Question

\(A=\dfrac{12n+1}{2n +3} \) Để A là một số nguyên

Answer 1

Ta có : \(A=\dfrac{12n+1}{2n+3}=\dfrac{6\left(2n+3\right)-17}{2n+3}=6-\dfrac{17}{2n+3}\)

\(\Rightarrow2n+3\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)

2n + 3	1	-1	17	-17
2n	-2	-4	14	-20
n	-1	-2	7	-10

 

Answer 2

Answer 3

Để A là số nguyên thì 12n+1⋮2n+3

12n+18-17⋮2n+3

12n+18⋮2n+3 ⇒17⋮2n+3

2n+3∈Ư(17)

Ư(17)={1;-1;17;-17}

n∈{-1;-2;7;-10}

 

Answer 4

Để A nguyên thì \(12n+1⋮2n+3\)

\(\Leftrightarrow-17⋮2n+3\)

\(\Leftrightarrow2n+3\inƯ\left(-17\right)\)

\(\Leftrightarrow2n+3\in\left\{1;-1;17;-17\right\}\)

\(\Leftrightarrow2n\in\left\{-2;-4;14-20\right\}\)

\(\Leftrightarrow n\in\left\{-1;-2;7;-10\right\}\)