a, \(3n+2⋮n-1\)

\(\Rightarrow3n-3+5⋮n-1\)

\(\Rightarrow3\left(n-1\right)+5⋮n-1\)

Vì : \(3\left(n-1\right)⋮n-1\Rightarrow5⋮n-1\)

\(\Rightarrow n-1\inƯ\left(5\right)\)

\(\Rightarrow n-1\in\left\{1;5\right\}\)

+) \(n-1=1\Rightarrow n=1+1\Rightarrow n=2\)

+) \(n-1=5\Rightarrow n=5+1\Rightarrow n=6\)

Vậy : \(n\in\left\{2;6\right\}\) thì \(3n+2⋮n-1\)

b, \(n+8⋮n+3\)

Vì : \(n+3⋮n+3\)

\(\Rightarrow\left(n+8\right)-\left(n+3\right)⋮n+3\)

\(\Rightarrow n+8-n-3⋮n+3\)

\(\Rightarrow5⋮n+3\)

\(\Rightarrow n+3\inƯ\left(5\right)\)

Mà : \(n+3\ge3\)

\(\Rightarrow n+3=5\Rightarrow n=5-3\Rightarrow n=2\)

Vậy n = 2 thì : \(n+8⋮n+3\)

c, \(n+6⋮n-1\)

Mà : \(n-1⋮n-1\)

\(\Rightarrow\left(n+6\right)-\left(n-1\right)⋮n-1\)

\(\Rightarrow n+6-n+1⋮n-1\)

\(\Rightarrow7⋮n-1\)

\(\Rightarrow n-1\inƯ\left(7\right)\)

\(\Rightarrow n-1\in\left\{1;7\right\}\)

+) \(n-1=1\Rightarrow n=1+1\Rightarrow n=2\)

+) \(n-1=7\Rightarrow n=7+1\Rightarrow n=8\)

Vậy \(n\in\left\{2;8\right\}\) thì \(n+6⋮n-1\)

d, \(4n-5⋮2n-1\)

\(\Rightarrow4n-2-3⋮2n-1\)

\(\Rightarrow2\left(2n-1\right)-3⋮2n-1\)

Vì : \(2\left(2n-1\right)⋮2n-1\)

\(\Rightarrow3⋮2n-1\)

\(\Rightarrow2n-1\inƯ\left(3\right)\)

\(\Rightarrow2n-1\in\left\{1;3\right\}\)

+) \(2n-1=1\Rightarrow2n=1+1\Rightarrow2n=2\Rightarrow n=2\div2\Rightarrow n=1\)

+) \(2n-1=3\Rightarrow2n=3+1\Rightarrow2n=4\Rightarrow n=4\div2\Rightarrow n=2\)

Vậy \(n\in\left\{1;2\right\}\) thì \(4n-5⋮2n-1\)

a) (n+2) \(⋮\) (n-1)

vì (n-1)\(⋮\) (n-1)

=>(n+2)-(n-1)\(⋮\left(n-1\right)\)

=>(n+2-n+1)\(⋮\) (n-1)

=> 3\(⋮\) (n-1)

=>(n-1)\(\in\) Ư(3) = { \(\pm\)1,\(\pm\)3}

ta có bảng

vậy n\(\in\) { 0;2;4}

b) \(\left(2n+7\right)⋮\left(n+1\right)\)

vì\(\left(n+1\right)⋮\left(n+1\right)\)

=>\(2\left(n+1\right)⋮\left(n+1\right)\)

=> \(\left(2n+2\right)⋮\left(n+1\right)\)

=>\(\left(2n+7\right)-\left(2n+2\right)⋮\left(n+1\right)\)

=>\(\left(2n+7-2n-2\right)⋮\left(n+1\right)\)

=>\(5⋮\left(n+1\right)\)

=> \(\left(n+1\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

TA CÓ BẢNG

vậy \(n\in\left\{0;4\right\}\)

các câu khác làm tương tự nha bạn

1.=> n+7-(n+2) chia hết cho n+2

=>n+7-n-2 chia hết cho n+2

=>5 chia hết cho n+2

=>n+2 thuộc Ư(5)=1;5

ta có bảng:

Vậy n=3

MÌNH MỚI NGHĨ ĐƯỢC TỚI ĐÂY THÔI XIN LỖI NHÉ

3.3n+15 chia hết cho n+1

=>3n+15-n+1 chia hết cho n+1

=>3n+15-3(n+1) chia hết cho n+1 

=>3n+15-3n-3 chia hết cho n+1 

=>12 chia hết cho n+1 

=>n+1 thuộc Ư(12)=1;2;3;4;6;12

ta có bảng:

Vậy n thuộc 0;1;2;3;11