Математика 6 класс учебник Виленкин, Жохов 2 часть ответы – задача на повторение 1

1) (1,8 + 1 2/7) ∶ (1,8 − 1 2/7) = (1 4⁽⁷/5 + 1 2⁽⁵/7) : (1 4⁽⁷/5 − 1 2⁽⁵/7) = (1 28/35 + 1 10/35) : (1 28/35 − 1 10/35) = 2 38/35 ∶ 18/35 = 108/35 · 35/18 = 108/18 = 6
2) (1,5 + 11/3) ∶ (7,6 – 51/3) = (11⁽³/2 + 11⁽²/3) : (7 3⁽³/5 – 51⁽⁵/3) = (1 3/6 + 1 2/6) : (7 9/15 − 5 5/15) = 2 5/6 ∶ 2 4/15 = 17/6 ∶ 34/15 = 17/6 · 15/34 = 15/6 · 2 = 15/12 = 5/4 = 1 1/4 = 1,25
3) (23 − 15 1/3) ∶ (− 5 1/9) = (22 3/3 – 15 1/3) ∶ (− 46/9) = 7 2/3 · (− 9/46) = − 23/3 · 9/46 = − 3/2 = − 1 1/2 = − 1,5
4) (19 − 13 2/3) ∶ (− 7 1/9) = (18 3/3 – 13 2/3) ∶ (− 64/9) = 5 1/3 · (− 9/64) = − 16/3 · 9/64 = − 3/4 = − 0,75