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Ta có: n+3 chia hết cho n-1
mà: n-1 chia hết cho n-1
suy ra:[(n+3)-(n-1)]chia hết cho n-1
(n+3-n+1)chia hết cho n-1
4 chia hết cho n-1
suy ra n-1 thuộc Ư(4)
Ư(4)={1;2;4}
suy ra n-1 thuộc {1;2;4}
Ta có bảng sau:
n-1 1 2 4
n 2 3 5
Vậy n=2 hoặc n=3 hoặc n=5
a,
Ta có: 4n-5 chia hết cho 2n-1
=>4n-2-3 chia hết cho 2n-1
=>2.(2n-1)-3 chia hết cho 2n-1
=>3 chia hết cho 2n-1
=>2n-1=Ư(3)=(-1,-3,1,3)
=>2n=(0,-2,2,4)
=>n=(0,-1,1,2)
Vậy n=0,-1,1,2
a) \(\left(n+6\right)⋮\left(n+1\right)\Rightarrow\left(n+1\right)+5⋮\left(n+1\right)\)
\(\Rightarrow\left(n+1\right)\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Do \(n\in N\)
\(\Rightarrow n\in\left\{0;4\right\}\)
b) \(\left(4n+9\right)⋮\left(2n+1\right)\Rightarrow2\left(2n+1\right)+7⋮\left(2n+1\right)\)
\(\Rightarrow\left(2n+1\right)\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
Do \(n\in N\)
\(\Rightarrow n\in\left\{0;3\right\}\)
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: \(n+19⋮18\)
=>\(n+1+18⋮18\)
=>\(n+1⋮18\)
=>\(n+1=18k\left(k\in Z\right)\)
=>\(n=18k-1\left(k\in Z\right)\)
\(a,\Rightarrow n+2+4⋮n+2\\ \Rightarrow n+2\inƯ\left(4\right)=\left\{1;2;4\right\}\\ \Rightarrow n\in\left\{0;2\right\}\\ b,\Rightarrow n-1+4⋮n-1\\ \Rightarrow n-1\inƯ\left(4\right)=\left\{1;2;4\right\}\\ \Rightarrow n\in\left\{2;3;5\right\}\)
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 \(n\in N\Rightarrow n\in\left\{0;3\right\}\)
\(a,\Rightarrow n+3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Rightarrow n\in\left\{-8;-4;-2;2\right\}\\ b,\Rightarrow n+3+5⋮n+3\\ \Rightarrow5⋮n+3\\ \Rightarrow n+3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Rightarrow n\in\left\{-8;-4;-2;2\right\}\\ c,\Rightarrow2\left(2n-1\right)-3⋮2n-1\\ \Rightarrow3⋮2n-1\\ \Rightarrow2n-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Rightarrow n\in\left\{-1;0;1;2\right\}\\ d,\Rightarrow8-n+4⋮8-n\\ \Rightarrow4⋮8-n\\ \Rightarrow8-n\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Rightarrow n\in\left\{12;10;9;7;6;4\right\}\)
a) 10 chia hết cho n-1
n-1 thuộc Ư của (10)={1,2,5,10}
n thuộc {2,3,7,11}
A)n+11\(⋮\)n-1
n-1\(⋮\)n-1
n+11-(n-1)\(⋮\)n-1
n+11-n-1\(⋮\)n-1
10\(⋮\)n-1
\(\Rightarrow\)n-1={1;2;5;10}
\(\Rightarrow\)n={2;3;6;11}
b)7.n\(⋮\)n-11
7n:\(⋮\)
n-1
7n-7n:n-1
0:n-1
Vậy n-1={0}
Vậy n={1}