Question

[nguyễn ngọc quyền linh]

Tim n sao cho : 3n + 1 chia het cho n - 2

                            4n - 3 chia het cho n + 1

Answer 1

[Nguyễn Phương Uyên]

3n + 1 ⋮ n - 2

=> (3n-6) + 6 + 1 ⋮ n - 2

=> (3.n-3.2) + 7 ⋮ n - 2

=> 3(n-2) + 7 ⋮ n - 2

có 3(n-2) ⋮ n - 2

=> 7 ⋮ n - 2

=> n - 2 ∈ Ư(7)

=> n-2 ∈ {-1;-7;1;7}

=> n ∈ {1;-5;3;9}

vậy___

4n - 3 ⋮ n + 1

=> (4n + 4) - 4 - 3 ⋮ n + 1

=> (4n + 4.1) - 7 ⋮ n + 1

=> 4(n+1) - 7 ⋮ n + 1

=> 4(n+1) ⋮ n + 1

=> -7 ⋮ n + 1

=> n + 1 ∈ Ư(-7)

=> n + 1 ∈ {-1;-7;1;7}

=> n ∈ {-2;-8;0;6}

vậy____

Answer 2

[Phùng Minh Quân]

Ta có :

\(\left(3n+1\right)⋮\left(n-2\right)\)

\(\Rightarrow\)\(\left(3n-6+7\right)⋮\left(n-2\right)\)

\(\Rightarrow\)\(3.\left(n-2\right)+7⋮\left(n-2\right)\)

\(\Rightarrow\)\(7⋮\left(n-2\right)\)

\(\Rightarrow\left(n-2\right)\inƯ\left(7\right)\)

Mà \(Ư\left(7\right)=\left\{1;-1;7;-7\right\}\)

Do đó :

\(n-2=1\Rightarrow n=1+2=3\)

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

\(n-2=7\Rightarrow n=7+2=9\)

\(n-2=-7\Rightarrow n=-7-2=-9\)

Vậy \(n\in\left\{3;1;9;-9\right\}\)