a: =>\(n+2\in\left\{1;-1;7;-7\right\}\)

=>\(n\in\left\{-1;-3;5;-9\right\}\)

b: =>n-3+4 chia hết cho n-3

=>\(n-3\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(n\in\left\{4;2;5;1;7;-1\right\}\)

c: =>3n^3+n^2+9n^2-1-4 chia hết cho 3n+1

=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(n\in\left\{0;-\dfrac{2}{3};\dfrac{1}{3};-1;1;-\dfrac{5}{3}\right\}\)

d: =>10n^2-10n+11n-11+1 chia hết cho n-1

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

=>\(n\in\left\{2;0\right\}\)

1: \(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;4;2;-2;-1;-4\right\}\)

\(\Leftrightarrow3n\in\left\{0;3;-3\right\}\)

hay \(n\in\left\{0;1;-1\right\}\)

1/

$10n+4\vdots 2n+7$

$\Rightarrow 5(2n+7)-31\vdots 2n+7$

$\Rightarrow 31\vdots 2n+7$

$\Rightarrow 2n+7\in Ư(31)$

$\Rightarrow 2n+7\in \left\{1; -1; 31; -31\right\}$

$\Rightarrow n\in \left\{-3; -4; 12; -19\right\}$

2/

$5n-4\vdots 3n+1$

$\Rightarrow 3(5n-4)\vdots 3n+1$

$\Rightarroq 15n-12\vdots 3n+1$

$\Rightarrow 5(3n+1)-17\vdots 3n+1$

$\Rightarrow 17\vdots 3n+1$

$\Rightarrow 3n+1\in Ư(17)$

$\Rightarrow 3n+1\in \left\{1; -1; 17; -17\right\}$

$\Rightarrow n\in \left\{0; \frac{-2}{3}; \frac{16}{3}; -6\right\}$

Do $n$ nguyên nên $n\in\left\{0; -6\right\}$

3/

$2n^2+n-6\vdots 2n+1$

$\Rightarrow n(2n+1)-6\vdots 2n+1$

$\Rightarrow 6\vdots 2n+1$

$\Rightarrow 2n+1\in Ư(6)$

Mà $2n+1$ lẻ nên: $2n+1\in \left\{1; -1; 3; -3\right\}$

$\Rightarrow n\in \left\{0; -1; 1; -2\right\}$

Ta có \(n^3+10n^2-5\)chia hết cho \(3n+1\)

Suy ra \(\frac{n^3+10n^2-5}{3n+1}\)là số nguyên

Ta thấy \(\frac{n^3+10n^2-5}{3n+1}\)

\(=\frac{n^3+9n^2+n^2-5}{3n+1}\)

\(=\frac{n^2.\left(1+3n\right)+n^2-5}{3n+1}\)

\(=n^2+\frac{n^2-5}{3n+1}\)

Tách tiếp cái phân số phía sau là ra nhé , lười @@

Ta có:  \(3n^3+10n^2-5=\left(3n+1\right).\left(n^2+3n-1\right)-4\)

để  \(3n^3+10n^2-5⋮3n+1\) thì  \(4⋮3n+1\)

Tức là \(3n+1\) là ước của 4

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

\(3n+1=-4\Rightarrow n=\frac{-5}{3}\left(loai\right)\)

\(3n+1=-2\Rightarrow n=-1\)

\(3n+1=-1\Rightarrow n=\frac{-2}{3}\left(loai\right)\)

\(3n+1=1\Rightarrow n=0\)

\(3n+1=2\Rightarrow n=\frac{1}{3}\left(loai\right)\)

\(3n+1=4\Rightarrow n=1\)

Vậy.....................

Ta có: \(3n^3+10n^2-5⋮3n+1\)

\(\Rightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

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

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

Vì \(3n+1⋮3n+1\) nên để \(\left(3n+1\right)\left(n^2+3n-1\right)-4⋮3n+1\) thì \(4⋮3n+1\)

\(\Rightarrow3n+1\inƯ\left(4\right)\)

\(\Rightarrow3n+1\in\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow3n\in\left\{0;1;3;-2;-3;-5\right\}\)

\(\Rightarrow n\in\left\{0;\frac{1}{3};1;-\frac{2}{3};-1;-\frac{5}{3}\right\}\)

Mà \(n\in Z\Rightarrow n\in\left\{0;1;-1\right\}\)

Vậy \(n\in\left\{0;1;-1\right\}\)

Bài 3:
a, Ta có: 3.n^3+10.n^2-5
= 3.+n^3+9.n^2+3n-3n-1-4
= n^2.(3n+1)+ 3n(3n+1)-(3n+1)-4
= (3n+1)(n^2+3n-1)-4
Để 3.+10.-5 chia hết cho 3n+1
=> (3n+1)(+3n-1)-4 chia hết cho 3n+1
=>  -4 chia hết cho 3n+1
mà Ư(-4) = {-4;-2;-1;1;2;4}
=> 3n+1 = {-4;-2;-1;1;2;4}
=> 3n = { -5;-3; -2; 0; 1; 3}
=> n={-5/3; -1;-2/3 ;0;1/3;1}
mà n thuộc Z
=> n = {-1; 0; 1}

cho minh hoi 9n^2 the con 1n^2 dau

ai biet