a, 4n - 7 ⋮ n - 1

=> 4n - 4 - 3 ⋮ n - 1

=> 4(n - 1) - 3 ⋮ n - 1

=> -3 ⋮ n - 1

=> n - 1 thuộc Ư(-3)

=> n - 1 thuộc {-1; 1; -3; 3}

=> n thuộc {0; 2; -2; 4}

a) \(2n+7⋮2n+1\)

\(\Rightarrow\left(2n+1\right)+6⋮2n+1\)

\(\Rightarrow6⋮2n+1\)(vì \(2n+1⋮2n+1\))

\(\Rightarrow2n+1\inƯ\left(6\right)\)

\(\Rightarrow2n+1\in\left\{1;2;3;6\right\}\)

\(\Rightarrow\)\(2n\in\left\{0;1;2;5\right\}\)

\(\Rightarrow n\in\left\{0;1\right\}\)

b) \(3m-9⋮3m-1\)

\(\Rightarrow\left(3m-1\right)-8⋮3m-1\)

\(\Rightarrow8⋮3m-1\)(vì \(3m-1⋮3m-1\))

\(\Rightarrow3m-1\inƯ\left(8\right)\)

\(\Rightarrow3m-1\in\left\{1;2;4;8\right\}\)

\(\Rightarrow3m\in\left\{2;3;5;9\right\}\)

\(\Rightarrow m\in\left\{1;3\right\}\)

Hok "tuốt" nha^^

Mk ra là 1.Thử hoài nhưng chỉ có 1 laf chia hết thôi à. Học giỏi nha, bạn Như

a) \(4\left(n-1\right)-3⋮\left(n-1\right)\)

\(\Rightarrow\left(n-1\right)\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Do \(n\in N\Rightarrow n\in\left\{0;2;4\right\}\)

b) \(-5\left(4-n\right)+12⋮\left(4-n\right)\)

\(\Rightarrow\left(4-n\right)\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)

Do \(n\in N\Rightarrow n\in\left\{16;10;8;7;6;5;3;2;1;0\right\}\)

c) \(-2\left(n-2\right)+6⋮\left(n-2\right)\)

\(\Rightarrow\left(n-2\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Do \(n\in N\Rightarrow n\in\left\{0;1;3;4;5;8\right\}\)

d) \(n\left(n+3\right)+6⋮\left(n+3\right)\)

\(\Rightarrow\left(n+3\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Do \(n\in N\Rightarrow n\in\left\{0;3\right\}\)

Gợi ý :

10 - 2n ⋮ n - 2

-2n + 4 + 6 ⋮ n - 2

-2 ( n - 2 ) + 6 ⋮ n - 2

Vì -2 ( n - 2 ) ⋮ n - 2

=> 6 ⋮ n - 2

=> n - 2 thuộc Ư(6) = { 1; 2; 3; 6; -1; -2; -3; -6 }

Đến đây tìm n nốt nhé bạn :))

Ta có:  ( n - 2)  \(⋮\)n - 2

      =>  2( n - 2) \(⋮\)n - 2

      =>  2n -4      \(⋮\)n - 2

Mà 10 - 2n \(⋮\)n - 2 ( gt )

 Nên ta có :    \([(\)10 - 2n \()\)\(-\)\((\)2n - 4\()]\)\(⋮\)n -2 

               =>    \([\)10 - 2n - 2n -4\(]\)\(⋮\)n - 2  

              =>        6 \(⋮\)n - 2

          =>  n - 2  \(\in\){ 1; 2; 3; 6 } 

          =>   n\(\in\){ 3; 4; 5; 8 }

Vậy  n\(\in\){ 3; 4; 5; 8 }

Ta có :

\(10-2n⋮n-2\)

\(\Rightarrow6+4-2n⋮n-2\)

\(\Rightarrow6-\left(2n-4\right)⋮n-2\)

\(\Rightarrow6-2\left(n-2\right)⋮n-2\)

Ta dễ thấy : \(2\left(n-2\right)⋮n-2\)

\(\Rightarrow6⋮n-2\)

\(\Rightarrow n-2\inƯ\left(6\right)=\left\{6;-6;3;-3;2;-2;1;-1\right\}\)

\(\Rightarrow n\in\left\{8;-4;5;-1;4;0;3;1\right\}\)

Vì \(n\inℕ\Rightarrow n\in\left\{0;1;3;4;5;8\right\}\)

Vậy \(n\in\left\{0;1;3;4;5;8\right\}\).