Để A là số nguyên thì 2n^2-n+4n-2+5 chia hết cho 2n-1

=>\(2n-1\in\left\{1;-1;5;-5\right\}\)

=>\(n\in\left\{1;0;3;-2\right\}\)

      `2n^2+3n+3 | 2n-1`

`-`   `2n^2-n`           `n+2`

     ------------------

                `4n+3`

          `-`   `4n-2`

              ------------

                       `5`

`<=> (2n^2+3n+3) : (2n-1)=5`

`<=> 5 ⋮ (2n-1)=> 2n-1 ∈ Ư(5)`\(=\left\{1,5\right\}\)

`+, 2n-1=1=>2n=2=>n=1`

`+, 2n-1=-1=>2n=0=>n=0`

`+, 2n-1=5=>2n=6=>n=3`

`+,2n-1=-5=>2n=-4=>n=-2`

vậy \(n\in\left\{1;0;3;-2\right\}\)

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

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

\(D=\frac{3n+5}{2n+3}\)

=> \(2D=\frac{6n+10}{2n+3}=\frac{6n+9+1}{2n+3}=\frac{3\left(2n+3\right)+1}{2n+3}\)

=> \(2D=3+\frac{1}{2n+3}\)

=> Để D là số nguyên thì 1 phải chia hết cho 2n+3 và \(\frac{1}{2n+3}\)phải là số lẻ

=> 2n+3 = {-1; 1}

+/ 2n+3=-1 => n=-2   => D=1

+/ 2n+3=1 => n=-1    => D=2

\(A=\dfrac{-\left(6-2n\right)+5}{3-n}=\dfrac{-2\left(3-n\right)+5}{3-n}=-2+\dfrac{5}{3-n}\)

Để A nguyên => 3-n = Ước của 5

\(\Rightarrow3-n=\left\{-5;-1;1;5\right\}\Rightarrow n=\left\{8;4;2;-2\right\}\)

2n^2 là 2. n^{2 à}

Để A nguyên => 3A nguyên

Khi đó \(3A=\frac{6n-9}{3n-1}=\frac{6n-2-7}{3n-1}=\frac{2\left(3n-1\right)-7}{3n-1}=2-\frac{7}{3n-1}\)

Vì \(2\inℤ\Rightarrow\frac{-6}{3n-1}\inℤ\Rightarrow-7⋮3n-1\Rightarrow3n-1\inƯ\left(-7\right)\)

=> \(3n-1\in\left\{1;7;-1;-7\right\}\)

=> \(3n\in\left\{2;8;0;-6\right\}\)

Vì n nguyên => \(3n\in\left\{0;-6\right\}\Rightarrow n\in\left\{0;-2\right\}\)

Vậy n \(\in\left\{0;-2\right\}\)