Tìm n ∈ Z:
a) n+6 chia hết cho n-3.
b) 3n+2 chia hết cho n-1
c) -n+5 chia hết cho n+2
d) n ² + 5 chia hết cho n+1
Tìm n ∈ Z:
a) n+6 chia hết cho n-3.
b) 3n+2 chia hết cho n-1
c) -n+5 chia hết cho n+2
d) n ² + 5 chia hết cho n+1
Đáp án:
Giải thích các bước giải:
$a$) `n+6 \vdots n-3`
`⇔ n-3 + 9 \vdots n-3`
`⇔ 9 \vdotss n-3`
`⇒ n-3` `∈` `Ư(9)={±1;±3;±9}`
$⇔$ $n$ $∈$ `{-6;0;2;4;6;12}`
Vậy $n$ $∈$ `{-6;0;2;4;6;12}`
$b$) ` 3n+2 \vdots n-1`
`⇔ 3n+2 – 3(n-1) \vdots n-1`
`⇔ 3n+2 – 3n + 3 \vdots n-1`
`⇔ 5 \vdots n-1`
`⇒` `n-1` `∈` `Ư(5)={±1;±5}`
$⇒$ $n$ $∈$ `{-4;0;2;6}`
Vậy $n$ $∈$ `{-4;0;2;6}`
$c$) `-n+5 \vdots n+2`
`⇔ -n + 5 + (n+2) \vdots n+2`
`⇔ 7 \vdots n+2`
`⇒ n+2` `∈` `Ư(7)={±1;±7}`
$⇔$ $n$ $∈$ `{-9;-3;-1;5}`
Vậy $n$ $∈$ `{-9;-3;-1;5}`
$d$) `n^2 + 5 \vdots n+1`
`⇔ n^2 + n – n + 5 \vdots n+1`
`⇔ n(n+1) – (n-5) \vdots n+1`
Vì : `n(n+1) \vdots n+1` `⇒` `n- 5 \vdots n+1`
$⇒ n + 1 – 6 \vdots n+1$
$⇒ n+1$ $∈$ `Ư(6)={±1;±2;±3;±6}`
`⇒ n ∈ {-7;-4;-3;-2;0;1;2;5}`
Vậy `n ∈ {-7;-4;-3;-2;0;1;2;5}`