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

thách ai cho mình làm đúng

Hello !!!!!!! I love you !!!!! Thanks you very much

a.1;6

b.1;5

c.n={1;2;19;38}

d.n={0;1;3}

e.n={2;8}

g.n=3

aha kết bạn đi mk fan hunhan đây!

\(n+3=\left(n+1\right)+2\)

mà \(n+1⋮n+1\)

\(\Rightarrow2⋮n+1\)

\(\Rightarrow n+1\inƯ\left(2\right)\)

\(\Rightarrow n+1\in\hept{ }1;2\)

TH1: \(n+1=1\Leftrightarrow n=1-1=0\)

Th2: \(n+1=2\Leftrightarrow n=2-1=1\)

Vậy \(n\in\hept{ }0;1\)

\(3n+5=3\left(n-1\right)+7\)

mà \(3\left(n-1\right)⋮n-1\)

\(\Rightarrow7⋮n-1\)

\(\Rightarrow n-1\inƯ\left(7\right)\)

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

TH1: \(n-1=1\Leftrightarrow n=1+1=2\)

TH2: \(n-1=7\Leftrightarrow n=7+1=8\)

Vậy \(n\in\hept{ }2;8\)

\(4n-6=4n-4-2\)

\(\Leftrightarrow4n+4-8-2\)

\(\Leftrightarrow4\left(n+1\right)-8-2\)

\(\Leftrightarrow4\left(n+1\right)-10\)

mà \(2n+2=2\left(n+1\right)\)

mà \(4\left(n+1\right)⋮2\left(n+1\right)\)

\(\Leftrightarrow10⋮2\left(n+1\right)\)

\(\Leftrightarrow2\left(n+1\right)\inƯ\left(10\right)\)

\(\Leftrightarrow2\left(n+1\right)\in\hept{ }1;2;5;10\)

TH1: \(2\left(n+1\right)=1\Leftrightarrow n=-0.5\notin N\)

TH2: \(2\left(n+1\right)=2\Leftrightarrow n=0\in N\)

TH3: \(2\left(n+1\right)=5\Leftrightarrow n=1.5\notin N\)

TH4: \(2\left(n+1\right)=10\Leftrightarrow n=4\in N\)

Vậy \(n\in\hept{ }0;4\)

Nhớ k cho mình nhé! Thank you!!!

n+3=(n+1)+2

mà n+1⋮n+1

⇒2⋮n+1

⇒n+1∈Ư(2)

⇒n+1∈{1;2

TH1: n+1=1⇔n=1−1=0

Th2: n+1=2⇔n=2−1=1

Vậy n∈{0;1

3n+5=3(n−1)+7

mà 3(n−1)⋮n−1

⇒7⋮n−1

⇒n−1∈Ư(7)

⇒n−1∈{1;7

TH1: n−1=1⇔n=1+1=2

TH2: n−1=7⇔n=7+1=8

Vậy n∈{2;8

4n−6=4n−4−2

⇔4n+4−8−2

⇔4(n+1)−8−2

⇔4(n+1)−10

mà 2n+2=2(n+1)

mà 4(n+1)⋮2(n+1)

⇔10⋮2(n+1)

⇔2(n+1)∈Ư(10)

⇔2(n+1)∈{1;2;5;10

TH1: 2(n+1)=1⇔n=−0.5∉N

TH2: 2(n+1)=2⇔n=0∈N

TH3: 2(n+1)=5⇔n=1.5∉N

TH4: 2(n+1)=10⇔n=4∈N

Vậy n∈{0;4