a) \(A=\frac{3-n}{n+1}=\frac{4-1-n}{n+1}=\frac{4}{n+1}-1\inℤ\)mà \(n\inℤ\)suy ra \(n+1\inƯ\left(4\right)=\left\{-4,-2,-1,1,2,4\right\}\)

\(\Leftrightarrow n\in\left\{-5,-3,-2,0,1,3\right\}\).

b) \(B=\frac{6n+5}{3n+2}=\frac{6n+4+1}{3n+2}=2+\frac{1}{3n+2}\inℤ\)mà \(n\inℤ\)suy ra \(3n+2\inƯ\left(1\right)=\left\{-1,1\right\}\)

\(\Rightarrow n\in\left\{-1\right\}\)

c) \(C\inℤ\Rightarrow3C=\frac{6n+3}{3n+2}=\frac{6n+4-1}{3n+2}=2-\frac{1}{3n+2}\inℤ\) mà \(n\inℤ\)suy ra 

.\(3n+2\inƯ\left(1\right)=\left\{-1,1\right\}\)\(\Rightarrow n\in\left\{-1\right\}\)

Thử lại thỏa mãn. 

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 \(\in\left\{0;-2\right\}\)

c) Để \(\dfrac{2n+5}{n-3}\) ∈ Z thì 2n+5⋮n-3

⇒ 2n-3+8⋮n-3

⇒ 8⋮n-3 ⇒ n-3∈Ư(8)

Ư(8)={...}

⇒n=...

;-------------------------------; làm hết đeeeeeeeeeeeeeeeeeeeeeeeeeeeee

Ta có \(A=\frac{2n-1}{n+3}\left(n\ne-3\right)\)

\(\Leftrightarrow A=\frac{2\left(n+3\right)-7}{n+3}=2-\frac{7}{n+3}\)

a) Để A đạt giá trị nguyên thì \(\frac{7}{n+3}\)đạt giá trị nguyên

=> 7 chia hết cho n+3

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

ta có bảng

\(A=\frac{2n-1}{n+3}=\frac{2\left(n+3\right)-7}{n+3}=2-\frac{7}{n+3}\)

A nguyên => \(\frac{7}{n+3}\)nguyên

=> \(n+3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

mình thua

bo tay

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

Để A nguyên => 3-n = Ước của 5

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

a, Để \(\dfrac{n+1}{n-2}\) có giá trị là một số nguyên thì n + 1 ⋮ n - 2

=> (n - 2) + 3 ⋮ n - 2

 Vì (n - 2) ⋮ n - 2 nên 3 ⋮ n - 2

=> n - 2 ∈ Ư(3) ∈ {-3;-1;1;3}

 => n ∈ {-1;1;3;5}

b, Để \(\dfrac{4n+5}{2n-1}\) có giá trị là một số nguyên thì 4n + 5 ⋮ 2n - 1

=> (4n - 2) + 7 ⋮ 2n - 1

=> 2(2n - 1) + 7 ⋮ 2n - 1

 Vì 2(2n - 1) ⋮ 2n -1 nên 7 ⋮ 2n - 1

=> 2n - 1 ∈ Ư(7) ∈ {-7;-1;1;7}

=> n ∈ {-3;0;1;4}

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

A nguyên <=>\(2+\frac{7}{n+3}\) nguyên

<=>7 chia hết cho n+3

<=>\(n+3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

<=>\(n\in\left\{-10;-4;-2;4\right\}\)

Vậy A nguyên khi \(n\in\left\{-10;-4;-2;4\right\}\)