Упр.94 ГДЗ Алимов 10-11 класс (Алгебра)
Вычислить:
- 1) $$48^0$$, $$10^{-2}$$, $$\left(\frac{2}{3}\right)^{-1}$$, $$\left(0{,}3\right)^{-3}$$, $$\left(-1{,}2\right)^{-2}$$, $$\left(2\cdot\frac{1}{4}\right)^{-2}$$;
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)
$$\sqrt[3]{27}=3$$
$$\sqrt[4]{81}=3$$
$$\sqrt[5]{32}=2$$
$$\sqrt[6]{8^2}=2$$
$$\sqrt[8]{16^2}=2$$
$$\sqrt[3]{27^2}=9$$
3)
$$8^{\frac{1}{3}}=2$$
$$27^{\frac{2}{3}}=9$$
$$10000^{\frac{1}{4}}=10$$
$$32^{\frac{2}{5}}=4$$
$$32^{-\frac{3}{5}}=\frac{1}{8}$$
$$\left(\frac{27}{64}\right)^{\frac{2}{3}}=\frac{9}{16}$$
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}
