Để \(\frac{2n^2+3n+3}{2n-1}\)là số nguyên
\(\Leftrightarrow2n^2+3n+3⋮2n-1\)
\(\Leftrightarrow n\left(2n-1\right)+4n+3⋮2n-1\)
Mà \(n\left(2n-1\right)⋮2n-1\)
\(\Rightarrow4n+3⋮2n-1\)
\(\Leftrightarrow2\left(2n-1\right)+5⋮2n-1\)
Mà \(2\left(2n-1\right)⋮2n-1\)
\(\Rightarrow5⋮2n-1\)
\(\Rightarrow2n-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng sau :
Vậy \(n\in\left\{0;1;3;-2\right\}\)
\(A=\frac{2n^2+3n+3}{2n-1}=n+2+\frac{5}{2n-1}\)
Để A nguyên \(\Leftrightarrow\frac{5}{2n-1}\) nguyên
\(\Rightarrow2n-1=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(\Rightarrow2n=\left\{-4;0;2;6\right\}\)
\(\Rightarrow n=\left\{-2;0;1;3\right\}\)
