Câu `15_b`:
`B = 1/5 - 3/7 + 5/9 - 1/11 + 7/13 - 9/16 - 7/13 + 2/11 - 5/9 + 3/7 - 1/5`
`= (1/5 - 1/5) + (-3/7 + 3/7) + (5/9 - 5/9) + (-1/11 + 2/11) + (7/13 - 7/13) - 9/16`
`= 0 + 0 + 0 + 1/11 + 0 - 9/16`
`= 1/11 - 9/16`
`= 16/176 - 99/176`
`= -83/176`
Bài `16`:
`x/6 - 1/y = 1/2`
`=> x/6 - 1/2 = 1/y`
`=> x/6 - 3/6 = 1/y`
`=> (x - 3)/6 = 1/y`
$\text{⇒ (x - 3)y = 6 = 1. 6 = (-1). (-6) = 2. 3 = (-2). (-3)}$
Vì `x, y in mathbb Z` nên ta có bảng sau:
Trogn ảnh
Vậy
`(x, y) ∈ {(4; 6), (9; 1), (2; -6), (-3; -1), (5; 3), (6; 2), (1; -3), (0; -2)}`
