Question

Tìm các số nguyên n sao cho biểu thức P = \(\dfrac{3n+2}{n-1}\) là số nguyên

Giúp mk vs mai mk thi r Thanks các bạn nhiều

Answer 1

Để \(P\in Z\) thì \(3n+2⋮n-1\)

\(\Rightarrow3\left(n-1\right)+5⋮n-1\)

mà \(3\left(n-1\right)⋮\left(n-1\right)\Rightarrow5⋮n-1\)

\(\Rightarrow n-1\inƯ\left(5\right)\)

\(\Rightarrow n-1\in\left\{\pm1;\pm5\right\}\)

\(\Rightarrow n\in\left\{2;0;6;-4\right\}\)

Vậy n \(\in\left\{2;0;6;-4\right\}\)

Answer 2

\(P=\dfrac{3n+2}{n-1}=\dfrac{3n-3+5}{n-1}=\dfrac{3n-3}{n-1}+\dfrac{5}{n-1}=3+\dfrac{5}{n-1}\)

\(\Rightarrow5⋮n-1\Leftrightarrow n-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow\left[{}\begin{matrix}n-1=1\\n-1=-1\\n-1=5\\n-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=2\\n=0\\n=6\\n=-4\end{matrix}\right.\left(tm\right)\)