Ta có
4n - 5 = 4n - 2 - 3 = 2x(2n - 1)-3
Vì 2x(2n-1)\(⋮\)2n - 1 \(\Rightarrow\)3\(⋮\)2n - 1
\(\Rightarrow2n-1\in\left(-3;-1;1;3\right)\)
mik bổ sung
\(\Rightarrow2n\in\left(-2;0;2;4\right)\)
\(\Rightarrow n\in\left(-1;0;1;2\right)\)
Ta có: \(4n-5⋮2n-1\)
<=> \(2\cdot\left(2n-1\right)-3⋮2n-1\)
Mà \(2\left(2n-1\right)⋮2n-1\)
\(\Rightarrow3⋮2n-1\)
\(\Rightarrow2n-1\inƯ\left(3\right)=\left\{-1;1;-3;3\right\}\)
Ta có bảng sau:
| \(2n-1\) | \(-1\) | \(1\) | \(-3\) | \(3\) |
| \(n\) | \(0\) | \(1\) | \(-1\) | \(2\) |
Vậy \(n\in\left\{0;1;-1;2\right\}\)thì \(4n-5⋮2n-1\)
Ta có: \(4n-5⋮2n-1\)
<=> \(2\left(2n-1\right)-3⋮2n-1\)
Mà \(2\left(2n-1\right)⋮2n-1\)
\(\Rightarrow3⋮2n-1\)
\(\Rightarrow2n-1\inƯ\left(3\right)=\left\{-1;1;-3;3\right\}\)
Ta có bảng sau:
| \(2n-1\) | \(-1\) | \(1\) | \(-3\) | \(3\) |
| \(n\) | \(0\) | \(1\) | \(-1\) | \(2\) |
Vì n là STN=> n=0 hoặc n = 2
Vậy n là STN \(n\in\left\{0;2\right\}\)thì\(4n-5⋮2n-1\)
