tK:

⇔3n−1∈{1;−1;2;−2;4;−4;8;−8;16;−16}⇔3n−1∈{1;−1;2;−2;4;−4;8;−8;16;−16}

hay n∈{0;1;−1;3;−5}

\(\Leftrightarrow n\in\left\{0;1;-1;3;-5\right\}\)

\(\Rightarrow6n+14⋮3n-1\\ \Rightarrow2\left(3n-1\right)+16⋮3n-1\\ \Rightarrow3n-1\inƯ\left(16\right)=\left\{-16;-8;-4;-2;-1;1;2;4;8;16\right\}\\ \Rightarrow3n\in\left\{-15;0;3;9\right\}\left(n\in Z\right)\\ \Rightarrow n\in\left\{-5;0;1;3\right\}\)

làm câu

a: Ta có: \(2n+1⋮n+2\)

\(\Leftrightarrow2n+4-3⋮n+2\)

\(\Leftrightarrow n+2\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{-1;-3;1;-5\right\}\)

b: Để B là số nguyên thì \(n+3⋮n-2\)

\(\Leftrightarrow n-2+5⋮n-2\)

\(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{3;1;7;-3\right\}\)

c: Để C là số nguyên thì \(3n+7⋮n-1\)

\(\Leftrightarrow3n-3+10⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

hay \(n\in\left\{2;0;3;-1;6;-4;11;-9\right\}\)

\(1,3n+7=3n+3+4=3\left(n+1\right)+4⋮\left(n+1\right)\\ =>n+1\inƯ\left(4\right)\\ Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\\ TH1,n+1=1\\ =>n=0\\ TH2,n+1=-1\\ =>n=-2\\ TH3,n+1=2\\ =>n=1\\ TH3,n+1=-2\\ =>n=-3\\ TH4,n+1=4\\ =>n=3\\ TH5,n+1=-4\\ =>n=-5\)

a) Ta có : n+7 \(⋮\)n+2

\(\Rightarrow\)n+2+5\(⋮\)n+2

mà n+2\(⋮\)n+2

\(\Rightarrow\)5\(⋮\)n+2

\(\Rightarrow n+2\in_{ }\){-5;-1;1;5}

\(\Rightarrow n\in\){-7;-3;-1;2}

b,c,d tương tự

giải hết ra giùm mk mk gấp lắm

cảm ơn bạn

ai làm đúng mk k cho

a)  \(n+7⋮n+2\)

=) \(\left[n+7-\left(n+2\right)\right]⋮n+2\)

=) \(n+7-n-2⋮n+2\)

=) \(5⋮n+2\)

=) \(n+2\inƯ\left(5\right)\)= \(\left\{+-1;+-5\right\}\)

=) \(n\in\left\{-3;-1;3;-7\right\}\)

đăng kí kênh V-I-S hộ mình nha !

b) \(9-n⋮n-3\)

=) \(\left[9-n+\left(n-3\right)\right]⋮n-3\)

=) \(9-n+n-3\)\(⋮n-3\)

=) \(6⋮n-3\)

=) \(n-3\inƯ\left(6\right)=\left\{+-1;+-2;+-3;+-6\right\}\)

=) \(n\in\left\{2;4;5;1;0;6;9;-3\right\}\)

Vì 2n + 1 chia hết cho 3n - 2

Nên 3( 2n + 1 ) chia hết cho 3n - 2

Suy ra 6n + 3 chia hết cho 3n - 2

Mà 3n - 2 chia hết cho 3n - 2

Nên 2( 3n - 2 ) chia hết cho 3n - 2

Duy ra 6n - 4 chia hết cho 3n - 2

Mà 6n + 3 cha hết cho 3n - 2

Suy ra ( 6n + 3 ) - ( 6n - 4 ) chia hết cho 3n - 2

Suy ra 7 chia hết cho 3n - 2

Mà n nguyên

Nên 3n - 2 = 7 ; 1 ; -1 ; -7

Suy ra 3n = 9 ; 3 ; 1 ; -5

Suy ra n = 3 ; 1

Vậy n = 1 ; 3 đẻ 2n + 1 chia hết cho 3n - 2
 

\(3n+7⋮2n+1\)

\(\Rightarrow2\left(3n+7\right)⋮2n+1\)

\(\Rightarrow6n+14⋮2n+1\)

\(\Rightarrow6n+3+11⋮2n+1\)

\(\Rightarrow3\left(2n+1\right)+11⋮2n+1\)

     \(3\left(2n+1\right)⋮2n+1\)

\(\Rightarrow11⋮2n+1\)

\(\Rightarrow2n+1\inƯ\left(11\right)\)

\(\Rightarrow2n+1\in\left\{-1;1;-11;11\right\}\)

\(\Rightarrow2n\in\left\{-2;0;-12;10\right\}\)

\(\Rightarrow n\in\left\{-1;0;-6;5\right\}\)

Ta có: 3n+7 chia hết 2n+1          <=>  2(3n+7) chia hết 2n+1        <=> 6n+14 chia hết 2n+1

          2n+1 chia hết 2n+1                   3(2n+1) chia hết 2n+1                6n+3 chia hết 2n+1

=>(6n+14)-(6n+3) chia hết 2n+1

<=> 6n+14-6n+3 chia hết 2n+1

<=> 17 chia hết 2n+1

=> 2n+1 thuộc Ư(17)={-1;-17;1;17}

<=> 2n thuộc {-2;-18;0;16}

<=> n thuộc {-1;-9;0;8}

Vậy.....................

K CHO MK NHA ~~~~

3n+7​\(⋮\)2n+1 => 3n+7-(2n+1) \(⋮\)2n+1

                       =>2(3n+7)-3(2n+1)\(⋮\)2n+1

                      =>6n+14 - 6n +3 \(⋮\)2n+1

                      => 11 \(⋮\)2n+1

=> 2n+1 \(\in\)Ư(11) =>2n+1\(\in\){ 1;11;-1;-11} => 2n\(\in\){0;10;-2;-12} => n\(\in\) {0;5;-1;-6}.