Question

Tìm số nguyên n biết : 2n+7 \(⋮\) n-2

Answer 1

\(\dfrac{2n+7}{n-2}=\dfrac{2\left(n-2\right)+11}{n-2}=\dfrac{2\left(n-2\right)}{n-2}+\dfrac{11}{n-2}=2+\dfrac{11}{n-2}\)

Để \(2n+7⋮n-2\) thì \(11⋮n-2\)

\(\Rightarrow n-2\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)

\(\Rightarrow n-2=\left\{3;1;13;-9\right\}\)

Answer 2

Ta có: 2n + 7 \(⋮\) n - 2 \(\Rightarrow\) 2n - 4 + 11 \(⋮\) n - 2

Mà 2n - 4 \(⋮\) n - 2 \(\Rightarrow\) 11 \(⋮\) n - 2

Suy ra: \(\Rightarrow\) n - 2 \(\in\) Ư(11) = \(\left\{\pm1;\pm11\right\}\)

+ n - 2 = -11 \(\Rightarrow\) n = -9

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

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

+ n - 2 = 11 \(\Rightarrow\) n = 13

Vậy n \(\in\left\{-9;1;3;13\right\}\)

Chúc bạn học tốt!!!

Answer 3

\(\dfrac{2n+7}{n-2}=\dfrac{2n-4+11}{n-2}=\dfrac{2\left(n-2\right)+11}{n-2}=\dfrac{2\left(n-2\right)}{n-2}+\dfrac{11}{n-2}=2+\dfrac{11}{n-2}\)\(\Rightarrow11⋮n-2\)

Xét Ư(11) là ra

Answer 4

Ta có:

\(2n+7⋮n-2\)

\(\Rightarrow\left(2n-4\right)+11⋮n-2\)

\(\Rightarrow11⋮n-2\left(Do2n-4⋮n-2\right)\)

Do \(n\in Z\) \(\Rightarrow n-2\inƯ\left\{11\right\}\)

\(\Rightarrow n-2\in\left\{\pm1;\pm11\right\}\)

\(\Rightarrow n\in\left\{-9;1;3;13\right\}\)