Đk: ` x ∈ Z `
` a ` ` (n + 3)(n^2 + 1) = 0 `
` => ` \(\left[ \begin{array}{l}x+3=0\\x^2 + 1=0\end{array} \right.\)
` => ` \(\left[ \begin{array}{l}x=-3\\x^2=-1(vô-lý)\end{array} \right.\)
` => x = -3 `
` b) ` ` (n - 1)(n^2 - 4) = 0 `
` => ` \(\left[ \begin{array}{l}x-1=0\\x^2-4=0\end{array} \right.\)
` => ` \(\left[ \begin{array}{l}x=1\\x^2=4\end{array} \right.\)
` => ` \(\left[ \begin{array}{l}x=1\\x=2\\x=-2\end{array} \right.\)
` c) ` ` n + 3 \vdots n - 1 `
` => n - 1 + 4 \vdots n - 1 `
Vì ` n - 1 \vdots n - 1 `
` => 4 \vdots n - 1 `
` => n - 1 ∈ Ư(4) = { ±1 ; ±2 ; ±4 } `
` => n ∈ {0 ; 2 ; -1 ; 3 ; -3 ; 5} `
` d) ` ` 2n - 1 \vdots n + 2 `
` => 2n + 2 - 3 \vdots n + 2 `
Vì ` 2n + 2 \vdots n + 2 `
` => -3 \vdots n + 2 `
` => n + 2 ∈ Ư(-3) = { ±1 ; ±3 } `
` => n ∈ {-3 ; -1 ; -5 ; 1} `