Question

Tìm n ∈ N sao cho 2n - 7 chia hết cho 2n + 1.

Answer 1

\(2n-7⋮2n+1\)

\(\Leftrightarrow2n+1-8⋮2n+1\)

\(\Leftrightarrow8⋮2n+1\)

\(\Leftrightarrow2n+1\inƯ\left(8\right)=\left\{1;2;4;8\right\}\) (Vì n thuộc N <=> n \(\ge\)0 <=> \(2n+1\ge1\))

<=> \(2n\in\left\{0;1;3;7\right\}\)

Vì n thuộc N => 2n \(⋮2\)

=> 2n = 0

=> n = 0

Vậy ...

 

Answer 2

2 n − 7 ⋮ 2 n + 1

=) 2 n + 1 − 8 ⋮ 2 n + 1

=) 8 ⋮ 2 n + 1

=) 2 n + 1  ∈ Ư ( 8 ) = { 1 ; 2 ; 4 ; 8 }  

=) 2 n ∈ { 0 ; 1 ; 3 ; 7 } 

Vì n thuộc N

=) n2 chỉ chia hết cho 2, 1, 0

Mà 2n ∈ { 0 ; 1 ; 3 ; 7 } 

=) 2n = 0

=) n = 0

 

Answer 3

để 2n-7⋮2n+1

thì 2n+1+8⋮2n+1

=>8⋮2n+1

=>2n+1∈ư(8)={1;2;4;8}

\(\Rightarrow\left[{}\begin{matrix}2n+1=1\\2n+1=2\\2n+1=4\\2n+1=8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2n=0\\2n=1\\2n=3\\2n=7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=0\\n=\dfrac{1}{2}\\n=\dfrac{3}{2}\\n=\dfrac{7}{2}\end{matrix}\right.\)

vì n∈N=>n=0

vậy n=0