4n-5⋮2n-1
⇒4n-2-3⋮2n-1
4n-2⋮2n-1 3⋮2n-1⇒2n-1∈Ư(3)
Ư(3)={1;3}
⇒n∈{1;2}
Vậy n∈{1;2}
Ta có: \(4n-5⋮2n-1\)
\(\Leftrightarrow4n-2-3⋮2n-1\)
mà \(4n-2⋮2n-1\)
nên \(-3⋮2n-1\)
\(\Leftrightarrow2n-1\inƯ\left(-3\right)\)
\(\Leftrightarrow2n-1\in\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow2n\in\left\{2;0;4;-2\right\}\)
\(\Leftrightarrow n\in\left\{0;1;2;-1\right\}\)
mà \(n\in N\)
nên \(n\in\left\{0;1;2\right\}\)
Vậy: \(n\in\left\{0;1;2\right\}\)