a)n=1

b)n=9

c)n=4

d)n=8

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don't no

aaaaaaaaaaa giúp vs 1 câu thui cũng đc

a) ta có: 3n + 2 chia hết cho n - 1

=> 3n - 3 + 5 chia hết cho n -1

3.(n-1) + 5 chia hết cho n - 1

mà 3.(n-1) chia hết cho n -1

=> 5 chia hết cho n - 1

=> n - 1 thuộc Ư(5)={1;-1;5;-5}

...

rùi bn tự lập bảng xét giá trị hộ mk nha!!!

b) ta có: n^2 + 2n + 7 chia hết cho n + 2

=> n.(n+2) + 7 chia hết cho n + 2

mà n.(n+2) chia hết cho n + 2

=> 7 chia hết cho n + 2

=>...

c) ta có: n^2 + 1 chia hết cho n - 1

=> n^2 - n + n -1 + 2 chia hết cho n - 1

n.(n-1) + (n-1) + 2 chia hết cho n -1

(n-1).(n+1) + 2 chia hết cho n - 1

mà (n-1).(n+1) chia hết cho n - 1

=> 2 chia hết cho n - 1

...

câu e;g bn dựa vào phần a mak lm nha!!!

\(d,n+8⋮n+3\)

\(\Leftrightarrow\left(n+3\right)+5⋮n+3\)

\(\Leftrightarrow n+3⋮n+3\Rightarrow5⋮n+3\)

\(\Leftrightarrow n+3\in\left(1;5\right)\)

\(\Leftrightarrow n+3=1\Rightarrow n=-2\left(l\right)\)

\(\Leftrightarrow n+3=5\Rightarrow n=2\left(c\right)\)

\(n+6⋮n-1\)

\(\Rightarrow\left(n-1\right)+7⋮n-1\)

\(\Rightarrow n-1⋮n-1\Leftrightarrow7⋮n-1\)

\(\Rightarrow n-1\in\left(1;7\right)\)

\(\rightarrow n-1=1\Rightarrow n=1+1\Rightarrow n=2\)

\(\rightarrow n-1=7\Rightarrow n=7+1\Rightarrow n=8\)

\(\Rightarrow n=2;8\)

\(4n-5⋮2n-1\)

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

\(\Rightarrow2.\left(2n-1\right)-3⋮2n-1\)

\(\Rightarrow2.\left(2n-1\right)⋮2n-1\)

\(\Rightarrow-3⋮2n-1\)

\(\Rightarrow2n-1\in\left(1;3\right)\)

\(\Rightarrow2n-1=1\Rightarrow n=\left(1+1\right):2=1\)

\(\Rightarrow2n-1=3\Rightarrow n=\left(3+1\right):2=2\)

\(\Rightarrow n=2;1\)

ai hiểu

a) n^2 chia hết cho n+3

b) 2n+6 chia hết cho 5

c) 5n+8 chia hết cho 11

Xin lỗi nha, mik ko bt làm

a, n+5 chia hết cho n-1 => n-1+6 chia hết cho n-1 => 6 chia hết cho n-1 hay n-1 thuộc Ư(6)

=> n-1={1,-1,2,-2,3,-3,6,-6} 

=>n={2,0,3,-1,4,-2,7,-5}

Các TH khác tương tự nk

b, 2n-4=2(n+2)-8

c, 6n+4=3(2n+1)+1

     \(n+5\)\(⋮\)\(n-1\)

\(\Leftrightarrow\)\(n-1+6\)\(⋮\)\(n-1\)

Ta thấy:    \(n-1\)\(⋮\)\(n-1\)

\(\Rightarrow\)\(6\)\(⋮\)\(n-1\)

\(\Rightarrow\)\(n-1\)\(\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

\(\Rightarrow\)\(n=\left\{-5;-2;-1;0;2;3;4;7\right\}\)