a, 

Ta có: 4n-5 chia hết cho 2n-1

=>4n-2-3 chia hết cho 2n-1

=>2.(2n-1)-3 chia hết cho 2n-1

=>3 chia hết cho 2n-1

=>2n-1=Ư(3)=(-1,-3,1,3)

=>2n=(0,-2,2,4)

=>n=(0,-1,1,2)

Vậy n=0,-1,1,2

b, 

11:

n^3-n^2+2n+7 chia hết cho n^2+1

=>n^3+n-n^2-1+n+8 chia hết cho n^2+1

=>n+8 chia hết cho n^2+1

=>(n+8)(n-8) chia hết cho n^2+1

=>n^2-64 chia hết cho n^2+1

=>n^2+1-65 chia hết cho n^2+1

=>n^2+1 thuộc Ư(65)

=>n^2+1 thuộc {1;5;13;65}

=>n^2 thuộc {0;4;12;64}

mà n là số tự nhiên

nên n thuộc {0;2;8}

Thử lại, ta sẽ thấy n=8 không thỏa mãn

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

a) 2n+1⋮n-3

2n-6+7⋮n-3

2n-6⋮n-3 ⇒7⋮n-3

n-3∈Ư(7)

Ư(7)={1;-1;7;-7}

⇒n∈{4;2;10;-4}

\(3n+1⋮11-2n\)

\(\Rightarrow2\times(3n+1)⋮11-2n\)

\(\Rightarrow6n+2⋮11-2n\)

\(\Rightarrow6n+2⋮-(11-2n)\)

\(\Rightarrow6n+2⋮2n-11\)

\(\Rightarrow(6n-33)+35⋮2n-11\)

\(\Rightarrow35⋮2n-11(6n-33⋮2n-11)\)

\(\Rightarrow2n-11\inƯ(35)=\left\{-35;-7;-5;-1;1;5;7;35\right\}\)

a, Để \(n\in N\)

\(3n+1⋮11-2n\)

\(\Rightarrow6n+2⋮11-2n\)

Ta có\(3.\left(11-2n\right)⋮2n\)

Vì  \(11-2n⋮11-2n\)

\(33-6n⋮11-2n\)

\(6n+2+33-6n⋮11-2n\)

\(35⋮11-2n\)

\(\Rightarrow11-2n\inƯ\left(35\right)=\left\{\mp1;\mp5;\mp7;\mp35\right\}\)

Ta có bảng 

phần b có gì sai sót ai đó sửa dùm ^^

\(2n+1⋮1-3n\)

\(\Rightarrow3(2n+1)⋮1-3n\)

\(\Rightarrow6n+3⋮1-3n\)

\(\Rightarrow6n+3⋮-(1-3n)\)

\(\Rightarrow6n+3⋮3n-1\)

\(\Rightarrow(6n-2)+5⋮3n-1\)

\(\Rightarrow5⋮3n-1\)

\(\Rightarrow3n-1\inƯ(5)=\left\{-5;-1;1;5\right\}\)

\(n\inℕ\Rightarrow n=0;2\)

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

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

\(3n+2⋮n-1\)

\(3\left(n-1\right)+1⋮n-1\)

\(1⋮n-1\)hay \(n-1\inƯ\left(1\right)=\left\{\pm1\right\}\)

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