Вопрос:
H) \(\frac{5}{63} \cdot 2\frac{3}{14} + \frac{5}{63} \cdot \frac{2}{21} + \frac{58}{63} : \frac{42}{97}\)
Ответ:
Решение:
- \(2\frac{3}{14} = \frac{2 \cdot 14 + 3}{14} = \frac{31}{14}\)
- \(\frac{5}{63} \cdot \frac{31}{14} = \frac{155}{882}\)
- \(\frac{5}{63} \cdot \frac{2}{21} = \frac{10}{1323}\)
- \(\frac{58}{63} : \frac{42}{97} = \frac{58}{63} \cdot \frac{97}{42} = \frac{58 \cdot 97}{63 \cdot 42} = \frac{5626}{2646}\)
- \(\frac{155}{882} + \frac{10}{1323} + \frac{5626}{2646}\)
- \(882 = 2 \cdot 3^2 \cdot 7^2\)
- \(1323 = 3^3 \cdot 7^2\)
- \(2646 = 2 \cdot 3^3 \cdot 7^2\)
- \(LCM(882, 1323, 2646) = 2 \cdot 3^3 \cdot 7^2 = 2646\)
- \(\frac{155 \cdot 3}{882 \cdot 3} + \frac{10 \cdot 2}{1323 \cdot 2} + \frac{5626}{2646} = \frac{465}{2646} + \frac{20}{2646} + \frac{5626}{2646} = \frac{465+20+5626}{2646} = \frac{6111}{2646}\)
- \(6111 = 3 \cdot 2037 = 3^2 \cdot 679 = 3^2 \cdot 7 \cdot 97\)
- \(2646 = 2 \cdot 3^3 \cdot 7^2\)
- \(\frac{6111}{2646} = \frac{3^2 \cdot 7 \cdot 97}{2 \cdot 3^3 \cdot 7^2} = \frac{97}{2 \cdot 3 \cdot 7} = \frac{97}{42}\)
Ответ: \(\frac{97}{42}\)
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