Вопрос:

14-\(49\frac{1}{3}:16-14:8\frac{1}{6}\)\(\cdot\) 7 / (1\(\frac{17}{18}\)\(\cdot\) \(1\frac{59}{70}+\frac{37}{42}+2\frac{19}{30}\)-10)

Ответ:

Решение:

Решим выражение по действиям.

  1. \(49\frac{1}{3} = \frac{49 \cdot 3 + 1}{3} = \frac{147+1}{3} = \frac{148}{3}\)
  2. \(8\frac{1}{6} = \frac{8 \cdot 6 + 1}{6} = \frac{48+1}{6} = \frac{49}{6}\)
  3. \(\frac{148}{3} : 16 = \frac{148}{3 \cdot 16} = \frac{37}{3 \cdot 4} = \frac{37}{12}\)
  4. \(14 : \frac{49}{6} = 14 \cdot \frac{6}{49} = \frac{2 \cdot 6}{7} = \frac{12}{7}\)
  5. \(\frac{37}{12} - \frac{12}{7} = \frac{37 \cdot 7 - 12 \cdot 12}{84} = \frac{259 - 144}{84} = \frac{115}{84}\)
  6. \(\frac{115}{84} \cdot 7 = \frac{115}{12}\)
  7. \(14 - \frac{115}{12} = \frac{14 \cdot 12 - 115}{12} = \frac{168 - 115}{12} = \frac{53}{12}\)
  8. \(1\frac{17}{18} = \frac{18+17}{18} = \frac{35}{18}\)
  9. \(1\frac{59}{70} = \frac{70+59}{70} = \frac{129}{70}\)
  10. \(\frac{129}{70} + \frac{37}{42} + 2\frac{19}{30} = \frac{129}{70} + \frac{37}{42} + \frac{60+19}{30} = \frac{129}{70} + \frac{37}{42} + \frac{79}{30}\)
  11. \(LCM(70, 42, 30) = LCM(2 \cdot 5 \cdot 7, 2 \cdot 3 \cdot 7, 2 \cdot 3 \cdot 5) = 2 \cdot 3 \cdot 5 \cdot 7 = 210\)
  12. \(\frac{129 \cdot 3}{210} + \frac{37 \cdot 5}{210} + \frac{79 \cdot 7}{210} = \frac{387 + 185 + 553}{210} = \frac{1125}{210} = \frac{75}{14}\)
  13. \(\frac{35}{18} \cdot \frac{75}{14} = \frac{5 \cdot 5}{18 \cdot 2} = \frac{25}{12}\)
  14. \(\frac{25}{12} - 10 = \frac{25 - 120}{12} = -\frac{95}{12}\)
  15. \(\frac{53}{12} / (-\frac{95}{12}) = \frac{53}{12} \cdot (-\frac{12}{95}) = -\frac{53}{95}\)

Ответ: -53/95

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