Question

Giúp em với 

 

Answer 1

\(\dfrac{4n-1}{n-1}=\dfrac{4n-4+3}{n-1}=\dfrac{4\left(n-1\right)+3}{n-1}=\dfrac{4\left(n-1\right)}{n-1}+\dfrac{3}{n-1}=4+\dfrac{3}{n-1}\)

Để nguyên thì \(n-1\in U\left(3\right)=\left\{\pm1;\pm3\right\}\)

n-1=1 => n=2

n-1=-1 =>n=0

n-1=3 => n=4

n-1=-3 => n=-2

Answer 2

ĐK:x\(\ne\)1

\(\dfrac{4n-1}{n-1}=\dfrac{4n-4+3}{n-1}=\dfrac{4\left(n-1\right)+3}{n-1}=4+\dfrac{3}{n-1}\)

Để \(\dfrac{4n-1}{n-1}\in Z\Rightarrow\dfrac{3}{n-1}\in Z\Rightarrow3⋮\left(n-1\right)\Rightarrow n-1\inƯ\left(3\right)\)

Ta có bảng:

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

Vậy \(n\in\left\{-2;0;2;4\right\}\)