a) Ta có: (x + 22) ⋮ (x + 1)
\(\Rightarrow\) (x + 1 + 21) ⋮ (x + 1)
mà (x + 1) ⋮ (x + 1) nên 21 ⋮ (x + 1)
Vậy (x + 1) \(\in\) Ư(21)
b) Ta có: (2x + 23) \(\in\) B(x – 1)
\(\Rightarrow\) (2x + 23) \(⋮\) (x - 1)
\(\Rightarrow\) (2x - 2 + 25) \(⋮\) (x - 1)
mà (2x - 2) \(⋮\) (x - 1) nên 25 \(⋮\) (x - 1)
Vậy (x - 1) \(\in\) Ư(25) = {\(\pm1;\pm5;\pm25\)}
\(\Rightarrow\) x \(\in\) {0; 2; -4; 6; -24; 26}
mà x \(\in\) N nên x \(\in\) {0; 2; 6; 26}
e) Ta có: xy + x + 2y = 5
\(\Rightarrow\) xy + x + 2y + 2 = 5 + 2
\(\Rightarrow\) x(y + 1) + 2(y + 1) = 7
\(\Rightarrow\) (x + 2)(y + 1) = 7
mà x, y \(\in\) N
Nên (x + 2) \(\in\) Ư(7); (y + 1) \(\in\) Ư(7)
Ta có bảng sau:
| x + 2 | 1 | -1 | 7 | -7 |
| x | -1 | -3 | 5 | -9 |
| y + 1 | 7 | -7 | 1 | -1 |
| y | 6 | -8 | 0 | -2 |
Vậy x = 5 ; y = 0.
c) Ta có: (3x + 1) ⋮ (2x – 1)
\(\Rightarrow\) 2(3x + 1) ⋮ (2x - 1)
\(\Rightarrow\) (6x + 2) ⋮ (2x - 1)
\(\Rightarrow\) (6x - 3 + 5) ⋮ (2x - 1)
mà (6x - 3) ⋮ (2x - 1) nên 5 ⋮ (2x - 1)
Vậy (2x - 1) \(\in\) Ư(5).
phần còn lại bn làm tương tự câu b nha :v
