a) \(\Rightarrow2\left(n+3\right)-38⋮\left(n+3\right)\)

Do \(n\in N\)

\(\Rightarrow\left(n+3\right)\inƯ\left(38\right)=\left\{19;38\right\}\)

\(\Rightarrow n\in\left\{16;35\right\}\)

b) \(\Rightarrow5\left(n+5\right)-74⋮\left(n+5\right)\)

Do \(n\in N\)

\(\Rightarrow\left(n+5\right)\inƯ\left(74\right)=\left\{37;74\right\}\)

\(\Rightarrow N\in\left\{32;69\right\}\)

a) ⇒2(n+3)−38⋮(n+3)⇒2(n+3)−38⋮(n+3)

Do n∈Nn∈N

⇒(n+3)∈Ư(38)={19;38}⇒(n+3)∈Ư(38)={19;38}

⇒n∈{16;35}⇒n∈{16;35}

b) ⇒5(n+5)−74⋮(n+5)⇒5(n+5)−74⋮(n+5)

Do n∈Nn∈N

⇒(n+5)∈Ư(74)={37;74}⇒(n+5)∈Ư(74)={37;74}

⇒N∈{32;69}

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

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

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

mình xin lỗi mình đánh máy sai câu hỏi như này

 A) n+7 chia hết cho n+2 ( với n khác 2 )

 B) 3n+1 chia hết cho 2n+3  

\(a,\Rightarrow n-1+7⋮n-1\)

Mà \(n-1⋮n-1\Rightarrow7⋮n-1\)

\(\Rightarrow n-1\inƯ\left(7\right)=\left\{1;7\right\}\\ \Rightarrow n\in\left\{2;8\right\}\)

\(b,\Rightarrow3\left(n+1\right)+2⋮n+1\)

Mà \(3\left(n+1\right)⋮n+1\Rightarrow2⋮n+1\)

\(\Rightarrow n+1\inƯ\left(2\right)=\left\{1;2\right\}\\ \Rightarrow n=1\left(n\ne0\right)\)