Ta có \(E=\frac{5n-4}{2n+5}\)
\(\Rightarrow2E=\frac{10n-8}{2n+5}=\frac{5\left(2n+5\right)-33}{2n+5}=5-\frac{33}{2n+5}\)
Để E nguyên => 2E nguyên => 5-\(\frac{33}{2n+5}\)nguyên
=> \(\frac{33}{2n+5}\)nguyên
=> \(33⋮2n+5\)
\(\Rightarrow2n+5=Ư_{\left(33\right)}=\left\{-33;-1;1;33\right\}\)
Ta có bảng
2n+5 | -33 | -1 | 1 | 33 |
2n | -38 | -6 | -4 | 28 |
n | -19 | -3 | -2 | 14 |
Vậy n={-19;-3;-2;14}