задание на каникулы.pdf

Постройте отрезок AB = 8 см. Отметьте на этом отрезке точки C, D и E так, чтобы:

1. AC был равен \(\frac{1}{4}\) от AB;
2. AD был равен \(\frac{7}{8}\) от AB;
3. DE был равен \(\frac{2}{7}\) от AD.

Решение:

1. \(AC = \frac{1}{4} \cdot 8 = 2\) см
2. \(AD = \frac{7}{8} \cdot 8 = 7\) см
3. \(DE = \frac{2}{7} \cdot 7 = 2\) см

Приведите:

1. дробь \(\frac{40}{49}\) к знаменателю 343;
2. дроби \(\frac{38}{57}\) и \(\frac{7}{6}\) к общему знаменателю;
3. дроби \(\frac{3}{7}\), \(\frac{1}{8}\) и \(\frac{11}{14}\) к общему знаменателю.

Решение:

1. \(\frac{40}{49} = \frac{40 \cdot 7}{49 \cdot 7} = \frac{280}{343}\)
2. НОЗ(57, 6) = 114, тогда \(\frac{38}{57} = \frac{38 \cdot 2}{57 \cdot 2} = \frac{76}{114}\), \(\frac{7}{6} = \frac{7 \cdot 19}{6 \cdot 19} = \frac{133}{114}\)
3. НОЗ(7, 8, 14) = 56, тогда \(\frac{3}{7} = \frac{3 \cdot 8}{7 \cdot 8} = \frac{24}{56}\), \(\frac{1}{8} = \frac{1 \cdot 7}{8 \cdot 7} = \frac{7}{56}\), \(\frac{11}{14} = \frac{11 \cdot 4}{14 \cdot 4} = \frac{44}{56}\)

Сократите дроби:

1. \(\frac{10}{15}\), \(\frac{12}{18}\), \(\frac{24}{48}\), \(\frac{28}{35}\)
2. \(\frac{8-11}{33-4}\)
3. \(\frac{18-25}{75-12}\)
4. \(\frac{6-7+7-5}{49}\)

Решение:

1. \(\frac{10}{15} = \frac{2}{3}\), \(\frac{12}{18} = \frac{2}{3}\), \(\frac{24}{48} = \frac{1}{2}\), \(\frac{28}{35} = \frac{4}{5}\)
2. \(\frac{8-11}{33-4} = \frac{-3}{29}\)
3. \(\frac{18-25}{75-12} = \frac{-7}{63} = -\frac{1}{9}\)
4. \(\frac{6-7+7-5}{49} = \frac{1}{49}\)

Расположите в порядке (чтобы выполнить задание, некоторые дроби нужно привести к общему знаменателю):

1. возрастания дроби: \(\frac{105}{15}\), \(\frac{3}{3}\), \(\frac{3}{8}\), \(\frac{5}{9}\), \(\frac{5}{3}\), \(\frac{100}{100}\), \(\frac{1}{4}\)
2. убывания дроби: \(\frac{1}{15}\), \(\frac{4}{15}\), \(\frac{2}{3}\), \(\frac{3}{8}\), \(\frac{2}{9}\), \(\frac{5}{3}\), \(\frac{5}{6}\)

Решение:

1. \(\frac{1}{4}\), \(\frac{5}{9}\), \(\frac{100}{100}\), \(\frac{3}{3}\), \(\frac{3}{8}\), \(\frac{5}{3}\), \(\frac{105}{15}\)
2. \(\frac{5}{3}\), \(\frac{5}{6}\), \(\frac{2}{3}\), \(\frac{3}{8}\), \(\frac{2}{9}\), \(\frac{4}{15}\), \(\frac{1}{15}\)

Выполните сложение:

1. \(\frac{3}{4} + \frac{2}{3}\)
2. \(\frac{4}{9} + \frac{1}{6}\)
3. \(\frac{6}{7} + \frac{9}{21}\)
4. \(\frac{3}{10} + \frac{2}{15}\)
5. \(\frac{7}{15} + \frac{3}{40}\)
6. \(\frac{11}{20} + \frac{6}{15}\)

Решение:

1. \(\frac{3}{4} + \frac{2}{3} = \frac{9}{12} + \frac{8}{12} = \frac{17}{12}\)
2. \(\frac{4}{9} + \frac{1}{6} = \frac{8}{18} + \frac{3}{18} = \frac{11}{18}\)
3. \(\frac{6}{7} + \frac{9}{21} = \frac{18}{21} + \frac{9}{21} = \frac{27}{21} = \frac{9}{7}\)
4. \(\frac{3}{10} + \frac{2}{15} = \frac{9}{30} + \frac{4}{30} = \frac{13}{30}\)
5. \(\frac{7}{15} + \frac{3}{40} = \frac{56}{120} + \frac{9}{120} = \frac{65}{120} = \frac{13}{24}\)
6. \(\frac{11}{20} + \frac{6}{15} = \frac{33}{60} + \frac{24}{60} = \frac{57}{60} = \frac{19}{20}\)

Выполните вычитание:

1. \(\frac{1}{11} - \frac{1}{33}\)
2. \(\frac{13}{30} - \frac{2}{45}\)
3. \(\frac{13}{60} - \frac{7}{40}\)
4. \(\frac{7}{11} - \frac{7}{12}\)
5. \(\frac{7}{15} - \frac{2}{5}\)
6. \(\frac{11}{21} - \(\frac{3}{14}\)

Решение:

1. \(\frac{1}{11} - \frac{1}{33} = \frac{3}{33} - \frac{1}{33} = \frac{2}{33}\)
2. \(\frac{13}{30} - \frac{2}{45} = \frac{39}{90} - \frac{4}{90} = \frac{35}{90} = \frac{7}{18}\)
3. \(\frac{13}{60} - \frac{7}{40} = \frac{26}{120} - \frac{21}{120} = \frac{5}{120} = \frac{1}{24}\)
4. \(\frac{7}{11} - \frac{7}{12} = \frac{84}{132} - \frac{77}{132} = \frac{7}{132}\)
5. \(\frac{7}{15} - \frac{2}{5} = \frac{7}{15} - \frac{6}{15} = \frac{1}{15}\)
6. \(\frac{11}{21} - \frac{3}{14} = \frac{22}{42} - \frac{9}{42} = \frac{13}{42}\)

Найдите значение выражений (выполнять по действиям):

1. \((1\frac{1}{3})^2 \cdot (3\frac{1}{16} + \frac{75}{100})\)
2. \((6 - 1\frac{5}{9}) \div \frac{7}{15} \div \frac{2}{3}\)
3. \(\frac{1\frac{1}{3} \cdot 3 \frac{3}{10} - 2\frac{2}{7}}{1\frac{2}{7} \cdot 1\frac{1}{10} \cdot 5\frac{1}{3}}\)

Решение:

1. \((1\frac{1}{3})^2 \cdot (3\frac{1}{16} + \frac{75}{100}) = (\frac{4}{3})^2 \cdot (\frac{49}{16} + \frac{3}{4}) = \frac{16}{9} \cdot (\frac{49}{16} + \frac{12}{16}) = \frac{16}{9} \cdot \frac{61}{16} = \frac{61}{9} = 6\frac{7}{9}\)
2. \((6 - 1\frac{5}{9}) \div \frac{7}{15} \div \frac{2}{3} = (6 - \frac{14}{9}) \div \frac{7}{15} \div \frac{2}{3} = (\frac{54}{9} - \frac{14}{9}) \div \frac{7}{15} \div \frac{2}{3} = \frac{40}{9} \cdot \frac{15}{7} \cdot \frac{3}{2} = \frac{40 \cdot 15 \cdot 3}{9 \cdot 7 \cdot 2} = \frac{1800}{126} = \frac{100}{7} = 14\frac{2}{7}\)
3. \(\frac{1\frac{1}{3} \cdot 3 \frac{3}{10} - 2\frac{2}{7}}{1\frac{2}{7} \cdot 1\frac{1}{10} \cdot 5\frac{1}{3}} = \frac{\frac{4}{3} \cdot \frac{33}{10} - \frac{16}{7}}{\frac{9}{7} \cdot \frac{11}{10} \cdot \frac{16}{3}} = \frac{\frac{4 \cdot 33}{3 \cdot 10} - \frac{16}{7}}{\frac{9 \cdot 11 \cdot 16}{7 \cdot 10 \cdot 3}} = \frac{\frac{132}{30} - \frac{16}{7}}{\frac{1584}{210}} = \frac{\frac{924}{210} - \frac{480}{210}}{\frac{1584}{210}} = \frac{\frac{444}{210}}{\frac{1584}{210}} = \frac{444}{1584} = \frac{37}{132}\)

Решите уравнения:

1. \(8\frac{7}{36} - x = 3\frac{2}{9}\)
2. \(y + 1\frac{75}{100} = 3\frac{2}{3} - 1\frac{1}{2}\)
3. \(8\frac{1}{3} - (5\frac{2}{15} - m) = 3\frac{2}{9}\)
4. \(1\frac{3}{14}q = 2\frac{5}{8} \cdot 2\frac{8}{10}\)
5. \(1\frac{1}{5} \cdot (\frac{2}{3}d + \frac{1}{6}) = 2\frac{6}{10}\)

Решение:

1. \(8\frac{7}{36} - x = 3\frac{2}{9}\) \(\Rightarrow\) \(x = 8\frac{7}{36} - 3\frac{2}{9}\) \(\Rightarrow\) \(x = 8\frac{7}{36} - 3\frac{8}{36}\) \(\Rightarrow\) \(x = 5\frac{-1}{36}\) \(\Rightarrow\) \(x = 4\frac{35}{36}\)
2. \(y + 1\frac{75}{100} = 3\frac{2}{3} - 1\frac{1}{2}\) \(\Rightarrow\) \(y + 1\frac{3}{4} = 3\frac{2}{3} - 1\frac{1}{2}\) \(\Rightarrow\) \(y + 1\frac{3}{4} = 3\frac{4}{6} - 1\frac{3}{6}\) \(\Rightarrow\) \(y + 1\frac{3}{4} = 2\frac{1}{6}\) \(\Rightarrow\) \(y = 2\frac{1}{6} - 1\frac{3}{4}\) \(\Rightarrow\) \(y = 2\frac{2}{12} - 1\frac{9}{12}\) \(\Rightarrow\) \(y = 1\frac{14}{12} - 1\frac{9}{12}\) \(\Rightarrow\) \(y = \frac{5}{12}\)
3. \(8\frac{1}{3} - (5\frac{2}{15} - m) = 3\frac{2}{9}\) \(\Rightarrow\) \(\frac{25}{3} - (\frac{77}{15} - m) = \frac{29}{9}\) \(\Rightarrow\) \(\frac{25}{3} - \frac{77}{15} + m = \frac{29}{9}\) \(\Rightarrow\) \(m = \frac{29}{9} - \frac{25}{3} + \frac{77}{15}\) \(\Rightarrow\) \(m = \frac{145}{45} - \frac{375}{45} + \frac{231}{45}\) \(\Rightarrow\) \(m = \frac{1}{45}\)
4. \(1\frac{3}{14}q = 2\frac{5}{8} \cdot 2\frac{8}{10}\) \(\Rightarrow\) \(\frac{17}{14}q = \frac{21}{8} \cdot \frac{28}{10}\) \(\Rightarrow\) \(\frac{17}{14}q = \frac{21 \cdot 28}{8 \cdot 10}\) \(\Rightarrow\) \(\frac{17}{14}q = \frac{588}{80}\) \(\Rightarrow\) \(q = \frac{588}{80} \cdot \frac{14}{17}\) \(\Rightarrow\) \(q = \frac{8232}{1360}\) \(\Rightarrow\) \(q = \frac{514.5}{85}\)
5. \(1\frac{1}{5} \cdot (\frac{2}{3}d + \frac{1}{6}) = 2\frac{6}{10}\) \(\Rightarrow\) \(\frac{6}{5} \cdot (\frac{2}{3}d + \frac{1}{6}) = \frac{26}{10}\) \(\Rightarrow\) \(\frac{2}{3}d + \frac{1}{6} = \frac{26}{10} \cdot \frac{5}{6}\) \(\Rightarrow\) \(\frac{2}{3}d + \frac{1}{6} = \frac{130}{60}\) \(\Rightarrow\) \(\frac{2}{3}d = \frac{130}{60} - \frac{1}{6}\) \(\Rightarrow\) \(\frac{2}{3}d = \frac{130}{60} - \frac{10}{60}\) \(\Rightarrow\) \(\frac{2}{3}d = \frac{120}{60}\) \(\Rightarrow\) \(\frac{2}{3}d = 2\) \(\Rightarrow\) \(d = 2 \cdot \frac{3}{2}\) \(\Rightarrow\) \(d = 3\)