Упр.23.22 ГДЗ Мерзляк Поляков 9 класс (Алгебра)
- Найдите значение выражения:
1) $$\left(343^{\frac{1}{2}}\cdot\left(\frac{1}{49}\right)^{\frac{3}{8}}\right)^{\frac{4}{3}}$$;
2) $$10^{\frac{1}{4}}\cdot40^{\frac{1}{4}}\cdot5^{\frac{1}{2}}$$;
3) $$0{,}0016^{-\frac{3}{4}}-0{,}04^{-\frac{1}{2}}+0{,}216^{-\frac{2}{3}}$$;
4) $$\frac{32^{0{,}24}\cdot4^{0{,}7}}{64^{0{,}6}\cdot16^{0{,}25}}$$;
5) $$\frac{12^{\frac{1}{2}}}{7^{\frac{2}{3}}\cdot8^{-\frac{1}{6}}}\cdot\frac{3^{\frac{1}{2}}\cdot7^{\frac{5}{3}}}{8^{\frac{1}{2}}}$$;
6) $$\left(\frac{5^{-\frac{2}{3}}\cdot3^{-\frac{2}{3}}}{15^{\frac{2}{3}}\cdot2^{-\frac{16}{3}}}\right)^{-1{,}5}$$.
$$\left(343^{\frac12}\cdot\left(\frac1{49}\right)^{\frac38}\right)^{\frac43}$$
$$=\left((7^3)^{\frac12}\cdot\left(\frac1{7^2}\right)^{\frac38}\right)^{\frac43}$$
$$=\left(7^{\frac32}\cdot 7^{-\frac34}\right)^{\frac43} =\left(7^{\frac34}\right)^{\frac43}=7.$$$$10^{\frac14}\cdot 40^{\frac14}\cdot 5^{\frac12} =(5\cdot 2)^{\frac14}\cdot(5\cdot 8)^{\frac14}\cdot 5^{\frac12}$$
$$=5^{\frac14}\cdot 2^{\frac14}\cdot 5^{\frac14}\cdot 2^{\frac34}\cdot 5^{\frac12} =5^{\frac14+\frac14+\frac12}\cdot 2^{\frac14+\frac34}$$
$$=5\cdot 2=10.$$$$0{,}0016^{-\frac34}-0{,}04^{-\frac12}+0{,}216^{-\frac23}$$
$$=\left(\frac{16}{10000}\right)^{-\frac34}-\left(\frac{4}{100}\right)^{-\frac12}+\left(\frac{216}{1000}\right)^{-\frac23}$$
$$=\left(\frac{2^4}{10^4}\right)^{-\frac34}-\left(\frac{2^2}{10^2}\right)^{-\frac12}+\left(\frac{6^3}{10^3}\right)^{-\frac23}$$
$$=\left(\frac{2}{10}\right)^{-3}-\left(\frac{2}{10}\right)^{-1}+\left(\frac{6}{10}\right)^{-2}$$
$$=\left(\frac{10}{2}\right)^3-\frac{10}{2}+\left(\frac{10}{6}\right)^2$$
$$=5^3-5+\left(\frac53\right)^2=125-5+\frac{25}{9}=120+\frac{25}{9}=122\frac79.$$$$\frac{32^{0{,}24}\cdot 4^{0{,}7}}{64^{0{,}6}\cdot 16^{0{,}25}} =\frac{(2^5)^{0{,}24}\cdot(2^2)^{0{,}7}}{(2^6)^{0{,}6}\cdot(2^4)^{0{,}25}}$$
$$=\frac{2^{1{,}2}\cdot 2^{1{,}4}}{2^{3{,}6}\cdot 2^1} =2^{1{,}2+1{,}4-3{,}6-1}=2^{-2}=\frac14=0{,}25.$$$$\frac{12^{\frac12}}{7^{\frac23}\cdot 8^{-\frac16}}\cdot \frac{3^{\frac12}\cdot 7^{\frac53}}{8^{\frac12}}$$
$$=\frac{(3\cdot 4)^{\frac12}\cdot 3^{\frac12}\cdot 7^{\frac53}}{7^{\frac23}\cdot (2^3)^{-\frac16}\cdot (2^3)^{\frac12}}$$
$$=\frac{3^{\frac12}\cdot 2\cdot 3^{\frac12}\cdot 7^{\frac53}}{7^{\frac23}\cdot 2^{-\frac12}\cdot 2^{\frac32}} =\frac{3\cdot 2\cdot 7}{2^1}=21.$$$$\left(\frac{5^{-\frac23}\cdot 3^{-\frac23}}{15^{\frac23}\cdot 2^{-\frac{16}{3}}}\right)^{-1{,}5}$$
$$=\left(\frac{5^{-\frac23}\cdot 3^{-\frac23}\cdot 2^{\frac{16}{3}}}{(5\cdot 3)^{\frac23}}\right)^{-\frac32} =\left(\frac{2^{\frac{16}{3}}}{5^{\frac23}\cdot 3^{\frac23}\cdot 5^{\frac23}\cdot 3^{\frac23}}\right)^{-\frac32}$$
$$=\left(\frac{2^{\frac{16}{3}}}{5^{\frac43}\cdot 3^{\frac43}}\right)^{-\frac32} =\left(\frac{5^{\frac43}\cdot 3^{\frac43}}{2^{\frac{16}{3}}}\right)^{\frac32}$$
$$=\frac{5^2\cdot 3^2}{2^8}=\frac{25\cdot 9}{256}=\frac{225}{256}.$$
Ответ
1) $$7$$; 2) $$10$$; 3) $$122\frac79$$; 4) $$0{,}25$$; 5) $$21$$; 6) $$\frac{225}{256}$$.
