Упр.10.16 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

Найдите значение выражения:

1. $$\left(343^{1/2}\cdot\left(\frac{1}{49}\right)^{3/8}\right)^{4/3}$$
2. $$10^{1/4}\cdot40^{1/4}\cdot5^{1/2}$$
3. $$0{,}0016^{-3/4}-0{,}04^{-1/2}+0{,}216^{-2/3}$$
4. $$\frac{32^{0{,}24}\cdot4^{0{,}7}}{64^{0{,}6}\cdot16^{0{,}25}}$$
5. $$\frac{12^{1/2}}{7^{2/3}\cdot8^{-1/6}}\cdot\frac{3^{1/2}\cdot7^{5/3}}{8^{1/2}}$$
6. $$\left(\frac{5^{-2/3}\cdot3^{-2/3}}{15^{2/3}\cdot2^{-16/3}}\right)^{-1{,}5}$$

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

2. $$10^{\frac14}\cdot40^{\frac14}\cdot5^{\frac12} =10^{\frac14}\cdot(5\cdot8)^{\frac14}\cdot5^{\frac12} =10^{\frac14}\cdot5^{\frac14}\cdot8^{\frac14}\cdot5^{\frac12}$$
   $$=(2\cdot5)^{\frac14}\cdot5^{\frac14}\cdot(2^3)^{\frac14}\cdot5^{\frac12} =2^{\frac14+\frac34}\cdot5^{\frac14+\frac14+\frac12} =2\cdot5=10.$$

3. $$0{,}0016^{-\frac34}-0{,}04^{-\frac12}+0{,}216^{-\frac23} =\left(\frac{16}{10000}\right)^{-\frac34}-\left(\frac4{100}\right)^{-\frac12}+\left(\frac{216}{1000}\right)^{-\frac23}$$
   $$=\left(\frac{2^4}{10^4}\right)^{-\frac34}-\left(\frac{2^2}{10^2}\right)^{-\frac12}+\left(\frac{6^3}{10^3}\right)^{-\frac23} =\left(\frac2{10}\right)^{-3}-\left(\frac2{10}\right)^{-1}+\left(\frac6{10}\right)^{-2}$$
   $$=\left(\frac{10}{2}\right)^3-\frac{10}{2}+\left(\frac{10}{6}\right)^2 =5^3-5+\frac{25}{9} =125-5+\frac{25}{9} =120+\frac{25}{9} =122\frac79.$$

4. $$\frac{32^{0{,}24}\cdot4^{0{,}7}}{64^{0{,}6}\cdot16^{0{,}25}} =\frac{(2^5)^{0{,}24}\cdot(2^2)^{0{,}7}}{(2^6)^{0{,}6}\cdot(2^4)^{0{,}25}} =\frac{2^{1{,}2}\cdot2^{1{,}4}}{2^{3{,}6}\cdot2^1}$$
   $$=2^{1{,}2+1{,}4-3{,}6-1} =2^{-2} =\frac14 =0{,}25.$$

5. $$\frac{12^{\frac12}}{7^{\frac23}\cdot8^{-\frac16}}\cdot\frac{3^{\frac12}\cdot7^{\frac53}}{8^{\frac12}} =\frac{(3\cdot4)^{\frac12}\cdot3^{\frac12}\cdot7^{\frac53}}{7^{\frac23}\cdot8^{-\frac16}\cdot8^{\frac12}}$$
   $$=\frac{3^{\frac12}\cdot2\cdot3^{\frac12}\cdot7^{\frac53}}{7^{\frac23}\cdot8^{\frac13}} =\frac{3\cdot2\cdot7^{\frac53}}{7^{\frac23}\cdot2} =3\cdot7^{\frac53-\frac23} =3\cdot7=21.$$

6. $$\left(\frac{5^{-\frac23}\cdot3^{-\frac23}}{15^{\frac23}\cdot2^{-\frac{16}{3}}}\right)^{-1{,}5} =\left(\frac{5^{-\frac23}\cdot3^{-\frac23}\cdot2^{\frac{16}{3}}}{(3\cdot5)^{\frac23}}\right)^{-\frac32}$$
   $$=\left(\frac{2^{\frac{16}{3}}}{5^{\frac23+ \frac23}\cdot3^{\frac23+\frac23}}\right)^{-\frac32} =\left(\frac{2^{\frac{16}{3}}}{5^{\frac43}\cdot3^{\frac43}}\right)^{-\frac32} =\left(\frac{2^{\frac{16}{3}}}{15^{\frac43}}\right)^{-\frac32}$$
   $$=\left(\frac{15^{\frac43}}{2^{\frac{16}{3}}}\right)^{\frac32} =\frac{15^2}{2^8} =\frac{225}{256}.$$

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