\(\dfrac{6n+1}{2n+1}\left(n\in Z\right)\\ =\dfrac{3\left(2n+1\right)-2}{2n+1}=3-\dfrac{2}{2n+1}\)

Để biểu thức nhận gt nguyên thì : \(\dfrac{2}{2n+1}\in Z\)

\(=>2n+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\\ =>2n\in\left\{0;-2;1;-3\right\}\\ =>n\in\left\{0;-1;\dfrac{1}{2};-\dfrac{3}{2}\right\}\)

Do n nguyên -> Kết luận : n = 0 hoặc n = -1

De \(\frac{6n+5}{2n-1}\)\(\in Z\)

=> 6n+5 chia het cho 2n-1

=> 6n-3+8 chia het cho 2n-1

=> 3(2n-1)+8 chia het cho 2n-1

=> 8 chia het cho 2n-1

=> 2n-1=-1;1;-2;2;-4;4;-8;8

Vi 2n-1 la so le

=> 2n-1=-1;1

=> 2n=0;2

=> n=0;1

A∈Z⇒\(\dfrac{2\left(x+1\right)}{x+3}\in Z\Rightarrow\left(2x+2\right)⋮\left(x+3\right)\)

\(\Rightarrow\left(2x+6-4\right)⋮\left(x+3\right)\\ \Rightarrow\left[2\left(x+3\right)-4\right]⋮\left(x+3\right)\)

 \(\text{Mà}2\left(x+3\right)⋮\left(x+3\right)\\ \Rightarrow-4⋮\left(x+3\right)\\ \Rightarrow x+3\inƯ\left(-4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Rightarrow x\in\left\{-7;-5;-4;-2;-1;1\right\}\)

- Bạn ơi lớp 6 cũng làm được nhé :)

x ∈{0;-6;-2;-4}

\(A=\dfrac{2\left(x+1\right)}{x+3}=\dfrac{2\left(x+3\right)-4}{x+3}=2-\dfrac{4}{x+3}\)

Để A nguyên \(\Rightarrow\dfrac{4}{x+3}\) nguyên

\(\Rightarrow x+3=Ư\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow x=\left\{-7;-5;-4;-2;-1;1\right\}\)