• Предмет: Алгебра
  • Автор: olenaalistratova7
  • Вопрос задан 6 лет назад

пж помогитееееееннннннннн​

Приложения:

Ответы

Ответ дал: Vopoxov
0

Объяснение:

14 \tfrac{7}{15}  - 3 \tfrac{3}{23} \cdot \tfrac{23}{27}  - 1 \tfrac{1}{5} \cdot \tfrac{1}{6} =  \\  = \tfrac{15 \cdot14 + 7}{15}  - \tfrac{3\cdot23 + 3}{23} \cdot \tfrac{23}{27}  - \tfrac{5 + 1}{5} \cdot \tfrac{1}{6} =   \\  = \tfrac{210 + 7}{15}  - \tfrac{69 + 3}{23} \cdot \tfrac{23}{27}  - \tfrac{5 + 1}{5} \cdot \tfrac{1}{6} =   \\  \large  = \tfrac{217}{15}  - \tfrac{72}{23} \cdot \tfrac{23}{27}  - \tfrac{6}{5} \cdot \tfrac{1}{6} =  \\ \large  = \tfrac{217}{15}  - \tfrac{ \cancel{9 \ }\cdot8}{\cancel{23}} \cdot \tfrac{\cancel{23}}{ \cancel{9 \ }\cdot3}  - \tfrac{ \cancel{6 \ }}{5} \cdot \tfrac{1}{ \cancel{6 \ }} =  \\ \large  = \tfrac{217}{15}  - \tfrac{ 8}{ 3}  - \tfrac{1}{5} =  \tfrac{217}{5 \cdot3}  - \tfrac{ 8 \cdot5}{ 3 \cdot5}  - \tfrac{1 \cdot3}{5 \cdot3} =  \\  \large \tfrac{217 - 40 - 3}{15}  =  \tfrac{174}{15}  = \tfrac{58 \cdot\cancel{3}}{5 \cdot\cancel{3}} =  \tfrac{58}{5}  =  \\  = 11 \tfrac{3}{5}  = 11.6

(5 \tfrac{8}{9} :1\tfrac{17}{36} + 1\tfrac{1}{4}) \cdot\tfrac{5}{21} = \\  =  (\tfrac{45 + 8}{9} :\tfrac{36 + 17}{36} + \tfrac{4  + 1}{4}) \cdot\tfrac{5}{21} =  \\  = (\tfrac{53}{9}  \cdot\tfrac{36}{53} + \tfrac{5}{4}) \cdot\tfrac{5}{21} =  \\ = (\tfrac{\cancel{53}}{\cancel{9 \:  }}  \cdot\tfrac{\cancel{9 \ } \cdot4}{\cancel{53}} + \tfrac{5}{4}) \cdot\tfrac{5}{21} = (4 +  \tfrac{5}{4}) \cdot\tfrac{5}{21} = \\  =  (\tfrac{16}{4}  +  \tfrac{5}{4}) \cdot\tfrac{5}{21} = \tfrac{16 + 5}{4} \cdot\tfrac{5}{21} = \tfrac{\cancel{21}}{4} \cdot\tfrac{5}{\cancel{21}} = \\  =  \frac{5}{4}  = 1 \frac{1}{4}  = 1.25

( - 1\tfrac{3}{8}- 2\tfrac{5}{12}):  5\tfrac{5}{12} =  \\  = ( -  \tfrac{8 + 3}{8}- \tfrac{24 + 5}{12}): \tfrac{60 + 5}{12} =  \\ = ( -  \tfrac{11}{8}- \tfrac{29}{12}) \cdot \tfrac{12}{65} = \\ = ( -  \tfrac{11}{2 \cdot4}- \tfrac{29}{ 3\cdot4}) \cdot \tfrac{12}{65} =  \\ = ( -  \tfrac{11 \cdot3}{2  \cdot3\cdot4}- \tfrac{29 \cdot2}{  2\cdot3\cdot4}) \cdot \tfrac{12}{65} =  \\  = -  \tfrac{33 + 58}{24} \cdot \tfrac{12}{65}  =   \\  = -  \tfrac{33 + 58}{2 \cdot \cancel{12}} \cdot \tfrac{ \cancel{12}}{65}  =   -  \tfrac{91}{130}  =  -  \tfrac{ \cancel{13} \cdot7}{ \cancel{13} \cdot10}  =  \\  =  -  \frac{7}{10}  =  - 0.7

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