Упр.565 ГДЗ Колягин Ткачёва 9 класс (Алгебра)
Вычислить:
1) $$\left(5\cdot10^{-2}-3\cdot5^{-1}\right):10^{-2}$$;
2) $$\left(1\frac{1}{2}\right)^{-3}:\left(\frac{2}{3}\right)^5+\left(1\frac{1}{3}\right)^{-2}\cdot\left(\frac{3}{4}\right)^{-1}$$;
3) $$\frac{3\cdot2^{-1}-2\cdot3^{-1}}{\left(1\frac{1}{5}\right)^{-1}}$$;
4) $$\frac{\left(3\frac{1}{7}\right)^{-1}+\left(4\frac{2}{5}\right)^{-1}}{11^{-1}}$$;
5) $$\frac{4^{-1}\cdot3^{-1}\cdot\left(\frac{2}{3}\right)^{-2}}{5-\left(\frac{1}{3}\right)^{-1}}$$;
6) $$3\cdot10^{-1}\left(8^0-\frac{1}{8}\right)^{-1}\cdot\left(\frac{1}{4}\right)^{-3}\cdot\left(\frac{1}{4}\right)^4\cdot\left(\frac{5}{7}\right)^{-1}$$;
7) $$\sqrt{6{,}8^2-3{,}2^2}$$.
$$\left(5\cdot 10^{-2}-3\cdot 5^{-1}\right):10^{-2}=\left(5\cdot \frac{1}{100}-3\cdot \frac{1}{5}\right):\frac{1}{100}$$
$$=\left(\frac{1}{20}-\frac{3}{5}\right):\frac{1}{100}=\left(\frac{1}{20}-\frac{12}{20}\right):\frac{1}{100}=-\frac{11}{20}\cdot 100=-55.$$$$\left(1\frac{1}{2}\right)^{-3}:\left(\frac{2}{3}\right)^5+\left(1\frac{1}{3}\right)^{-2}\cdot \left(\frac{3}{4}\right)^{-1}$$
$$=\left(\frac{3}{2}\right)^{-3}:\left(\frac{2}{3}\right)^5+\left(\frac{4}{3}\right)^{-2}\cdot \left(\frac{3}{4}\right)^{-1}$$
$$=\left(\frac{2}{3}\right)^3\cdot \left(\frac{3}{2}\right)^5+\left(\frac{3}{4}\right)^2\cdot \frac{4}{3}$$
$$=\left(\frac{2}{3}\right)^3\cdot \left(\frac{3}{2}\right)^5+\left(\frac{3}{4}\right)$$
$$=\left(\frac{3}{2}\right)^2+\frac{3}{4}=\frac{9}{4}+\frac{3}{4}=3.$$$$\frac{3\cdot 2^{-1}-2\cdot 3^{-1}}{\left(1\frac{1}{5}\right)^{-1}}=\frac{\frac{3}{2}-\frac{2}{3}}{\left(\frac{6}{5}\right)^{-1}}=\frac{\frac{9-4}{6}}{\frac{5}{6}}=\frac{5}{6}\cdot \frac{6}{5}=1.$$
$$\frac{\left(3\frac{1}{7}\right)^{-1}+\left(4\frac{2}{5}\right)^{-1}}{11^{-1}}=\frac{\left(\frac{22}{7}\right)^{-1}+\left(\frac{22}{5}\right)^{-1}}{\frac{1}{11}}$$
$$=11\left(\frac{7}{22}+\frac{5}{22}\right)=11\cdot \frac{12}{22}=11\cdot \frac{6}{11}=6.$$$$\frac{4^{-1}\cdot 3^{-1}\cdot \left(\frac{2}{3}\right)^{-2}}{5-\left(\frac{1}{3}\right)^{-1}}=\frac{\frac{1}{4}\cdot \frac{1}{3}\cdot \left(\frac{3}{2}\right)^2}{5-3}$$
$$=\frac{\frac{1}{4}\cdot \frac{1}{3}\cdot \frac{9}{4}}{2}=\frac{3}{16}\cdot \frac{1}{2}=\frac{3}{32}.$$$$3\cdot 10^{-1}\left(8^0-\frac{1}{8}\right)^{-1}\cdot \left(\frac{1}{4}\right)^{-3}\cdot \left(\frac{1}{4}\right)^4\cdot \left(\frac{5}{7}\right)^{-1}$$
$$=3\cdot \frac{1}{10}\cdot \left(1-\frac{1}{8}\right)^{-1}\cdot 4^3\cdot \frac{1}{4}\cdot \frac{7}{5}$$
$$=\frac{3}{10}\cdot \left(\frac{7}{8}\right)^{-1}\cdot \frac{1}{4}\cdot \frac{7}{5}=\frac{3}{10}\cdot \frac{8}{7}\cdot \frac{1}{4}\cdot \frac{7}{5}$$
$$=\frac{3}{10}\cdot \frac{2}{5}=\frac{3}{25}.$$$$\sqrt{6{,}8^2-3{,}2^2}=\sqrt{(6{,}8-3{,}2)(6{,}8+3{,}2)}=\sqrt{3{,}6\cdot 10}=\sqrt{36}=6.$$
Ответ
1) $$-55$$; 2) $$3$$; 3) $$1$$; 4) $$6$$; 5) $$\frac{3}{32}$$; 6) $$\frac{3}{25}$$; 7) $$6$$.
