1: =>3n-12+17 chia hết cho n-4

=>\(n-4\in\left\{1;-1;17;-17\right\}\)

hay \(n\in\left\{5;3;21;-13\right\}\)

2: =>6n-2+9 chia hết cho 3n-1

=>\(3n-1\in\left\{1;-1;3;-3;9;-9\right\}\)

hay \(n\in\left\{\dfrac{2}{3};0;\dfrac{4}{3};-\dfrac{2}{3};\dfrac{10}{3};-\dfrac{8}{3}\right\}\)

4: =>2n+4-11 chia hết cho n+2

=>\(n+2\in\left\{1;-1;11;-11\right\}\)

hay \(n\in\left\{-1;-3;9;-13\right\}\)

5: =>3n-4 chia hết cho n-3

=>3n-9+5 chia hết cho n-3

=>\(n-3\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{4;2;8;-2\right\}\)

6: =>2n+2-7 chia hết cho n+1

=>\(n+1\in\left\{1;-1;7;-7\right\}\)

hay \(n\in\left\{0;-2;6;-8\right\}\)

a: =>4n-2-3 chia hết cho 2n-1

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

=>\(n\in\left\{1;0;2\right\}\)

b: =>6n-4+11 chia hết cho 3n-2

=>\(3n-2\in\left\{1;-1;11;-11\right\}\)

=>\(n\in\left\{1\right\}\)

Chứng minh hay sao

Đề bài bài này là : Tìm n thuộc N, biết

a) Ta có:

\(4n+7⋮3n-2\)

\(\Rightarrow3\left(4n+7\right)⋮3n-2\)

\(\Rightarrow12n+21⋮3n-2\)

\(\Rightarrow\left(12n-8\right)+29⋮3n-2\)

\(\Rightarrow4\left(3n-2\right)+29⋮3n-2\)

\(\Rightarrow29⋮3n-2\)

\(\Rightarrow3n-2\in U\left(29\right)=\left\{1;29\right\}\) ( Vì \(n\in N\) )

\(\Rightarrow\left\{{}\begin{matrix}3n-2=1\Rightarrow n=1\\3n-2=29\Rightarrow n=\dfrac{31}{3}\left(loai\right)\end{matrix}\right.\)

Vậy \(n=1\)

b) Ta có:

\(6n+8⋮5n-2\)

\(\Rightarrow5\left(6n+8\right)⋮5n-2\)

\(\Rightarrow30n+40⋮5n-2\)

\(\Rightarrow\left(30n-12\right)+52⋮5n-2\)

\(\Rightarrow6\left(5n-2\right)+52⋮5n-2\)

\(\Rightarrow52⋮5n-2\)

\(\Rightarrow5n-2\in U\left(52\right)=\left\{1;2;4;13;26;52\right\}\) ( Vì \(n\in N\) )

+) \(5n-2=1\Rightarrow n=\dfrac{3}{5}\left(loai\right)\)

+) \(5n-2=2\Rightarrow n=\dfrac{4}{5}\left(loai\right)\)

+) \(5n-2=4\Rightarrow n=\dfrac{6}{5}\left(loai\right)\)

+) \(5n-2=13\Rightarrow n=3\left(thoa\right)\)

+) \(5n-2=26\Rightarrow n=\dfrac{28}{5}\left(loai\right)\)

+) \(5n-2=52\Rightarrow n=\dfrac{54}{5}\left(loai\right)\)

Vậy \(n=3\)

c) Ta có:

\(2n+1⋮3n-4\)

\(\Rightarrow3\left(2n+1\right)⋮3n-4\)

\(\Rightarrow6n+3⋮3n-4\)

\(\Rightarrow\left(6n-8\right)+11⋮3n-4\)

\(\Rightarrow2\left(3n-4\right)+11⋮3n-4\)

\(\Rightarrow11⋮3n-4\)

\(\Rightarrow3n-4\in U\left(11\right)=\left\{1;11\right\}\) ( Vì \(n\in N\) )

\(\Rightarrow\left\{{}\begin{matrix}3n-4=1\Rightarrow n=\dfrac{5}{3}\left(loai\right)\\3n-4=11\Rightarrow n=5\end{matrix}\right.\)

Vậy \(n=5\)

a, \(n+8⋮n\)

\(\Rightarrow8⋮n\)(vì \(n⋮n\))

\(\Rightarrow n\inƯ\left(8\right)\)

\(\Rightarrow n\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

b, \(3n+5⋮n\)

\(\Rightarrow5⋮n\)(vì \(3n⋮n\))

\(\Rightarrow n\inƯ\left(5\right)\)

\(\Rightarrow n\in\left\{\pm1;\pm5\right\}\)

c, \(n+7⋮n+1\)

\(\Rightarrow\left(n+1\right)+6⋮n+1\)

\(\Rightarrow6⋮n+1\)(vì \(n+1⋮n+1\))

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

\(\Rightarrow n\in\left\{-7;-4;-3;-2;0;1;2;5\right\}\)

Hok tốt nha^^

a) \(3n+19⋮n+1\)

\(\Rightarrow\)\(3\left(n+1\right)+16⋮n+1\)

mà \(3\left(n+1\right)⋮n+1\)\(\Rightarrow\)\(16⋮n+1\)

\(\Rightarrow\)\(n+1\in\left\{1,-1,2,-2,4,-4,8,-8,16,-16\right\}\)

\(\Rightarrow n\in\left\{0,-2,1,-3,3,-5,7,-9,15,-17\right\}\)

b) \(2n+7⋮n+2\)

\(\Rightarrow2\left(n+2\right)+3⋮n+2\)

mà \(2\left(n+2\right)⋮n+2\Rightarrow3⋮n+2\)

\(\Rightarrow n+2\in\left\{1,3,-1,-3\right\}\)

\(\Rightarrow n\in\left\{-1,1,-3,-5\right\}\)

c)\(6n+39⋮2n+1\Rightarrow3\left(2n+1\right)+36⋮2n+1\)

mà\(3\left(2n+1\right)⋮2n+1\)\(\Rightarrow36⋮2n+1\)

\(\Rightarrow2n+1\in\left\{1,-1,2,-2,3,-3,4,-4,6,-6,9,-9,12,-12,18,-18,36,-36\right\}\)

\(\Rightarrow2n\in\left\{0,-2,1,-3,2,-4,3,-5,5,-7,8,-10,11,-13,17,-19,35,-37\right\}\)

\(\Rightarrow\)\(n\in\left\{0,-1,1,-2,4,-5\right\}\)