\(C=\frac{n+2}{2n-1}\)
Do C là nguyên
=> \(n+2⋮2n-1\)
=> \(n+n-\left(n-2\right)⋮2n-1\)
=> \(2n-\left(n-2\right)⋮2n-1\)
=> \(n-2⋮1\)
=> \(n-2\in\left\{\pm1\right\}\)
=> \(\left[{}\begin{matrix}n-2=1=>n=3\\n-2=-1=>n=1\end{matrix}\right.\)
Vậy để C là ngyên thì
\(n\in\left\{3;1\right\}\)
\(D=\frac{2n+1}{3n+1}\)
Do D là nguyên
=> \(2n+1⋮3n+1\)
=> \(2n+n-\left(n-1\right)⋮3n+1\)
=> \(3n-\left(n-1\right)⋮3n+1\)
=> \(n-1⋮1\)
=> \(n-1\in\left\{\pm1\right\}\)
=> \(\left[{}\begin{matrix}n-=1=>n=2\\n-1=-1=>n=0\end{matrix}\right.\)
Vậy để D là nguyên thì
\(n\in\left\{2;0\right\}\)