Ta có:
\(\frac{4n-5}{2n-1}=\frac{2\left(2n-1\right)-3}{2n-1}=2-\frac{3}{2n-1}\)
Để (4n - 5) \(⋮\) (2n - 1) thì 3 \(⋮\) (2n - 1)
\(\Rightarrow\) 2n - 1 = -1; 2n - 1 = 1; 2n - 1 = 3; 2n - 1 = -3
*) 2n - 1 = 1
2n = 2
n = 1
*) 2n - 1 = -1
2n = 0
n = 0
*) 2n - 1 = 3
2n = 4
n = 2
*) 2n - 1 = -3
2n = -2
n = -1
Vậy n = -1; n = 2; n = 0; n = 1
4n-5⋮2n-1
4n-2-3⋮2n-1
2.2n-2.1-3⋮2n-1
2(2n-1)-3⋮2n-1
Vì 2(2n-1)-3⋮2n-1
2n-1⋮2n-1 => 2(2n-1)⋮2n-1
=>3⋮2n-1
=>2n-1∈ Ư(3)
=>2n-1=1; -1; 3; -3
Ta có bảng sau:
2n-1 | 1 | -1 | 3 | -3 |
n | 1 | 0 | 2 | -1 |
Vậy n=1; 0; 2; -1.