Question

K+8/K+1 să aparțină mulțimii Z, K=?

Answer 1

$\it \dfrac{k+8}{k+1}\in\mathbb{Z} \Rightarrow \dfrac{k+8}{k+1}-1\in\mathbb{Z} \Rightarrow \dfrac{7}{k+1}\in\mathbb{Z} \Rightarrow k+1\Big |7 \Rightarrow \\ \\ \Rightarrow k+1\in\{-7.\ -1,\ 1,\ 7\}|_{-1} \Rightarrow k\in\{-8,\ -2,\ 0,\ 6\}$

Answer 2

k  + 1 / k + 8

k  + 1 / k + 1

=> k + 1 / (k + 8 - k - 1)

k + 1 / 7 => k + 1 apartine {-7,-1,1,7}

k apartine {-8,-2,0,6}