Question

Tìm các số nguyên n sao cho 2n³ + n² + 7n + 1 chia hết cho 2n - 1

 

 

Answer 1

Ta có:

\(2n^3+n^2+7n+1⋮2n-1\)

\(\Rightarrow2n^3-n^2+2n^2-n+8n-4+5⋮2n-1\)

\(\Rightarrow\left(2n-1\right)\left(n^2+n+4\right)+5⋮2n-1\)

\(\Rightarrow5⋮2n-1\)

\(\Rightarrow2n-1\inƯ\left(5\right)=1;-1;5;-5\)

Với:

\(2n-1=1\Rightarrow2n=2\Rightarrow n=1\)

\(2n-1=-1\Rightarrow2n=0\Rightarrow n=0\)

\(2n-1=5\Rightarrow2n=6\Rightarrow n=3\)

\(2n-1=-5\Rightarrow2n=-4\Rightarrow n=-2\)

Vậy \(n=1;0;3;-2\)

Answer 2

thank