\(6n+8⋮3n-1\)
\(\Rightarrow6n+8-2\left(3n-1\right)⋮3n-1\)
\(\Rightarrow6n+8-6n+2⋮3n-1\)
\(\Rightarrow10⋮3n-1\)
\(\Rightarrow3n-1\inƯ\left(10\right)\)
\(\Rightarrow3n-1\in\left\{1;2;5;10\right\}\)
| 3n - 1 | 1 | 2 | 5 | 10 |
| n | \(\frac{2}{3}\) (loại) | 1 (chọn) | 2 (chọn) | \(\frac{11}{3}\) (loại) |
Vậy n \(\in\) {1;2}
Do \(6n+8⋮3n-1\)mà 3n-1 chia hết cho 3n-1
nên \(2.\left(3n-1\right)+10⋮3n+1\)
Mà \(2.\left(3n-1\right)⋮3n-1\)
\(\Rightarrow10⋮3n-1\)
hay 3n-1\(\inƯ\left(10\right)\)
\(Ư\left(10\right)\in\left\{1;2;5;10\right\}\)
\(\Rightarrow3n-1\in\left\{1;2;5;10\right\}\)
Ta xét các trường hợp sau:
TH1: 3n-1=1
\(\Rightarrow3n=2\Rightarrow n=\frac{2}{3}\left(loai\right)\)
TH2 : 3n-1=2
\(\Rightarrow3n=3\Rightarrow n=1\left(TM\right)\)
TH3: 3n-1=5
\(\Rightarrow3n=6\Rightarrow n=2\left(TM\right)\)
TH4: 3n-1=10
\(\Rightarrow3n=11\Rightarrow n=\frac{11}{3}\left(loai\right)\)
Vậy n=2 hoặc n=1
