Question

[nguyễn hương thảo]

tìm n thuộc N để 

n + 3 chia hết cho n + 1

3n +5 chia hết  n -1 

4n -6 chia hết cho 2n + 2

Answer 1

[My Nguyễn Thị Trà]

\(n+3=\left(n+1\right)+2\)

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!!!

Answer 2

[lê đình nam]

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