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)B=3(n+1)/n+1 - 3/n+1

      =3 - 3/n+1

để B nguyên thì n+1 thuộc ước của 3 (1;3)

suy ra n =(0;2)

câu b tương tự

cj ko rõ đề câu a lắm e ghi lại nhé

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

\(\Rightarrow\dfrac{3}{n+1}\in Z\Rightarrow n+1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\Rightarrow n\in\left\{0;-2;2;-4\right\}\)

b) \(\dfrac{6n+5}{2n-1}=\dfrac{3\left(2n-1\right)}{2n-1}+\dfrac{8}{2n-1}=3+\dfrac{8}{2n-1}\in Z\)

\(\Rightarrow\dfrac{8}{2n-1}\in Z\Rightarrow2n-1\inƯ\left(8\right)=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

Vì \(n\in Z\Rightarrow n\in\left\{1;0\right\}\)

\(N=\frac{6n+5}{2n-1}=\frac{6n-3+8}{2n-1}=3+\frac{8}{2n-1}\inℤ\Leftrightarrow\frac{8}{2n-1}\inℤ\)

mà \(n\)là số nguyên nên \(2n-1\inƯ\left(8\right)\)mà \(2n-1\)là số lẻ nên

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