Question

Tìm n thuộc Z để 2n²-n+2 chia hết cho n+1 

Answer 1

\(2n^2-n+2=2n^2+n-2n-1+3=n\left(2n+1\right)-\left(2n+1\right)+3\)

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

\(\Rightarrow2n+1⋮2n+1\)

\(\Rightarrow3⋮2n+1\)

\(\Rightarrow2n+1\in\text{Ư}\left(3\right)=\left\{1;-1;3;-3\right\}\)

Xét: \(2n+1=1\Rightarrow n=0\)

Xét: \(2n+1=-1\Rightarrow n=-1\)

Xét: \(2n+1=3\Rightarrow n=1\)

Xét: \(2n+1=-3\Rightarrow n=-2\)

Vậy: \(n\in\left\{-2;-1;0;1\right\}\)

Answer 2

Xin lỗi mình viết nhầm:

Tìm n thuộc Z để 2n²-n+2 chia hết cho 2n+1