Упр.23.21 ГДЗ Мерзляк Поляков 9 класс (Алгебра)

Задача

Вычислите значение выражения:
1) $$12^{1/3}\cdot6^{2/3}\cdot(0{,}5)^{1/3}$$;
2) $$25^{1{,}5}+(0{,}25)^{-0{,}5}-81^{0{,}75}$$;
3) $$(1/16)^{-3/4}+(1/8)^{-2/3}\cdot(0{,}81)^{-0{,}5}$$;
4) $$16^{1/8}\cdot8^{-5/6}\cdot4^{1{,}5}$$;
5) $$\frac{10000^{0{,}4}\cdot10^{0{,}5}}{100^{0{,}3}\cdot1000^{1/6}}$$;
6) $$\frac{5^{3/2}\cdot8^{1/12}}{9^{1/6}}\cdot\frac{8^{1/4}}{5^{5/2}\cdot9^{1/3}}$$;
7) $$(72^{2/3})^{1/2}\cdot2^{-4/3}:36^{-1/6}$$;
8) $$\left(\frac{3^{-5/6}\cdot7^{-5/6}}{21^{-1}\cdot5^{1/3}}\right)^{-6}$$.

Подробный ответ

$$12^{\frac13}\cdot 6^{\frac23}\cdot (0{,}5)^{\frac13}=(6\cdot 2)^{\frac13}\cdot 6^{\frac23}\cdot \left(\frac12\right)^{\frac13}$$
$$=6^{\frac13}\cdot 2^{\frac13}\cdot 6^{\frac23}\cdot 2^{-\frac13}=6^{\frac13+\frac23}\cdot 2^{\frac13-\frac13}=6\cdot 1=6.$$
$$25^{1{,}5}+(0{,}25)^{-0{,}5}-81^{0{,}75}=(5^2)^{\frac32}+\left(\frac14\right)^{-\frac12}-(3^4)^{\frac34}$$
$$=5^3+4^{\frac12}-3^3=125+2-27=100.$$
$$\left(\frac1{16}\right)^{-\frac34}+\left(\frac18\right)^{-\frac23}\cdot (0{,}81)^{-0{,}5}=16^{\frac34}+8^{\frac23}\cdot \left(\frac{81}{100}\right)^{-\frac12}$$
$$=(2^4)^{\frac34}+(2^3)^{\frac23}\cdot \left(\frac{100}{81}\right)^{\frac12}=2^3+2^2\cdot \frac{10}{9}$$
$$=8+\frac{40}{9}=12\frac49.$$
$$16^{\frac18}\cdot 8^{-\frac56}\cdot 4^{1{,}5}=(2^4)^{\frac18}\cdot (2^3)^{-\frac56}\cdot (2^2)^{\frac32}$$
$$=2^{\frac12}\cdot 2^{-\frac52}\cdot 2^3=2^{\frac12-\frac52+3}=2^2=4.$$
$$\frac{10\,000^{0{,}4}\cdot 10^{0{,}5}}{100^{0{,}3}\cdot 1000^{\frac16}}=\frac{(10^4)^{0{,}4}\cdot 10^{0{,}5}}{(10^2)^{0{,}3}\cdot (10^3)^{\frac16}}$$
$$=\frac{10^{1{,}6}\cdot 10^{0{,}5}}{10^{0{,}6}\cdot 10^{0{,}5}}=10^{1{,}6-0{,}6}=10.$$
$$\frac{5^{\frac32}\cdot 8^{\frac1{12}}}{9^{\frac16}}\cdot \frac{8^{\frac14}}{5^{\frac52}\cdot 9^{\frac13}}= \frac{5^{\frac32-\frac52}\cdot 8^{\frac1{12}+\frac14}}{9^{\frac16+\frac13}}$$
$$=\frac{5^{-1}\cdot 8^{\frac13}}{9^{\frac12}}=\frac{5^{-1}\cdot 2}{3}=\frac{2}{15}.$$
$$\left(72^{\frac23}\right)^{\frac12}\cdot 2^{-\frac43}:36^{-\frac16}=72^{\frac13}\cdot 2^{-\frac43}\cdot 36^{\frac16}$$
$$=(36\cdot 2)^{\frac13}\cdot 2^{-\frac43}\cdot 36^{\frac16}=36^{\frac13}\cdot 2^{\frac13-\frac43}\cdot 36^{\frac16}$$
$$=36^{\frac12}\cdot 2^{-1}=6\cdot \frac12=3.$$
$$\left(\frac{3^{-\frac56}\cdot 7^{-\frac56}}{21^{-1}\cdot 5^{\frac13}}\right)^{-6} =\left(\frac{21^{-1}\cdot 5^{\frac13}}{3^{-\frac56}\cdot 7^{-\frac56}}\right)^6$$
$$=\left(\frac{3^{\frac56}\cdot 7^{\frac56}\cdot 5^{\frac13}}{21}\right)^6 =\left(\frac{21^{\frac56}\cdot 5^{\frac13}}{21}\right)^6$$
$$=\left(21^{-\frac16}\cdot 5^{\frac13}\right)^6 =21^{-1}\cdot 5^2=\frac{25}{21}=1\frac4{21}.$$