\(\dfrac{3n-5}{4n+1}\)ϵ z =>\(\dfrac{4\left(3n-5\right)}{4n+1}\)ϵ z
Ta có :
\(\dfrac{4\left(3n-5\right)}{4n+1}\)=\(\dfrac{3\left(4n+1\right)-23}{4n+1}\)=3-\(\dfrac{23}{4n+1}\)
Để \(\dfrac{4\left(3n-5\right)}{4n+1}\)ϵ Z=>4n+1ϵ Ư(23)=(1;-1;23;-23)
4n+1=1=>n=0
4n+1=-1=>n=\(\dfrac{-1}{2}\)(loại)
4n+1=23=>n=\(\dfrac{11}{2}\)(loại)
4n+1=-23=>n=-6
Vậy n ϵ 0;-6
\(\dfrac{3n-5}{4n+1}\) là số nguyên khi :
\(3n-5⋮4n+1\)
\(\Rightarrow4\left(3n-5\right)-3\left(4n+1\right)⋮4n+1\)
\(\Rightarrow12n-20-12n-3⋮4n+1\)
\(\Rightarrow-23⋮4n+1\)
\(\Rightarrow4n+1\in\left\{-1;1;-23;23\right\}\)
\(\Rightarrow n\in\left\{\dfrac{1}{2};0;-6;\dfrac{11}{2}\right\}\Rightarrow n\in\left\{0;-6\right\}\left(n\in Z\right)\)
Ta có: 3n - 5 ⋮ n + 4
=> 3n + 12 - 17 ⋮ n + 4
=> 3.(n + 4) - 17 ⋮ n + 4
Mà 3.(n + 4) ⋮ n + 4
=> 17 ⋮ n + 4
=> n + 4 ϵ Ư(17) = {-17; -1; 1; 17}
=> n ϵ {-21; -5; -3; 13}.
