Question

Tìm số nguyên n biết \(\dfrac{2n-1}{n+1}\) là số nguyên

Answer 1

\(\dfrac{2n-1}{n+1}=\dfrac{2\left(n+1\right)-3}{n+1}=....\)

3 chia hết n+1

Answer 2

Ta có:

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

\(\Rightarrow\left(n+1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Ta có bảng sau:

n+1	1	-1	3	-3
n	0	-2	2	-4

Vậy \(n\in\left\{0;-2;2;-4\right\}\)