Решите пожалуйста!

Question 1

Решите пожалуйста!

Question 2

Какой номер

Question 3

все

Question 4

А ответы есть?

Question 5

Нужно решить и дать ответ

Question 6

А в учебнике нет ответов?

Question 7

В конце учебника

Question 8

???

Question 9

это не из учебника

Question 10

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

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

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

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

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

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

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

<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctt%202.%20%5C%20%5C%20%5Cfrac%7B3%7D%7B5%7D%3A%5Cfrac%7B7%7D%7B8%7D%3D%5Cfrac%7B3%7D%7B5%7D%5Ccdot%5Cfrac%7B8%7D%7B7%7D%3D%5Cfrac%7B24%7D%7B35%7D%3B%5C%5C%5C%5C3.%20%5C%20%5C%2048%3A%5Cfrac%7B24%7D%7B35%7D%3D48%5Ccdot%5Cfrac%7B35%7D%7B24%7D%3D2%5Ccdot35%3D70%3B%5C%5C%5C%5C4.%20%5C%20%5C%20%5Cfrac%7B1%7D%7B7%7D%5Ccdot70%3D10%3B%5C%5C%5C%5C5.%20%5C%20%5C%20%5Cfrac%7B19%7D%7B26%7D%2B%5C

Question 11

Ответ:

Пошаговое объяснение:

Question 12

Ошибка в первом примере. Восьмое действие. В числителе умножаем 29 на 181. Получаем 5336. Должно быть 5249.

Question 13

Спасибо сейчас исправлю

Question 14

Да не за что..)))

Regent1828_zn · Answer 1 · 2020-05-29T17:15:07+0000

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

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

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

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

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

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

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

<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctt%202.%20%5C%20%5C%20%5Cfrac%7B3%7D%7B5%7D%3A%5Cfrac%7B7%7D%7B8%7D%3D%5Cfrac%7B3%7D%7B5%7D%5Ccdot%5Cfrac%7B8%7D%7B7%7D%3D%5Cfrac%7B24%7D%7B35%7D%3B%5C%5C%5C%5C3.%20%5C%20%5C%2048%3A%5Cfrac%7B24%7D%7B35%7D%3D48%5Ccdot%5Cfrac%7B35%7D%7B24%7D%3D2%5Ccdot35%3D70%3B%5C%5C%5C%5C4.%20%5C%20%5C%20%5Cfrac%7B1%7D%7B7%7D%5Ccdot70%3D10%3B%5C%5C%5C%5C5.%20%5C%20%5C%20%5Cfrac%7B19%7D%7B26%7D%2B%5C

Решите пожалуйста! ​

Решите пожалуйста!