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a, \(3n+2⋮n-1\)
\(\Rightarrow3n-3+5⋮n-1\)
\(\Rightarrow3\left(n-1\right)+5⋮n-1\)
Vì : \(3\left(n-1\right)⋮n-1\Rightarrow5⋮n-1\)
\(\Rightarrow n-1\inƯ\left(5\right)\)
\(\Rightarrow n-1\in\left\{1;5\right\}\)
+) \(n-1=1\Rightarrow n=1+1\Rightarrow n=2\)
+) \(n-1=5\Rightarrow n=5+1\Rightarrow n=6\)
Vậy : \(n\in\left\{2;6\right\}\) thì \(3n+2⋮n-1\)
b, \(n+8⋮n+3\)
Vì : \(n+3⋮n+3\)
\(\Rightarrow\left(n+8\right)-\left(n+3\right)⋮n+3\)
\(\Rightarrow n+8-n-3⋮n+3\)
\(\Rightarrow5⋮n+3\)
\(\Rightarrow n+3\inƯ\left(5\right)\)
Mà : \(n+3\ge3\)
\(\Rightarrow n+3=5\Rightarrow n=5-3\Rightarrow n=2\)
Vậy n = 2 thì : \(n+8⋮n+3\)
c, \(n+6⋮n-1\)
Mà : \(n-1⋮n-1\)
\(\Rightarrow\left(n+6\right)-\left(n-1\right)⋮n-1\)
\(\Rightarrow n+6-n+1⋮n-1\)
\(\Rightarrow7⋮n-1\)
\(\Rightarrow n-1\inƯ\left(7\right)\)
\(\Rightarrow n-1\in\left\{1;7\right\}\)
+) \(n-1=1\Rightarrow n=1+1\Rightarrow n=2\)
+) \(n-1=7\Rightarrow n=7+1\Rightarrow n=8\)
Vậy \(n\in\left\{2;8\right\}\) thì \(n+6⋮n-1\)
d, \(4n-5⋮2n-1\)
\(\Rightarrow4n-2-3⋮2n-1\)
\(\Rightarrow2\left(2n-1\right)-3⋮2n-1\)
Vì : \(2\left(2n-1\right)⋮2n-1\)
\(\Rightarrow3⋮2n-1\)
\(\Rightarrow2n-1\inƯ\left(3\right)\)
\(\Rightarrow2n-1\in\left\{1;3\right\}\)
+) \(2n-1=1\Rightarrow2n=1+1\Rightarrow2n=2\Rightarrow n=2\div2\Rightarrow n=1\)
+) \(2n-1=3\Rightarrow2n=3+1\Rightarrow2n=4\Rightarrow n=4\div2\Rightarrow n=2\)
Vậy \(n\in\left\{1;2\right\}\) thì \(4n-5⋮2n-1\)
a) (n+2) \(⋮\) (n-1)
vì (n-1)\(⋮\) (n-1)
=>(n+2)-(n-1)\(⋮\left(n-1\right)\)
=>(n+2-n+1)\(⋮\) (n-1)
=> 3\(⋮\) (n-1)
=>(n-1)\(\in\) Ư(3) = { \(\pm\)1,\(\pm\)3}
ta có bảng
| n-1 | -1 | 1 | -3 |
3 |
| n | 0 | 2 | -2 | 4 |
| loại |
vậy n\(\in\) { 0;2;4}
b) \(\left(2n+7\right)⋮\left(n+1\right)\)
vì\(\left(n+1\right)⋮\left(n+1\right)\)
=>\(2\left(n+1\right)⋮\left(n+1\right)\)
=> \(\left(2n+2\right)⋮\left(n+1\right)\)
=>\(\left(2n+7\right)-\left(2n+2\right)⋮\left(n+1\right)\)
=>\(\left(2n+7-2n-2\right)⋮\left(n+1\right)\)
=>\(5⋮\left(n+1\right)\)
=> \(\left(n+1\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
TA CÓ BẢNG
| n+1 | -5 | -1 | 1 | 5 |
| n | -6 | -2 | 0 | 4 |
| loại | loại |
vậy \(n\in\left\{0;4\right\}\)
\(n+3=\left(n+1\right)+2\)
mà \(n+1⋮n+1\)
\(\Rightarrow2⋮n+1\)
\(\Rightarrow n+1\inƯ\left(2\right)\)
\(\Rightarrow n+1\in\hept{ }1;2\)
TH1: \(n+1=1\Leftrightarrow n=1-1=0\)
Th2: \(n+1=2\Leftrightarrow n=2-1=1\)
Vậy \(n\in\hept{ }0;1\)
\(3n+5=3\left(n-1\right)+7\)
mà \(3\left(n-1\right)⋮n-1\)
\(\Rightarrow7⋮n-1\)
\(\Rightarrow n-1\inƯ\left(7\right)\)
\(\Rightarrow n-1\in\hept{ }1;7\)
TH1: \(n-1=1\Leftrightarrow n=1+1=2\)
TH2: \(n-1=7\Leftrightarrow n=7+1=8\)
Vậy \(n\in\hept{ }2;8\)
\(4n-6=4n-4-2\)
\(\Leftrightarrow4n+4-8-2\)
\(\Leftrightarrow4\left(n+1\right)-8-2\)
\(\Leftrightarrow4\left(n+1\right)-10\)
mà \(2n+2=2\left(n+1\right)\)
mà \(4\left(n+1\right)⋮2\left(n+1\right)\)
\(\Leftrightarrow10⋮2\left(n+1\right)\)
\(\Leftrightarrow2\left(n+1\right)\inƯ\left(10\right)\)
\(\Leftrightarrow2\left(n+1\right)\in\hept{ }1;2;5;10\)
TH1: \(2\left(n+1\right)=1\Leftrightarrow n=-0.5\notin N\)
TH2: \(2\left(n+1\right)=2\Leftrightarrow n=0\in N\)
TH3: \(2\left(n+1\right)=5\Leftrightarrow n=1.5\notin N\)
TH4: \(2\left(n+1\right)=10\Leftrightarrow n=4\in N\)
Vậy \(n\in\hept{ }0;4\)
Nhớ k cho mình nhé! Thank you!!!
n+3=(n+1)+2
mà n+1⋮n+1
⇒2⋮n+1
⇒n+1∈Ư(2)
⇒n+1∈{1;2
TH1: n+1=1⇔n=1−1=0
Th2: n+1=2⇔n=2−1=1
Vậy n∈{0;1
3n+5=3(n−1)+7
mà 3(n−1)⋮n−1
⇒7⋮n−1
⇒n−1∈Ư(7)
⇒n−1∈{1;7
TH1: n−1=1⇔n=1+1=2
TH2: n−1=7⇔n=7+1=8
Vậy n∈{2;8
4n−6=4n−4−2
⇔4n+4−8−2
⇔4(n+1)−8−2
⇔4(n+1)−10
mà 2n+2=2(n+1)
mà 4(n+1)⋮2(n+1)
⇔10⋮2(n+1)
⇔2(n+1)∈Ư(10)
⇔2(n+1)∈{1;2;5;10
TH1: 2(n+1)=1⇔n=−0.5∉N
TH2: 2(n+1)=2⇔n=0∈N
TH3: 2(n+1)=5⇔n=1.5∉N
TH4: 2(n+1)=10⇔n=4∈N
Vậy n∈{0;4