2,4+5,6×(13 целых 3/4-12 целых 13/14)-Помогите пожалуйста запуталась в решении этого примера!!!!!!!!

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Ответы и объяснения

2014-01-21T20:46:02+04:00
А) 24и(4/5) - [19,5:(7и2/9)] = (124/5) - [(19и5/10):(7и2/9)] = (124/5) - [(195/10)*(9/65)]=(248/10) - (27/10) = (221/10) = 22и(1/10) 

б) 2,4 + 5,6*[(13и3/4) - (12и13/14)] = (24/10) + (56/10)*[(55/4) - (181/14)] = (24/10) + (56/10)*[(385/28) - (362/28)] = (24/10) + (56/10)*(23/28) = (24/10) + (46/10) = (70/10) = 7 
2014-01-21T20:48:28+04:00
А

) 24и(4/5) - [19,5:(7и2/9)] = (124/5) - [(19и5/10):(7и2/9)] = (124/5) - [(195/10)*(9/65)]=(248/10) - (27/10) = (221/10) = 22и(1/10) 

б
) 2,4 + 5,6*[(13 
и 3/4) - (12 и 13/14)] = (24/10) + (56/10)*[(55/4) - (181/14)] = (24/10) + (56/10)*[(385/28) - (362/28)] = (24/10) + (56/10)*(23/28) = (24/10) + (46/10) = (70/10) = 7