=>(9n-6)+6+1 chia hết cho 3n-2

=>3(3n-2)+7 chia hết cho 3n-2

Mà 3(3n-2) chia hết cho 3n-2

=>7 chia hết cho 3n-2

=>3n-2 thuộc Ư(7)={1;7}

=>3n thuộc {3;9}

=>n thuộc {1;3}

a) ta có: n + 10 chia hết cho n + 1

=> n + 1 + 9 chia hết cho  n + 1

Do n + 1 chia hết cho  n + 1

=> 9 chia hết cho  n + 1

\(\Rightarrow n+1\in U_{\left(9\right)}=\left\{1;-1;3;-3;9;-9\right\}\)

....

r bn tu xet gia tri nha

b) ta  có: 3n + 40 chia hết cho n + 2

=> 3n + 6 + 34 chia hết cho n + 2

3.(n+2) + 34 chia hết cho n + 2

Do 3.(n+2) chia hết cho n + 2

=> 34 chia hết cho n + 2

\(\Rightarrow n+2\in U_{\left(34\right)}=\left\{\pm1;\pm2;\pm17;\pm34\right\}\)

...

a, ta có n+10 = (n+1)+9

do (n+1)chia hết cho(n+1) =>9 phải chia hết cho (n+1) => (n+1)thuộc Ư(9)=(-1,1,3,-3,9,-9)

do n thuộc N nên (n+1)>hoặc bằng 1 => (n+1)=(1,3,9)

  nếu n+1=1=>n=0

        n+1=3=>n=2

        n+1=9=>n=8

 vậy ......

b, ta có : 3n+40=(3n +6)+34=3(n+2)+34

do 3(n+2) chia hết cho (n+2) => 34 phải chia hết cho n+2 => (n+2) thuộc Ư(34)=(1,-1,2,-2,17,-17,34,-34)

do n thuộc N nên (n+2)>hoặc bằng 2 => (n+2)=(2,17,34)

nếu n+2=2=>n=0

     n+2=17=>n=15

      n+2=34=>n=32

vậy .......

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đợi lát

cùng nhau  ko phải bằng nhau

b: Vì 2n+3 là số lẻ

mà 4n+8 là số chẵn

nên 2n+3 và 4n+8 là hai số nguyên tố cùng nhau