Предмет: Математика, автор: unicornvera

Решите пожалуйста! ​

Приложения:

tatsach: Какой номер
unicornvera: все
tatsach: А ответы есть?
unicornvera: Нужно решить и дать ответ
tatsach: А в учебнике нет ответов?
tatsach: В конце учебника
tatsach: ???
unicornvera: это не из учебника

Ответы

Автор ответа: tatsach
3

Ответ:

Пошаговое объяснение:

Приложения:

Regent1828: Ошибка в первом примере. Восьмое действие. В числителе умножаем 29 на 181. Получаем 5336. Должно быть 5249.
tatsach: Спасибо сейчас исправлю
Regent1828: Да не за что..)))
Автор ответа: Regent1828
13

\displaystyle \tt 1). \ \ \frac{\bigg(9-5\dfrac{3}{8}\bigg)\cdot{\bigg[4\dfrac{5}{12}-4:2\dfrac{2}{3}+\bigg(\dfrac{3}{10}-\dfrac{1}{2}:4\bigg)\cdot\dfrac{4}{7}\bigg]}}{\dfrac{1}{24}+\dfrac{1}{4}:13\dfrac{1}{3}}=181\\\\\\1. \ \ 9-5\frac{3}{8}=8\frac{8}{8}-5\frac{3}{8}=3\frac{5}{8};\\\\2. \ \ 4:2\frac{2}{3}=4:\frac{8}{3}=4\cdot\frac{3}{8}=\frac{3}{2}=1\frac{1}{2};\\\\3. \ \ \frac{1}{2}:4=\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8};\\\\4. \ \ \frac{3}{10}-\frac{1}{8}=\frac{12-5}{40}=\frac{7}{40};

\displaystyle \tt 5. \ \ \frac{7}{40}\cdot\frac{4}{7}=\frac{1}{10};\\\\6. \ \ 4\frac{5}{12}-1\frac{1}{2}=3\frac{17}{12}-1\frac{6}{12}=2\frac{11}{12};\\\\7. \ \ 2\frac{11}{12}+\frac{1}{10}=2\frac{55}{60}+\frac{6}{60}=2\frac{61}{60}=3\frac{1}{60};\\\\8. \ \ 3\frac{5}{8}\cdot3\frac{1}{60}=\frac{29}{8}\cdot\frac{181}{60}=\frac{5249}{480}=10\frac{449}{480};\\\\9. \ \ \frac{1}{4}:13\frac{1}{3}=\frac{1}{4}:\frac{40}{3}=\frac{1}{4}\cdot\frac{3}{40}=\frac{3}{160};

\displaystyle \tt 10. \ \frac{1}{24}+\frac{3}{160}=\frac{20+9}{480}=\frac{29}{480};\\\\11. \ 10\frac{449}{480}:\frac{29}{480}=\frac{5249}{480}\cdot\frac{480}{29}=\frac{5249}{29}=181;

\displaystyle \tt 2). \ \ \left[\frac{\bigg(3\dfrac{2}{5}+1\dfrac{5}{7}\bigg)\cdot11\dfrac{2}{3}}{1\dfrac{2}{9}-1\dfrac{1}{18}}-\frac{\bigg(10\dfrac{3}{4}-1\dfrac{5}{6}\bigg)\cdot6}{\bigg(5\dfrac{3}{20}-4\dfrac{1}{4}\bigg)\cdot1\dfrac{1}{9}}\right]:42\frac{1}{2}=7\frac{14}{85}\\\\\\1. \ \ 3\frac{2}{5}+1\frac{5}{7}=4+\frac{14+25}{35}=4+1\frac{4}{35}=5\frac{4}{35};\\\\2. \ \ 5\frac{4}{35}\cdot11\frac{2}{3}=\frac{179}{35}\cdot\frac{35}{3}=\frac{179}{3}=59\frac{2}{3};

\displaystyle \tt 3. \ \ 1\frac{2}{9}-1\frac{1}{18}=1\frac{4}{18}-1\frac{1}{18}=\frac{3}{18}=\frac{1}{6};\\\\4. \ \ 59\frac{2}{3}:\frac{1}{6}=\frac{179}{3}\cdot6=179\cdot2=358;\\\\5. \ \ 10\frac{3}{4}-1\frac{5}{6}=\frac{43}{4}-\frac{11}{6}=\frac{129-22}{12}=\frac{107}{12}= 8\frac{11}{12};\\\\6. \ \ 8\frac{11}{12}\cdot6=\frac{107}{12}\cdot6=\frac{107}{2}=53\frac{1}{2};\\\\7. \ \ 5\frac{3}{20}-4\frac{1}{4}=4\frac{23}{20}-4\frac{5}{20}=\frac{18}{20}=\frac{9}{10};

\displaystyle \tt 8. \ \ \frac{9}{10}\cdot1\frac{1}{9}=\frac{9}{10}\cdot\frac{10}{9}=1;\\\\9. \ \ 358-53\frac{1}{2}=304\frac{1}{2};\\\\10. \ 304\frac{1}{2}:42\frac{1}{2}=\frac{609}{2}\cdot\frac{2}{85}=\frac{609}{85}=7\frac{14}{85};

\displaystyle \tt 3). \ \ \frac{\bigg[\bigg(\dfrac{23}{36}+\dfrac{31}{63}\bigg)-\bigg(\dfrac{3}{4}+\dfrac{5}{21}\bigg)\bigg]\cdot48:\bigg(\dfrac{3}{5}:\dfrac{7}{8}\bigg)}{\bigg(\dfrac{19}{26}+\dfrac{14}{39}-\dfrac{1}{6}\bigg)\cdot54\dfrac{1}{6}:\bigg(8\dfrac{4}{7}:\dfrac{12}{35}\bigg)}=5\\\\\\1. \ \bigg(\frac{23}{36}+\frac{31}{63}\bigg)-\bigg(\frac{3}{4}+\frac{5}{21}\bigg)=\frac{23}{36}-\frac{3}{4}+\frac{31}{63}-\frac{5}{21}=\frac{16}{63}-\frac{4}{36}=\frac{64-28}{252}=\frac{1}{7};

\displaystyle \tt 2. \ \ \frac{3}{5}:\frac{7}{8}=\frac{3}{5}\cdot\frac{8}{7}=\frac{24}{35};\\\\3. \ \ 48:\frac{24}{35}=48\cdot\frac{35}{24}=2\cdot35=70;\\\\4. \ \ \frac{1}{7}\cdot70=10;\\\\5. \ \ \frac{19}{26}+\frac{14}{39}=\frac{57+28}{78}=\frac{85}{78}=1\frac{7}{78};\\\\6. \ \ 1\frac{7}{78}-\frac{1}{6}=\frac{85-13}{78}=\frac{72}{78}=\frac{12}{13};\\\\7. \ \ \frac{12}{13}\cdot54\frac{1}{6}=\frac{12}{13}\cdot\frac{325}{6}=\frac{650}{13}=50;

\displaystyle \tt 8. \ \ 8\frac{4}{7}:\frac{12}{35}=\frac{60}{7}\cdot\frac{35}{12}=5\cdot5=25;\\\\9. \ \ 50:25=2;\\\\10. \ 10:2=5;\\\\\\4). \ \ 3\frac{1}{4}-\left[\frac{6:\dfrac{3}{5}-1\dfrac{1}{6}\cdot\dfrac{6}{7}}{4\dfrac{1}{5}\cdot\dfrac{10}{11}+5\dfrac{2}{11}}-\frac{\bigg(\dfrac{3}{20}+\dfrac{1}{2}-\dfrac{1}{15}\bigg)\cdot\dfrac{12}{49}}{3\dfrac{1}{3}+\dfrac{2}{9}}\right]\cdot2\frac{1}{3}=1\frac{1}{96}\\\\\\1. \ \ 6:\frac{3}{5}=6\cdot\frac{5}{3}=10;

\displaystyle \tt 2. \ \ 1\frac{1}{6}\cdot\frac{6}{7}=\frac{7}{6}\cdot\frac{6}{7}=1;\\\\3. \ \ 10-1=9;\\\\4. \ \ 4\frac{1}{5}\cdot\frac{10}{11}=\frac{21}{5}\cdot\frac{10}{11}=\frac{42}{11}=3\frac{9}{11};\\\\5. \ \ 3\frac{9}{11}+5\frac{2}{11}=8\frac{11}{11}=9;\\\\6. \ \ 9:9=1\\\\7. \ \ \frac{3}{20}+\frac{1}{2}=\frac{3+10}{20}=\frac{13}{20};\\\\8. \ \ \frac{13}{20}-\frac{1}{15}=\frac{39-4}{60}=\frac{35}{60}=\frac{7}{12};\\\\9. \ \ \frac{7}{12}\cdot\frac{12}{49}=\frac{1}{7};

\displaystyle \tt 10. \ 3\frac{1}{3}+\frac{2}{9}=3\frac{3}{9}+\frac{2}{9}=3\frac{5}{9};\\\\11. \ \frac{1}{7}:3\frac{5}{9}=\frac{1}{7}\cdot\frac{9}{32}=\frac{9}{224};\\\\12. \ 1-\frac{9}{224}=\frac{215}{224};\\\\13. \ \frac{215}{224}\cdot2\frac{1}{3}=\frac{215}{224}\cdot\frac{7}{3}=\frac{215}{96}=2\frac{23}{96};\\\\14. \ 3\frac{1}{4}-2\frac{23}{96}=3\frac{24}{96}-2\frac{23}{96}=1\frac{1}{96};

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