Thôi tự trả lời vậy:

Vì (3n-2;3)=1 nên (4n-3).3 chia hết cho 3n-2.

Ta có: 12n-9 = 12n-8-1 chia hết cho 3n-2

Suy ra 1 chia hết cho 3n-2

Suy ra n=1. Vậy n = 1

3n+3 chia hết cho 4n+1

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

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

1: \(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;4;2;-2;-1;-4\right\}\)

\(\Leftrightarrow3n\in\left\{0;3;-3\right\}\)

hay \(n\in\left\{0;1;-1\right\}\)

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) ta có    \(\frac{3n-2}{n-1}=\frac{3\left(n-1\right)+5}{n-1}=3+\frac{5}{n-1}\)

Để 3n+2 chia hết cho n-1 thì n-1\(\varepsilon\)Ư(5)={1;5}

=> n thuộc { 2;6}

b)\(\frac{4n-5}{2n-1}=\frac{2\left(2n-1\right)-3}{2n-1}=2-\frac{3}{2n-1}\)

Để 4n-2 chia hết cho 2n-1 thì 2n-1\(\varepsilon\)Ư(3)={1;3}

=> n thuộc { 1;2}

sdgaef