ta co: 2n-3 chia het cho n+1

n+1 chia het cho n+1

=>2(n+1) chia het cho n+1

hay 2n+2 chia het cho n+1

=>(2n+2)-(2n-3) chia het cho n-1

         5 chia het cho n-1

=> n-1 thuoc uoc cua 5  ={1;5;-1;-5}

=> n thuoc{2;6;0;-4}

sai rồi đoạn cuối là n+1 chứ
 

\(\left(3n-4\right)⋮\left(2n-1\right)\Rightarrow n\in\left\{3\right\}\)

+-4,-10,-2

Tick nha 

{-10;-4;-2;4} , tick nha

\(2n-3⋮n+1\)

\(\Rightarrow\left(2n+2\right)-2-3⋮n+1\)

\(\Rightarrow2\left(n+1\right)-5⋮n+1\)

      \(2\left(n+1\right)⋮n+1\)

\(\Rightarrow-5⋮n+1\)

\(\Rightarrow\) \(n+1\inƯ\left(-5\right)\)

đến đây dễ r`, bn tự lm tiếp đi!

\(2n-3⋮n+1\)

\(\Rightarrow2n+2-5⋮n+1\)

mà \(2n+2⋮n+1\)

\(\Rightarrow5⋮n+1\)

\(\Rightarrow n+1\inƯ\left(5\right)\)

\(\Rightarrow n+1\in\left\{1;\left(-1\right);5;\left(-5\right)\right\}\)

\(\Rightarrow n\in\left\{0;\left(-2\right);\left(-6\right);4\right\}\)

vậy :n  =  0

       n  =  -2

       n  = -6

       n = 4

2n - 3 \(⋮\)n + 1

=> 2n + 2 - 5 \(⋮\)n + 1

=> 2 ( n + 1 ) - 5 \(⋮\)n + 1

Ta thấy 2 ( n + 1 ) \(⋮\)n + 1

=> 5 \(⋮\)n + 1

=> n + 1 \(\in\)Ư ( 5 )

Ư ( 5 ) = { 1 ; - 1 ; 5 ; - 5 )

Ta có bảng sau :

Vậy .....

a) \(a\left(b+1\right)=3\left(a;b\inℤ\right)\)

\(\Rightarrow a;\left(b+1\right)\in U\left(3\right)=\left\{-1;1;-3;3\right\}\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(-1;-4\right);\left(1;2\right);\left(-3;-2\right);\left(3;0\right)\right\}\)

b) \(2n+7⋮n+1\left(n\inℤ\right)\)

\(\Rightarrow2n+7-2\left(n+1\right)⋮n+1\)

\(\Rightarrow2n+7-2n-2⋮n+1\)

\(\Rightarrow5⋮n+1\)

\(\Rightarrow n+1\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)

\(\Rightarrow n\in\left\{-2;0;-6;4\right\}\)

c) \(xy+x-y=6\left(x;y\inℤ\right)\)

\(\Rightarrow x\left(y+1\right)-y-1+1=6\)

\(\Rightarrow x\left(y+1\right)-\left(y+1\right)=5\)

\(\Rightarrow\left(x-1\right)\left(y+1\right)=5\)

\(\Rightarrow\left(x-1\right);\left(y+1\right)\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-0;-6\right);\left(2;4\right);\left(-4;-2\right);\left(6;0\right)\right\}\)

Ta có: \(2n-3⋮n+1\)

\(\Leftrightarrow-5⋮n+1\)

\(\Leftrightarrow n+1\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{0;-2;4;-6\right\}\)

`2n-3 vdots n+1`

`=>2n+2-5 vdots n+1`

`=>2(n+1)-5 vdots n+1`

`=>5 vdots n+1` do `2(n+1) vdots n+1`

`=>n+1 in Ư(5)={+-1,+-5}`

`=>n in {0,-2,4,-6}`

Vậy `n in {0,-2,4,-6}` thì `2n-3 vdots n+1`

Để \(2n-3⋮n+1\)

<=> \(2n-3-2\left(n+1\right)⋮n+1\)

<=> \(-5⋮n+1\)

<=> \(n+1\inƯ\left(5\right)\)

<=> \(n+1\in\left\{-5;-1;1;5\right\}\)

<=> \(n\in\left\{-6;-2;0;4\right\}\)

Giải:

\(2n-3⋮n+1\) 

\(\Rightarrow2n+2-5⋮n+1\) 

\(\Rightarrow5⋮n+1\) 

\(\Rightarrow n+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\) 

Ta có bảng giá trị:

Tham khảo:

2n-3 chia hết cho n+1

=> 2n+2-5  chia hết cho n+1

=> 2(n+1)-5  chia hết cho n+1

Mà 2(n+1)  chia hết cho n+1 => 5  chia hết cho n+1

=> n+1 thuộc Ư(5) ={1;-1;5;-5}

TH1: n+1=1 => n=0 thuộc Z

TH2: n+1=-1 => n=-2 thuộc Z

TH3: n+1=5 => n=4 thuộc Z

TH4: n+1=-5 => n=-6 thuộc Z

=> n thuộc {0;-2;4;6}