Упр.444 Часть 2 ГДЗ Дорофеев Петерсон 5 класс (Математика)
Вычисли:
а) $$\frac{30}{\frac{5}{12}+\frac{1}{4}}$$;
в) $$\frac{4-1\frac{3}{5}}{7\frac{1}{2}\cdot\frac{9}{25}}$$;
д) $$\frac{\left(7\frac{1}{3}:1\frac{1}{6}\right):\left(8:7\right)}{9:\left(1\frac{3}{11}+2\right)}$$;
б) $$\frac{1\frac{3}{8}\cdot2}{11}$$;
г) $$\frac{8\frac{1}{3}:2}{1\frac{9}{10}+4\frac{7}{20}}$$;
е) $$\frac{3\frac{4}{7}:\left(6\frac{1}{28}-3\frac{3}{4}\right)}{\left(1\frac{5}{6}\cdot1\frac{5}{22}\right):18\cdot5}$$.
а) $$30:\left(\frac{5}{12}+\frac{1}{4}\right)=30:\left(\frac{5}{12}+\frac{3}{12}\right)=30:\frac{8}{12}=30:\frac{2}{3}=30\cdot\frac{3}{2}=45$$
б) $$\left(1\frac{3}{8}\cdot 2\right):11=\left(\frac{11}{8}\cdot 2\right):11=\frac{11}{4}:11=\frac{11}{4}\cdot\frac{1}{11}=\frac{1}{4}$$
в) $$\left(4-1\frac{3}{5}\right):\left(7\frac{1}{2}\cdot\frac{9}{25}\right)=\frac{12}{5}:\frac{27}{10}=\frac{12}{5}\cdot\frac{10}{27}=\frac{8}{9}$$
г) $$\left(8\frac{1}{3}:2\right):\left(1\frac{9}{10}+4\frac{7}{20}\right)=\left(\frac{25}{3}:2\right):\left(\frac{19}{10}+\frac{87}{20}\right)$$
$$=\frac{25}{6}:\frac{25}{4}=\frac{25}{6}\cdot\frac{4}{25}=\frac{2}{3}$$
д) $$\left(\left(7\frac{1}{3}:1\frac{1}{6}\right):(8:7)\right):\left(9:\left(1\frac{3}{11}+2\right)\right)$$
$$=\left(\frac{22}{3}:\frac{7}{6}\right):\frac{8}{7}:\left(9:\frac{36}{11}\right)=\frac{22}{3}\cdot\frac{6}{7}\cdot\frac{7}{8}:\frac{9\cdot 11}{36}$$
$$=\frac{11}{2}:\frac{11}{4}=\frac{11}{2}\cdot\frac{4}{11}=2$$
е) $$\left(3\frac{4}{7}:\left(6\frac{1}{28}-3\frac{3}{4}\right)\right):\left(\left(1\frac{5}{6}\cdot 1\frac{5}{22}\right):18\cdot 5\right)$$
$$=\left(\frac{25}{7}:\frac{15}{28}\right):\left(\frac{11}{6}\cdot\frac{27}{22}:18\cdot 5\right)=\frac{25}{7}:\frac{15}{28}:\frac{5}{8}$$
$$=\frac{25}{7}\cdot\frac{28}{15}\cdot\frac{8}{5}=\frac{5}{2}=2\frac{1}{2}$$
Ответ
а) 45; б) $$\frac{1}{4}$$; в) $$\frac{8}{9}$$; г) $$\frac{2}{3}$$; д) 2; е) $$2\frac{1}{2}$$.
