Question

Tìm số nguyên n biết: 3n+7 chia hết cho 2n+1

Answer 1

\(3n+7⋮2n+1\)

\(\Rightarrow2\left(3n+7\right)⋮2n+1\)

\(\Rightarrow6n+14⋮2n+1\)

\(\Rightarrow6n+3+11⋮2n+1\)

\(\Rightarrow3\left(2n+1\right)+11⋮2n+1\)

     \(3\left(2n+1\right)⋮2n+1\)

\(\Rightarrow11⋮2n+1\)

\(\Rightarrow2n+1\inƯ\left(11\right)\)

\(\Rightarrow2n+1\in\left\{-1;1;-11;11\right\}\)

\(\Rightarrow2n\in\left\{-2;0;-12;10\right\}\)

\(\Rightarrow n\in\left\{-1;0;-6;5\right\}\)

Answer 2

Ta có: 3n+7 chia hết 2n+1          <=>  2(3n+7) chia hết 2n+1        <=> 6n+14 chia hết 2n+1

          2n+1 chia hết 2n+1                   3(2n+1) chia hết 2n+1                6n+3 chia hết 2n+1

=>(6n+14)-(6n+3) chia hết 2n+1

<=> 6n+14-6n+3 chia hết 2n+1

<=> 17 chia hết 2n+1

=> 2n+1 thuộc Ư(17)={-1;-17;1;17}

<=> 2n thuộc {-2;-18;0;16}

<=> n thuộc {-1;-9;0;8}

Vậy.....................

K CHO MK NHA ~~~~

Answer 3

3n+7​\(⋮\)2n+1 => 3n+7-(2n+1) \(⋮\)2n+1

                       =>2(3n+7)-3(2n+1)\(⋮\)2n+1

                      =>6n+14 - 6n +3 \(⋮\)2n+1

                      => 11 \(⋮\)2n+1

=> 2n+1 \(\in\)Ư(11) =>2n+1\(\in\){ 1;11;-1;-11} => 2n\(\in\){0;10;-2;-12} => n\(\in\) {0;5;-1;-6}.