Упр.23.22 ГДЗ Мерзляк Поляков 9 класс (Алгебра)

1) (343^(1/2)·(1/49)^(3/8))^(4/3);
2) 10^(1/4)·40^(1/4)·5^(1/2);
3) 0,0016^(-3/4)-0,04^(-1/2)+0,216^(-2/3);
4) (32^0,24·4^0,7)/(64^0,6·16^0,25);
5) (12^(1/2))/(7^(2/3)·8^(-1/6))·(3^(1/2)·7^(5/3))/8^(1/2);
6) ((5^(-2/3)·3^(-2/3))/(15^(2/3)·2^(-16/3)))^(-1,5).

1. $$\left(343^{\frac12}\cdot\left(\frac1{49}\right)^{\frac38}\right)^{\frac43}$$
   $$=\left((7^3)^{\frac12}\cdot\left(\frac1{7^2}\right)^{\frac38}\right)^{\frac43}$$
   $$=\left(7^{\frac32}\cdot 7^{-\frac34}\right)^{\frac43} =\left(7^{\frac34}\right)^{\frac43}=7.$$

2. $$10^{\frac14}\cdot 40^{\frac14}\cdot 5^{\frac12} =(5\cdot 2)^{\frac14}\cdot(5\cdot 8)^{\frac14}\cdot 5^{\frac12}$$
   $$=5^{\frac14}\cdot 2^{\frac14}\cdot 5^{\frac14}\cdot 2^{\frac34}\cdot 5^{\frac12} =5^{\frac14+\frac14+\frac12}\cdot 2^{\frac14+\frac34}$$
   $$=5\cdot 2=10.$$

3. $$0{,}0016^{-\frac34}-0{,}04^{-\frac12}+0{,}216^{-\frac23}$$
   $$=\left(\frac{16}{10000}\right)^{-\frac34}-\left(\frac{4}{100}\right)^{-\frac12}+\left(\frac{216}{1000}\right)^{-\frac23}$$
   $$=\left(\frac{2^4}{10^4}\right)^{-\frac34}-\left(\frac{2^2}{10^2}\right)^{-\frac12}+\left(\frac{6^3}{10^3}\right)^{-\frac23}$$
   $$=\left(\frac{2}{10}\right)^{-3}-\left(\frac{2}{10}\right)^{-1}+\left(\frac{6}{10}\right)^{-2}$$
   $$=\left(\frac{10}{2}\right)^3-\frac{10}{2}+\left(\frac{10}{6}\right)^2$$
   $$=5^3-5+\left(\frac53\right)^2=125-5+\frac{25}{9}=120+\frac{25}{9}=122\frac79.$$

4. $$\frac{32^{0{,}24}\cdot 4^{0{,}7}}{64^{0{,}6}\cdot 16^{0{,}25}} =\frac{(2^5)^{0{,}24}\cdot(2^2)^{0{,}7}}{(2^6)^{0{,}6}\cdot(2^4)^{0{,}25}}$$
   $$=\frac{2^{1{,}2}\cdot 2^{1{,}4}}{2^{3{,}6}\cdot 2^1} =2^{1{,}2+1{,}4-3{,}6-1}=2^{-2}=\frac14=0{,}25.$$

5. $$\frac{12^{\frac12}}{7^{\frac23}\cdot 8^{-\frac16}}\cdot \frac{3^{\frac12}\cdot 7^{\frac53}}{8^{\frac12}}$$
   $$=\frac{(3\cdot 4)^{\frac12}\cdot 3^{\frac12}\cdot 7^{\frac53}}{7^{\frac23}\cdot (2^3)^{-\frac16}\cdot (2^3)^{\frac12}}$$
   $$=\frac{3^{\frac12}\cdot 2\cdot 3^{\frac12}\cdot 7^{\frac53}}{7^{\frac23}\cdot 2^{-\frac12}\cdot 2^{\frac32}} =\frac{3\cdot 2\cdot 7}{2^1}=21.$$

6. $$\left(\frac{5^{-\frac23}\cdot 3^{-\frac23}}{15^{\frac23}\cdot 2^{-\frac{16}{3}}}\right)^{-1{,}5}$$
   $$=\left(\frac{5^{-\frac23}\cdot 3^{-\frac23}\cdot 2^{\frac{16}{3}}}{(5\cdot 3)^{\frac23}}\right)^{-\frac32} =\left(\frac{2^{\frac{16}{3}}}{5^{\frac23}\cdot 3^{\frac23}\cdot 5^{\frac23}\cdot 3^{\frac23}}\right)^{-\frac32}$$
   $$=\left(\frac{2^{\frac{16}{3}}}{5^{\frac43}\cdot 3^{\frac43}}\right)^{-\frac32} =\left(\frac{5^{\frac43}\cdot 3^{\frac43}}{2^{\frac{16}{3}}}\right)^{\frac32}$$
   $$=\frac{5^2\cdot 3^2}{2^8}=\frac{25\cdot 9}{256}=\frac{225}{256}.$$

Ответ

1) $$7$$; 2) $$10$$; 3) $$122\frac79$$; 4) $$0{,}25$$; 5) $$21$$; 6) $$\frac{225}{256}$$.