А)37/48 : 5 2/7-19/42*7 /8 +4 1/5 б)(3*2/9 -4 *1/12): - вопрос №25023073748527194278415329411210336258 от rezistol228 05.10.2021 16:19

Question

Математика: А)37/48 : 5 2/7-19/42*7 /8 +4 1/5 б)(3*2/9 -4 *1/12): (10/3-3 в)6 2/5 : 8/125 -11 1/9 : 1 1/9+3 1/7 г )(3/5 *1 7/11) - 1 7 /11 : 10 ! !

rezistol228 · Answer

А) 37/48 : 5 2/7 - 19/42 * 7/8 + 4 1/5 = 37/48 : 37/7 - 19/48 + 4 1/5 = 37/48 * 7/37 -19/48 + 21/5 = 7/48 - 19/48 + 21/5 = - 12/48 + 21/5 = - 1/4 + 21/5 = - 5/20 + 84/20 = 79/20 =
= 3 19/20
б) (3 * 2/9 - 4* 1/1) : (10/3 - 3) = (2/3 - 1/3) : (10/3 - 9/3) = 1/3 : 1/3 = 1/3 * 3/1 = 1
в) 6 2/5 : 8/125 - 11 1/9 : 1 1/9 + 3 1/7 = 32/5 * 125/8 - 100/9 : 10/9 + 3 1/7 = 100 - 100/9 * 9/10 + 3 1/7 = 100 - 10 + 3 1/7 = 93 1/7
г) (3/5 * 1 7/11) - 1 7/11 : 10 = 3/5 * 1 7/11 - 1 7/11 : 10/1 = 3/5 *1 7/11 -1 7/11 * 1/10 =
1 7/11 * (3/5 - 1/10) = 1 7/11 * (6/10 - 1/10) = 18/11 * 5/10 = 9/11