\(\dfrac{n+2}{n+1}:\dfrac{3}{4}=Z\) \(\Leftrightarrow\dfrac{n+2}{n+1}.\dfrac{4}{3}=Z\)
\(\dfrac{4\left(n+2\right)}{3\left(n+1\right)}=Z\)\(\Leftrightarrow\dfrac{4n+8}{3n+3}=Z\)
\(\Leftrightarrow4n+8⋮3n+3\)
\(\Leftrightarrow3\left(4n+8\right)⋮3n+3\)
\(12n+24⋮3n+3\)
\(12n+12+12⋮3n+3\)
\(4\left(3n+3\right)+12⋮3n+3\)
\(\Leftrightarrow12⋮3n+3\)
\(\Leftrightarrow3n+3\inƯ\left(12\right)\)
\(Ư\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
\(n\in\left\{-2;1;-5\right\}\)
Ta có :
\(\dfrac{n+2}{n+1}:\dfrac{3}{4}=\dfrac{4\left(n+2\right)}{3\left(n+1\right)}=\dfrac{4n+8}{3n+3}\)
Để \(\dfrac{n+2}{n+1}:\dfrac{3}{4}\in Z\) thì :
\(4n+8⋮3n+3\)
Mà \(3n+3⋮3n+3\)
\(\Rightarrow\left\{{}\begin{matrix}12n+24⋮3n+3\\12n+12⋮3n+3\end{matrix}\right.\)
\(\Rightarrow12⋮3n+3\)
\(\Rightarrow3n+3\inƯ\left(12\right)\)
Ta có bảng :
| \(3n+3\) | \(1\) | \(12\) | \(3\) | \(4\) | \(-1\) | \(-12\) | \(-3\) | \(-4\) |
| \(n\) | \(-1\) | \(3\) | \(0\) | \(\dfrac{1}{3}\) | \(\dfrac{-4}{3}\) | \(-5\) | \(3\) | \(-\dfrac{7}{3}\) |
| \(Đk\) \(n\in Z\) | TM | TM | TM | loại | loại | TM | TM | loại |
Vậy \(n\in\left\{-1;3;0;-5;3\right\}\) là giá trị cần tìm
