\(\Rightarrow\frac{3n+2}{n-1}=\frac{3n-3+5}{n-1}=\frac{3n-3}{n-1}+\frac{5}{n-1}\)
\(\Rightarrow3+\frac{5}{n-1}\)
\(\Rightarrow n-1\inƯ_5\left\{-5;-1;1;5\right\}\)
\(\Rightarrow\left[\begin{array}{nghiempt}n-1=-5\\n-1=-1\\n-1=1\\n-1=5\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}n=-4\\n=0\\n=2\\n=6\end{array}\right.\)
Vậy: Các giá trị nguyên tập hợp của n là:
\(n=-4;0;2;6\)
Đặt \(A=\frac{3n+2}{n-1}=\frac{3n-3+5}{n-1}=\frac{3\left(n-1\right)+5}{n-1}=3+\frac{5}{n-1}\)
\(\Rightarrow A\in Z\Leftrightarrow3+\frac{5}{n-1}\in Z\Leftrightarrow\frac{5}{n-1}\in Z\Leftrightarrow5⋮n-1\Leftrightarrow n-1\inƯ\left(5\right)\)
\(\Rightarrow n-1\in\left\{-1;-5;1;5\right\}\)
\(\Rightarrow n\in\left\{0;-4;2;6\right\}\)
Để \(\frac{3n+2}{n-1}\) là số nguyên \(\Rightarrow3n+2⋮n-1\)
Vì \(3n+2⋮n-1\) và \(n-1⋮n-1\Rightarrow\left(n-1\right).3⋮n-1\Rightarrow3n-3⋮n-1\)
\(\Rightarrow3n+2-\left(3n-3\right)⋮n-1\Rightarrow3n+2-3n+3⋮n-1\)
\(\Rightarrow5⋮n-1\Rightarrow n-1\in\left\{-1;1;-5;5\right\}\Rightarrow n\in\left\{0;2;-4;6\right\}\)
Vậy \(n\in\left\{0;2;-4;6\right\}\)
để \(\frac{3n+2}{n-1}\)đạt giá trị nguyên thì 3n+2\(⋮n-1\)
ta có : 3n-3+5\(⋮\)n-1
3(n-1)+5\(⋮\)n-1
vì 3(n-1)\(⋮\)n-1 nên 5\(⋮\)n-1
n-1\(\in\left\{\pm1,\pm5\right\}\)
ta xét bảng :
| n-1 | -5 | -1 | 1 | 5 |
| n | -4 | 0 | 2 | 6 |
vậy n\(\in\left\{-4;0;2;6\right\}\)