Question

Cho biểu thức B= 2n-1/ n+3 . Tìm n để B ϵZ

Answer 1

Để B là số nguyên thì \(2n-1⋮n+3\)

\(\Leftrightarrow2n+6-7⋮n+3\)

\(\Leftrightarrow n+3\in\left\{1;-1;7;-7\right\}\)

hay \(n\in\left\{-2;-4;4;-10\right\}\)

Answer 2

\(B=\dfrac{2n-1}{n+3}=\dfrac{2n+6-7}{n+3}=\dfrac{2\left(n+3\right)-7}{n+3}=2-\dfrac{7}{n+3}\)

Để \(B\in Z\Rightarrow\dfrac{7}{n+3}\in Z\Rightarrow7⋮\left(n+3\right)\Rightarrow n+3\inƯ\left(7\right)\)

Ta có bảng:
 

n+3	-7	-1	1	7
n	-10	-4	-2	4

Vậy \(n\in\left\{-10;-4;-2;4\right\}\)