Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(4\left(n-1\right)-3⋮\left(n-1\right)\)
\(\Rightarrow\left(n-1\right)\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
Do \(n\in N\Rightarrow n\in\left\{0;2;4\right\}\)
b) \(-5\left(4-n\right)+12⋮\left(4-n\right)\)
\(\Rightarrow\left(4-n\right)\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
Do \(n\in N\Rightarrow n\in\left\{16;10;8;7;6;5;3;2;1;0\right\}\)
c) \(-2\left(n-2\right)+6⋮\left(n-2\right)\)
\(\Rightarrow\left(n-2\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
Do \(n\in N\Rightarrow n\in\left\{0;1;3;4;5;8\right\}\)
d) \(n\left(n+3\right)+6⋮\left(n+3\right)\)
\(\Rightarrow\left(n+3\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
Do \(n\in N\Rightarrow n\in\left\{0;3\right\}\)
\(a,\Rightarrow n\inƯ\left(5\right)=\left\{1;5\right\}\\ b,\Rightarrow n\inƯ\left(4\right)=\left\{1;2;4\right\}\\ c,\Rightarrow n\inƯ\left(27\right)=\left\{1;3\right\}\left(n< 7\right)\)
a: 7n chia hết cho 3
mà 7 không chia hết cho 3
nên \(n⋮3\)
=>\(n=3k;k\in Z\)
b: \(-22⋮n\)
=>\(n\inƯ\left(-22\right)\)
=>\(n\in\left\{1;-1;2;-2;11;-11;22;-22\right\}\)
c: \(-16⋮n-1\)
=>\(n-1\inƯ\left(-16\right)\)
=>\(n-1\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)
=>\(n\in\left\{2;0;3;-1;5;-3;9;-7;17;-15\right\}\)
d: \(n+19⋮18\)
=>\(n+1+18⋮18\)
=>\(n+1⋮18\)
=>\(n+1=18k\left(k\in Z\right)\)
=>\(n=18k-1\left(k\in Z\right)\)
Bài 5:
b: Ta có: \(n+6⋮n+2\)
\(\Leftrightarrow n+2\in\left\{2;4\right\}\)
hay \(n\in\left\{0;2\right\}\)
c: Ta có: \(3n+1⋮n-2\)
\(\Leftrightarrow n-2\in\left\{-1;1;7\right\}\)
hay \(n\in\left\{1;3;9\right\}\)
a) n \(\in\text{ }\text{ }\left\{2;7;12;17;22;27;...\right\}\)
b) \(n\in\left\{3;10;17;24;31;39;46;...\right\}\)
c) \(n\in\left\{14;27;40;53;66;79;...\right\}\)
a)
6+n chia hết cho n
<=> 6+n - n chia hết cho n
<=> 6 chia hết cho n
<=> \(n\inƯ_6\)
\(\Rightarrow n\in\left\{1;2;3;6;-1;-2;-3;-6\right\}\)
b)
5n+4 chia hết cho n
<=> 5n+4 - 5n chia hết cho n
<=> 4 chia hết cho n
<=> n\(\inƯ_4\)
\(\Rightarrow n\in\left\{1;2;4;-1;-2;-4\right\}\)
c)
20+7n chia hết cho n
<=> 20+7n - 7n chia hết cho n
<=> 20 chia hết cho n
<=> n\(\inƯ_{20}\)
\(\Rightarrow n\in\left\{1;2;4;5;10;20;-1;-2;-4;-5;-10;-20\right\}\)
đúng đó