Упр.565 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

1) (5·10^(-2)-3·5^(-1)):10^(-2);
2) (1 1/2)^(-3):(2/3)^5+(1 1/3)^(-2)·(3/4)^(-1);
3) (3·2^(-1)-2·3^(-1))/(1 1/5)^(-1);
4) ((3 1/7)^(-1)+(4 2/5)^(-1))/11^(-1);
5) (4^(-1)·3^(-1)·(2/3)^(-2))/(5-(1/3)^(-1));
6) 3·10^(-1)(8^0-1/8)^(-1)·(1/4)^(-3)·(1/4)^4·(5/7)^(-1);
7) v(6,8^2-3,2^2).

1. $$\left(5\cdot 10^{-2}-3\cdot 5^{-1}\right):10^{-2}=\left(5\cdot \frac{1}{100}-3\cdot \frac{1}{5}\right):\frac{1}{100}$$
   $$=\left(\frac{1}{20}-\frac{3}{5}\right):\frac{1}{100}=\left(\frac{1}{20}-\frac{12}{20}\right):\frac{1}{100}=-\frac{11}{20}\cdot 100=-55.$$

2. $$\left(1\frac{1}{2}\right)^{-3}:\left(\frac{2}{3}\right)^5+\left(1\frac{1}{3}\right)^{-2}\cdot \left(\frac{3}{4}\right)^{-1}$$
   $$=\left(\frac{3}{2}\right)^{-3}:\left(\frac{2}{3}\right)^5+\left(\frac{4}{3}\right)^{-2}\cdot \left(\frac{3}{4}\right)^{-1}$$
   $$=\left(\frac{2}{3}\right)^3\cdot \left(\frac{3}{2}\right)^5+\left(\frac{3}{4}\right)^2\cdot \frac{4}{3}$$
   $$=\left(\frac{2}{3}\right)^3\cdot \left(\frac{3}{2}\right)^5+\left(\frac{3}{4}\right)$$
   $$=\left(\frac{3}{2}\right)^2+\frac{3}{4}=\frac{9}{4}+\frac{3}{4}=3.$$

3. $$\frac{3\cdot 2^{-1}-2\cdot 3^{-1}}{\left(1\frac{1}{5}\right)^{-1}}=\frac{\frac{3}{2}-\frac{2}{3}}{\left(\frac{6}{5}\right)^{-1}}=\frac{\frac{9-4}{6}}{\frac{5}{6}}=\frac{5}{6}\cdot \frac{6}{5}=1.$$

4. $$\frac{\left(3\frac{1}{7}\right)^{-1}+\left(4\frac{2}{5}\right)^{-1}}{11^{-1}}=\frac{\left(\frac{22}{7}\right)^{-1}+\left(\frac{22}{5}\right)^{-1}}{\frac{1}{11}}$$
   $$=11\left(\frac{7}{22}+\frac{5}{22}\right)=11\cdot \frac{12}{22}=11\cdot \frac{6}{11}=6.$$

5. $$\frac{4^{-1}\cdot 3^{-1}\cdot \left(\frac{2}{3}\right)^{-2}}{5-\left(\frac{1}{3}\right)^{-1}}=\frac{\frac{1}{4}\cdot \frac{1}{3}\cdot \left(\frac{3}{2}\right)^2}{5-3}$$
   $$=\frac{\frac{1}{4}\cdot \frac{1}{3}\cdot \frac{9}{4}}{2}=\frac{3}{16}\cdot \frac{1}{2}=\frac{3}{32}.$$

6. $$3\cdot 10^{-1}\left(8^0-\frac{1}{8}\right)^{-1}\cdot \left(\frac{1}{4}\right)^{-3}\cdot \left(\frac{1}{4}\right)^4\cdot \left(\frac{5}{7}\right)^{-1}$$
   $$=3\cdot \frac{1}{10}\cdot \left(1-\frac{1}{8}\right)^{-1}\cdot 4^3\cdot \frac{1}{4}\cdot \frac{7}{5}$$
   $$=\frac{3}{10}\cdot \left(\frac{7}{8}\right)^{-1}\cdot \frac{1}{4}\cdot \frac{7}{5}=\frac{3}{10}\cdot \frac{8}{7}\cdot \frac{1}{4}\cdot \frac{7}{5}$$
   $$=\frac{3}{10}\cdot \frac{2}{5}=\frac{3}{25}.$$

7. $$\sqrt{6{,}8^2-3{,}2^2}=\sqrt{(6{,}8-3{,}2)(6{,}8+3{,}2)}=\sqrt{3{,}6\cdot 10}=\sqrt{36}=6.$$

Ответ

1) $$-55$$; 2) $$3$$; 3) $$1$$; 4) $$6$$; 5) $$\frac{3}{32}$$; 6) $$\frac{3}{25}$$; 7) $$6$$.