Упр.94 ГДЗ Алимов 10-11 класс (Алгебра)

1. Вычислить:
   1) $$48^0,\ 10^{-2},\ \left(\frac{2}{3}\right)^{-1},\ (0{,}3)^{-3},\ (-1{,}2)^{-2},\ \left(2\frac{1}{4}\right)^{-2}$$;
   2) $$\sqrt[3]{27},\ \sqrt[4]{81},\ \sqrt[5]{32},\ \sqrt[6]{8^2},\ \sqrt[8]{16^2},\ \sqrt[3]{27^2}$$;
   3) $$8^{\frac{1}{3}},\ 27^{\frac{2}{3}},\ 10000^{\frac{1}{4}},\ 32^{\frac{2}{5}},\ 32^{-\frac{3}{5}},\ \left(\frac{27}{64}\right)^{\frac{2}{3}}$$.

1)

$$48^0=1$$
$$10^{-2}=\frac{1}{10^2}=\frac{1}{100}=0{,}01$$
$$\left(\frac{2}{3}\right)^{-1}=\frac{3}{2}=1{,}5$$
$$\left(0{,}3\right)^{-3}=\left(\frac{3}{10}\right)^{-3}=\left(\frac{10}{3}\right)^3=\frac{1000}{27}=37\frac{1}{27}$$
$$(-1{,}2)^{-2}=\left(-\frac{12}{10}\right)^{-2}=\left(-\frac{6}{5}\right)^{-2}=\left(\frac{5}{6}\right)^2=\frac{25}{36}$$
$$\left(2\frac{1}{4}\right)^{-2}=\left(\frac{9}{4}\right)^{-2}=\left(\frac{4}{9}\right)^2=\frac{16}{81}$$

2)

$$\sqrt[3]{27}=3$$
$$\sqrt[4]{81}=3$$
$$\sqrt[5]{32}=2$$
$$\sqrt[6]{8^2}=\sqrt[6]{2^6}=2$$
$$\sqrt[8]{16^2}=\sqrt[8]{(2^4)^2}=\sqrt[8]{2^8}=2$$
$$\sqrt[3]{27^2}=\sqrt[3]{(3^3)^2}=\sqrt[3]{3^6}=3^2=9$$

3)

$$8^{\frac{1}{3}}=(2^3)^{\frac{1}{3}}=2$$
$$27^{\frac{2}{3}}=(3^3)^{\frac{2}{3}}=3^2=9$$
$$10000^{\frac{1}{4}}=(10^4)^{\frac{1}{4}}=10$$
$$32^{\frac{2}{5}}=(2^5)^{\frac{2}{5}}=2^2=4$$
$$32^{-\frac{3}{5}}=(2^5)^{-\frac{3}{5}}=2^{-3}=\frac{1}{8}$$
$$\left(\frac{27}{64}\right)^{\frac{2}{3}}=\left(\frac{3^3}{4^3}\right)^{\frac{2}{3}}=\left(\frac{3}{4}\right)^2=\frac{9}{16}$$

Ответ: 1; 0,01; 1,5; $$37\frac{1}{27}$$; $$\frac{25}{36}$$; $$\frac{16}{81}$$; 3; 3; 2; 2; 2; 9; 2; 9; 10; 4; $$\frac{1}{8}$$; $$\frac{9}{16}$$.