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 \(n\in\left\{0;1;-1\right\}\)

Ta có: \(\frac{4n^3+11n^2+5n+5}{n+2}=\frac{\left(n+2\right)\left(4n^2+3n-1\right)+7}{n+2}=4n^2+3n-1+\frac{7}{n+2}\)

Để 4n³ + 11n² + 5n + 5 chia hết cho n + 2 thì \(\frac{7}{n+2}\inℤ\Rightarrow7⋮n+2\Rightarrow n+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Ta lập bảng giá trị:

Vậy \(n\in\left\{-1;-3;5;-9\right\}\)thì 4n³ + 11n² + 5n + 5 chia hết cho n + 2

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

\(\frac{n^3-4n^2+5n-1}{n-3}=n^2-n+2+\frac{5}{n-3}\)

Để \(n^3-4n^2+5n-1⋮n-3\Rightarrow5⋮n-3\)

\(\Rightarrow n-3=\left\{-5;-1;1;5\right\}\Rightarrow n=\left\{-2;2;4;8\right\}\)

13 chia hết cho 4n - 15

=> 4n - 15 thuộc Ư(13) = {1;13}

=> 4n = 16;28

=> n = 4;7

toán học thêm nhà ai đấy

Lời giải:
$5n+17\vdots 4n+13$

$\Rightarrow 4(5n+17)\vdots 4n+13$

$\Rightarrow 20n+68\vdots 4n+13$

$\Rightarrow 5(4n+13)+3\vdots 4n+13$

$\Rightarrow 3\vdots 4n+13$

$\Rightarrow 4n+13\left\{\pm 1; \pm 3\right\}$

$\Rightarrow n\in \left\{-3; \frac{-7}{2}; \frac{-5}{2}; -4\right\}$

Mà $n$ nguyên nên $n\in \left\{-3; -4\right\}$

a)n=4,5

d) n =0