Question

Câu hỏi: Tìm n \(\in\) Z để 2n² - n + 2 chia hết cho 2n + 1.

Answer 1

2n2−n+2=2n2+n−2n−1+3=n(2n+1)−(2n+1)+32n2−n+2=2n2+n−2n−1+3=n(2n+1)−(2n+1)+3

⇒n(2n+1)⋮(2n+1)⇒n(2n+1)⋮(2n+1)

⇒2n+1⋮2n+1⇒2n+1⋮2n+1

⇒3⋮2n+1⇒3⋮2n+1

⇒2n+1∈Ư(3)={1;−1;3;−3}⇒2n+1∈Ư(3)={1;−1;3;−3}

Xét: 2n+1=1⇒n=02n+1=1⇒n=0

Xét: 2n+1=−1⇒n=−12n+1=−1⇒n=−1

Xét: 2n+1=3⇒n=12n+1=3⇒n=1

Xét: 2n+1=−3⇒n=−22n+1=−3⇒n=−2

Vậy: n∈{−2;−1;0;1}

Answer 2

Ta có:

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

Để đa thức trên chia hết cho \(2n+1\Leftrightarrow2n+2⋮2n+1\)

\(\Rightarrow2n+1+1⋮2n+1\)

\(\Rightarrow1⋮2n+1\Rightarrow1\in U\left(1\right)=\left\{1;-1\right\}\)

Với \(2n+1=1\Leftrightarrow n=0\)

Với \(2n+1=-1\Leftrightarrow n=-1\)

Vậy,...