Домашнее задание по теме: «Приведение подобных слагаем №1. Найдите значение выражения. 7 -2:5 3 10 1 15 2 2)1.1 2.+(-):4. 53 10 √2.(5-11)+11.(2-5). 13 13 13 13 19 -3,8: (2+(-2,5))-2. 95 3 7 (-0,42): (1,75 -3,5 (-0,1)) : (-0,05). 5,25 (-0,1 + 3,05) - 3,05. (5,25 - 0,1). №2. Приведите подобные слагаемые

1. \[\sqrt{5} + \frac{2}{3} \cdot (-2\frac{1}{2}) = \sqrt{5} + \frac{2}{3} \cdot (-\frac{5}{2}) = \sqrt{5} - \frac{2 \cdot 5}{3 \cdot 2} = \sqrt{5} - \frac{5}{3} = \sqrt{5} - 1\frac{2}{3}\]

2. \[2\frac{1}{5} \cdot \frac{1}{3} + (-\frac{3}{10}) : 4\frac{1}{2} = \frac{11}{5} \cdot \frac{1}{3} + (-\frac{3}{10}) : \frac{9}{2} = \frac{11}{15} - \frac{3}{10} \cdot \frac{2}{9} = \frac{11}{15} - \frac{3 \cdot 2}{10 \cdot 9} = \frac{11}{15} - \frac{1}{15} = \frac{10}{15} = \frac{2}{3}\]

3. \[\sqrt{2\frac{1}{13}} \cdot (5 - 11\frac{2}{13}) + 11\frac{2}{13} \cdot (2\frac{1}{13} - 5) = \sqrt{\frac{27}{13}} \cdot (5 - \frac{145}{13}) + \frac{145}{13} \cdot (\frac{27}{13} - 5) = \sqrt{\frac{27}{13}} \cdot (\frac{65}{13} - \frac{145}{13}) + \frac{145}{13} \cdot (\frac{27}{13} - \frac{65}{13}) = \sqrt{\frac{27}{13}} \cdot (-\frac{80}{13}) + \frac{145}{13} \cdot (-\frac{38}{13}) = \sqrt{\frac{27}{13}} \cdot (-\frac{80}{13}) - \frac{145 \cdot 38}{13 \cdot 13} = \sqrt{\frac{27}{13}} \cdot (-\frac{80}{13}) - \frac{5510}{169}\]

4. \[-3.8 : (2\frac{3}{7} + \frac{3}{7} \cdot (-2.5)) - 2\frac{19}{95} = -3.8 : (\frac{17}{7} + \frac{3}{7} \cdot (-2.5)) - \frac{209}{95} = -3.8 : (\frac{17}{7} - \frac{3}{7} \cdot 2.5) - \frac{209}{95} = -3.8 : (\frac{17}{7} - \frac{7.5}{7}) - \frac{209}{95} = -3.8 : \frac{9.5}{7} - \frac{209}{95} = -3.8 \cdot \frac{7}{9.5} - \frac{209}{95} = -\frac{3.8 \cdot 7}{9.5} - \frac{209}{95} = -2.8 - 2.2 - \frac{9}{95} = -5.09\]

5. \[(-0.42) : (1.75 - 3.5 \cdot (-0.1)) : (-0.05) = (-0.42) : (1.75 + 0.35) : (-0.05) = (-0.42) : 2.1 : (-0.05) = -0.2 : (-0.05) = 4\]

6. \[5.25 \cdot (-0.1 + 3.05) - 3.05 \cdot (5.25 - 0.1) = 5.25 \cdot 2.95 - 3.05 \cdot 5.15 = 15.4875 - 15.6975 = -0.21\]

7. \[-2 : 5\frac{1}{3} - \frac{7}{10} : 1\frac{1}{15} = -2 : \frac{16}{3} - \frac{7}{10} : \frac{16}{15} = -2 \cdot \frac{3}{16} - \frac{7}{10} \cdot \frac{15}{16} = -\frac{6}{16} - \frac{105}{160} = -\frac{60}{160} - \frac{105}{160} = -\frac{165}{160} = -1\frac{5}{16}\]

8. \[\frac{3}{7} - (1\frac{1}{4} - 1\frac{1}{8} \cdot \frac{2}{9}) = \frac{3}{7} - (\frac{5}{4} - \frac{9}{8} \cdot \frac{2}{9}) = \frac{3}{7} - (\frac{5}{4} - \frac{1}{4}) = \frac{3}{7} - \frac{4}{4} = \frac{3}{7} - 1 = -\frac{4}{7}\]