=2(3n+2)/3n+2 -5/3n+2
=2 -5/3n+2
để ps trên nguyên thì (3n+2) thuộc ước của 5=(+-1;+-5)
=>n=(-1;-1/3;1;-7/3)
Đề là \(n\in Z\) đúng không?
\(\dfrac{6n-1}{3n+2}\in Z\Leftrightarrow\dfrac{2\left(3n+2\right)-5}{3n+2}\in Z\)
\(\Leftrightarrow2-\dfrac{5}{3n+2}\in Z\)
\(\Leftrightarrow3n+2\inƯ_5=\left\{\pm1;\pm5\right\}\)
\(\Leftrightarrow3n\in\left\{-7;-3;-1;3\right\}\)
\(\Leftrightarrow n\in\left\{-1;1\right\}\)
\(A=\dfrac{6n-1}{3n+2}=\dfrac{2\left(3n+2\right)-5}{3n+2}=2+\dfrac{-5}{3n+2}\)
Để A nguyên \(-5\text{ }⋮\text{ }\left(3n+2\right)\)
\(3n+2\inƯ\left(-5\right)=\left\{-5;-1;1;5\right\}\)
\(n\in\left\{-\dfrac{7}{3};-1;-\dfrac{1}{3};1\right\}\)
Để A có giá trị nguyên hay \(\dfrac{6n-1}{3n+2}\) là số nguyên
Ta có: A=\(\dfrac{6n-1}{3n+2}=\dfrac{6n+4}{3n+2}=\dfrac{2.\left(3n+2\right)}{3n+2}=\dfrac{5}{3n+2}\)
A=2-\(\dfrac{5}{3n+2}\)
=>3n+2∈Ư(5)={\(\pm1;\pm5\)}
ta có bảng:
3n+2 :1 ; -1 ; 5 ; -5
n : \(\dfrac{-1}{3}\)(loại) ;-1 ;1 ;\(\dfrac{-7}{3}\)(loại)
=>n∈{-1;1}
Để A là số nguyên thì \(6n-1⋮3n+2\)
\(\Leftrightarrow3n+2\in\left\{1;-1;5;-5\right\}\)
\(\Leftrightarrow3n\in\left\{-1;-3;3;-7\right\}\)
hay \(n\in\left\{-\dfrac{1}{3};-1;1;-\dfrac{7}{3}\right\}\)