Vì 6\(⋮\)x-1=>x-1ϵƯ(6)={1;2;3;6}

Với x-1=1=>x=2

x-1=2=>x=3

x-1=3=>x=4

x-1=6=>x=7

Vậy xϵ{2;3;4;7}

\(\left(n+5\right)⋮\left(n+1\right)\)

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

Vì n+1\(⋮\)n+1

Buộc 4\(⋮\)n+1=>n+1ϵƯ(4)={1;2;4}

Với n+1=1=>n=0

n+1=2=>n=1

n+1=4=>n=3

Vậy nϵ{0;1;3}

\(V\text{ì}45⋮\Rightarrow x\in\text{Ư}\left(45\right)\)={1;3;5;9;15;45}

2n+15\(⋮\)n+3

2n+6+9\(⋮\)n+3

2(n+3)+9\(⋮\)n+3

Vì n+3\(⋮\)n+3

Buộc 9\(⋮\)n+3=>n+3ϵƯ(9)={1;3;9}

Với n+3=1=>n= -2

n+3=3=>n=0

n+3=9=>n=6

Vậy nϵ{-2;0;6}

6\(⋮\)x-1=>x-1ϵƯ(6)={1;2;3;6}

Với x-1=1=>x=2

x-1=2=>x=3

x-1=3=>x=4

x-1=6=>x=7

Vậy xϵ{2;3;4;7}

Vì -10<3x<5=>3x ϵ {-9;-8;-7;-6;-5;-4;-3;-2;-1;0;1;2;3;4}

=>x ϵ {-3;-2;-1;0;3}