Упражнение 641

Решение:

• $${\largeа)}\ \frac{3}{4}:\frac{5}{6}+2\frac{1}{2}\cdot\frac{2}{5}-1:1\frac{1}{6}=\frac{3\phantom{0}}{\cancel4^{\backslash2}}\cdot\frac{\cancel6^{\backslash3}}{5\phantom{0}}+\frac{\cancel5^{\backslash1}}{\cancel2^{\backslash1}}\cdot\frac{\cancel2^{\backslash1}}{\cancel5^{\backslash1}}-1\cdot\frac{6}{7}=\frac{9^{\backslash7}}{10\phantom{^{\backslash7}}}+1-\frac{6^{\backslash10}}{7\phantom{^{\backslash10}}}=1+\frac{63}{70}-\frac{60}{70}=1\frac{3}{70}$$

• $${\largeб)}\ 2\frac{3}{4}:\left(1\frac{1^{\backslash5}}{2\phantom{^{\backslash5}}}-\frac{2^{\backslash2}}{5\phantom{^{\backslash2}}}\right)+\left(\frac{3^{\backslash3}}{4\phantom{^{\backslash3}}}+\frac{5^{\backslash2}}{6\phantom{^{\backslash2}}}\right):3\frac{1}{6}=\frac{11}{4}:\left(1\frac{5}{10}-\frac{4}{10}\right)+\left(\frac{9}{12}+\frac{10}{12}\right):\frac{19}{6}=\frac{11}{4}:1\frac{1}{10}+\frac{19}{12}:\frac{19}{6}=\frac{\cancel{11}^{\backslash1}}{\cancel4^{\backslash2}}\cdot\frac{\cancel{10}^{\backslash5}}{\cancel{11}^{\backslash1}}+\frac{\cancel{19}^{\backslash1}}{\cancel{12}^{\backslash2}}\cdot\frac{\cancel6^{\backslash1}}{\cancel{19}^{\backslash1}}=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3$$

• $${\largeв)}\ \left(\frac{2^{\backslash4}}{15\phantom{^{\backslash4}}}+\frac{7^{\backslash5}}{12\phantom{^{\backslash5}}}\right)\cdot\frac{30}{43}-2:2\frac{1}{2}\cdot\frac{5}{32}=\left(\frac{8}{60}+\frac{35}{60}\right)\cdot\frac{30}{43}-2:\frac{5}{2}\cdot\frac{5}{32}=\frac{\cancel{43}^{\backslash1}}{\cancel{60}^{\backslash2}}\cdot\frac{\cancel{30}^{\backslash1}}{\cancel{43}^{\backslash1}}-\frac{\cancel2^{\backslash1}}{1\phantom{0}}\cdot\frac{\cancel2^{\backslash1}}{\cancel5^{\backslash1}}\cdot\frac{\cancel5^{\backslash1}}{\cancel{32}^{\backslash8}}=\frac{1^{\backslash4}}{2\phantom{^{\backslash4}}}-\frac{1^{\backslash1}}{8\phantom{^{\backslash1}}}=\frac{4}{8}-\frac{1}{8}=\frac{3}{8}$$

• $${\largeг)}\ \left(3\frac{1}{2}:4\frac{2}{3}+4\frac{2}{3}:3\frac{1}{2}\right)\cdot4\frac{4}{5}=\left(\frac{7}{2}:\frac{14}{3}+\frac{14}{3}:\frac{7}{2}\right)\cdot\frac{24}{5}=\left(\frac{\cancel7^{\backslash1}}{2\phantom{0}}\cdot\frac{3\phantom{0}}{\cancel{14}^{\backslash2}}+\frac{\cancel{14}^{\backslash2}}{3\phantom{0}}\cdot\frac{2\phantom{0}}{\cancel7^{\backslash1}}\right)\cdot\frac{24}{5}=\left(\frac{3^{\backslash3}}{4\phantom{^{\backslash3}}}+\frac{4^{\backslash4}}{3\phantom{^{\backslash4}}}\right)\cdot\frac{24}{5}=\left(\frac{9}{12}+\frac{16}{12}\right)\cdot\frac{24}{5}=\frac{\cancel{25}^{\backslash5}}{\cancel{12}^{\backslash1}}\cdot\frac{\cancel{24}^{\backslash2}}{\cancel5^{\backslash1}}=10$$

• $${\largeд)}\ \left(11\frac{5^{\backslash2}}{11\phantom{^{\backslash2}}}-8\frac{21^{\backslash1}}{22\phantom{^{\backslash1}}}\right):1\frac{2}{3}=\left(11\frac{10}{22}-8\frac{21}{22}\right):\frac{5}{3}=\left(10\frac{32}{22}-8\frac{21}{22}\right)\cdot\frac{3}{5}=2\frac{11}{22}\cdot\frac{3}{5}=2\frac{1}{2}\cdot\frac{3}{5}=\frac{\cancel5^{\backslash1}}{2\phantom{0}}\cdot\frac{3\phantom{0}}{\cancel5^{\backslash1}}=\frac{3}{2}=1\frac{1}{2}$$

• $${\largeе)}\ \left(\left(1\frac{1}{2}\right)^3-\frac{3}{4}\right):\frac{7}{8}=\left(\left(\frac{3}{2}\right)^3-\frac{3}{4}\right):\frac{7}{8}=\left(\frac{27^{\backslash1}}{8\phantom{^{\backslash1}}}-\frac{3^{\backslash2}}{4\phantom{^{\backslash2}}}\right)\cdot\frac{8}{7}=\left(\frac{27}{8}-\frac{6}{8}\right)\cdot\frac{8}{7}=\frac{\cancel{21}^{\backslash3}}{\cancel8^{\backslash1}}\cdot\frac{\cancel8^{\backslash1}}{\cancel7^{\backslash1}}=3$$