Question

Tìm các số nguyên n sao cho biểu thức sau là nguyên :

\(P=\frac{2n-1}{n-1}\)

Hiếu cần gấp ạ

Answer 1

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

\(\Rightarrow P\in Z\Leftrightarrow2+\frac{1}{n-1}\in Z\Leftrightarrow\frac{1}{n-1}\in Z\Leftrightarrow1⋮n-1\Leftrightarrow n-1\inƯ\left(1\right)\)

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

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

Answer 2

\(\frac{2n-1}{n-1}\in Z\)

\(\Rightarrow2n-1⋮n-1\)

\(\Rightarrow\left(2n-1\right)-\left(n-1\right)⋮\left(n-1\right)\)

\(\Rightarrow2⋮\left(n-1\right)\)

Bảng:

n-1	-1	1	2	-2
n	0	2	3	-1

 

Vậy \(n\in\left\{0;-1;2;3\right\}\)