\(C\in Z\Leftrightarrow\left\{{}\begin{matrix}\dfrac{n-2}{2n+3}\in Z\left(1\right)\\\dfrac{n^2+2n+4}{2n+3}\in Z\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow n-2⋮2n+3\\ \Leftrightarrow2n-4⋮2n+3\\ \Leftrightarrow2n+3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow n\in\left\{-5;-2;-1;2\right\}\left(3\right)\\ \left(2\right)\Leftrightarrow2n^2+4n+8⋮2n+3\\ \Leftrightarrow2n\left(n+3\right)-\left(2n+3\right)+11⋮2n+3\\ \Leftrightarrow2n+3\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\\ \Leftrightarrow n\in\left\{-7;-2;-1;4\right\}\left(4\right)\\ \left(3\right)\left(4\right)\Leftrightarrow n\in\left\{-2;-1\right\}\)
Vậy ...
Tìm x là số nguyên để mỗi phân số sau nhận giá trị nguyên 
