Ta có: \(4n-15⋮2n-4\)
\(\Leftrightarrow4n-8-7⋮2n-4\)
mà \(4n-8⋮2n-4\)
nên \(-7⋮2n-4\)
\(\Leftrightarrow2n-4\inƯ\left(-7\right)\)
\(\Leftrightarrow2n-4\in\left\{1;-1;7;-7\right\}\)
\(\Leftrightarrow2n\in\left\{5;3;11;-3\right\}\)
hay \(n\in\left\{\dfrac{5}{2};\dfrac{3}{2};\dfrac{11}{2}-\dfrac{3}{2}\right\}\)
Vậy: \(n\in\left\{\dfrac{5}{2};\dfrac{3}{2};\dfrac{11}{2}-\dfrac{3}{2}\right\}\)