Question

Tìm số nguyên n để \(\dfrac{2}{3}.\dfrac{6}{n+2}\)cũng là một số nguyên

Answer 1

ta có \(\dfrac{2}{3}=\dfrac{6}{9}\)

=>\(\dfrac{6}{9}=\dfrac{6}{n+2}\)

=>n+2 =9

n = 9-2

n=7

Vậy n=7

Answer 2

Ta có:

\(\dfrac{2}{3}.\dfrac{6}{n+2}\)= \(\dfrac{12}{3\left(n+2\right)}\)= \(\dfrac{12:3}{3\left(n+2\right):3}\)= \(\dfrac{4}{n+2}\)

Để \(\dfrac{2}{3}.\dfrac{6}{n+2}\) \(\in\) Z thì

\(\dfrac{4}{n+2}\) \(\in\) Z

\(\Leftrightarrow\) 4 \(⋮\) n + 2

\(\Leftrightarrow\) n + 2 \(\in\) Ư₍₄₎

\(\Leftrightarrow\) n + 2 \(\in\) \(\left\{\pm1;\pm2;\pm4\right\}\)

\(\Leftrightarrow\) n \(\in\) \(\left\{-3;-1;-4;0;-6;2\right\}\)

Vậy n \(\in\) \(\left\{-3;-1;-4;0;-6;2\right\}\)