left(\frac12\right)^{\frac13}=6^{\frac13}\cdot 2^{\frac13}\cdot 6^{\frac23}\cdot 2^{-\frac13}=6^{\frac13+\frac23}\cdot 2^{\frac13-\frac13}=6^1\cdot 2^0=6.$$
2. $$ 25^{1{,}5}+(0{,}25)^{-0{,}5}-81^{0{,}75} =(5^2)^{\frac32}+\left(\frac14\right)^{-\frac12}-(3^4)^{\frac34} =5^3+4^{\frac12}-3^3 =125+2-27=100. $$
3. $$ \left(\frac1{16}\right)^{-\frac34}+\left(\frac18\right)^{-\frac23}\cdot (0{,}81)^{-0{,}5} =(2^4)^{\frac34}+(2^3)^{\frac23}\cdot \left(\frac{81}{100}\right)^{-\frac12} =2^3+2^2\cdot \frac{10}{9} =8+\frac{40}{9} =\frac{112}{9}=12\frac49. $$
4. $$ 16^{\frac18}\cdot 8^{-\frac56}\cdot 4^{1{,}5} =(2^4)^{\frac18}\cdot (2^3)^{-\frac56}\cdot (2^2)^{\frac32} =2^{\frac12}\cdot 2^{-\frac52}\cdot 2^3 =2^{\frac12-\frac52+3}=2. $$
5. $$ \frac{10\,000^{0{,}4}\cdot 10^{0{,}5}}{100^{0{,}3}\cdot 1000^{\frac16}} =\frac{(10^4)^{0{,}4}\cdot 10^{0{,}5}}{(10^2)^{0{,}3}\cdot (10^3)^{\frac16}} =\frac{10^{1{,}6}\cdot 10^{0{,}5}}{10^{0{,}6}\cdot 10^{0{,}5}} =10^{1{,}6-0{,}6}=10. $$
6. $$ \frac{5^{\frac32}\cdot 8^{\frac1{12}}}{9^{\frac16}}\cdot \frac{8^{\frac14}}{5^{\frac52}\cdot 9^{\frac13}} =\frac{5^{\frac32-\frac52}\cdot 8^{\frac1{12}+\frac14}}{9^{\frac16+\frac13}} =\frac{5^{-1}\cdot 8^{\frac13}}{9^{\frac12}} =\frac{1}{5}\cdot \frac{2}{3} =\frac{2}{15}. $$
7. $$ \left(72^{\frac23}\right)^{\frac12}\cdot 2^{-\frac43}:36^{-\frac16} =72^{\frac13}\cdot 2^{-\frac43}\cdot 36^{\frac16}. $$
   Представим числа в виде степеней двойки и тройки:
   $$ 72=2^3\cdot 3^2,\qquad 36=2^2\cdot 3^2. $$
   Тогда
   $$ 72^{\frac13}=(2^3\cdot 3^2)^{\frac13}=2\cdot 3^{\frac23},\qquad 36^{\frac16}=(2^2\cdot 3^2)^{\frac16}=2^{\frac13}\cdot 3^{\frac13}. $$
   Получаем
   $$ 2\cdot 3^{\frac23}\cdot 2^{-\frac43}\cdot 2^{\frac13}\cdot 3^{\frac13} =2^{1-\frac43+\frac13}\cdot 3^{\frac23+\frac13} =2^0\cdot 3^1=3. $$
8. $$ \left(\frac{3^{-\frac56}\cdot 7^{-\frac56}}{21^{-1}\cdot 5^{\frac13}}\right)^{-6} =\left(\frac{(3\cdot 7)^{-\frac56}}{21^{-1}\cdot 5^{\frac13}}\right)^{-6} =\left(\frac{21^{-\frac56}}{21^{-1}\cdot 5^{\frac13}}\right)^{-6} =\left(\frac{21^{\frac16}}{5^{\frac13}}\right)^{-6}. $$
   Тогда
   $$ \left(\frac{21^{\frac16}}{5^{\frac13}}\right)^{-6} =21^{-1}\cdot 5^2 =\frac{25}{21} =1\frac{4}{21}. $$

Ответ

1) $$6$$; 2) $$100$$; 3) $$12\frac49$$; 4) $$2$$; 5) $$10$$; 6) $$\frac{2}{15}$$; 7) $$3$$; 8) $$1\frac{4}{21}$$.