3n2+5/n-1=6n+5/n-1=6(n-1)+11/n-1=6(n-1)/n-1+11/n-1=6+11/n-1

để n nguyên thì 6+11/n-1 nguyên mà 6+11/n-1 nguyên khi n-1 thuộc ước 11

mà ước 11 thuộc (-1;1;11;-11)

ta có bảng;.............................

<=>4n-5=4n-2+7

<=>2.(2n-1)+7

vì 2.(2n-1) chia hết cho 2n-1

Nên 2n-1 thuộc Ư(7)={1;7;-1;-7}

do đó 2n-1=1=>n=1

2n-1=7=>n=8

2n-1=-1=>n=0

2n-1=-7=>n=-3

Vậy n ={1;8;0;-3}

\(\frac{4n-5}{2n-1}=\frac{4n-2}{2n-1}-\frac{3}{2n-1}\)\(=\frac{2\left(2n-1\right)}{2n-1}-\frac{3}{2n-1}\)\(=2-\frac{3}{2n-1}\)

=> \(\frac{3}{2n-1}\in Z=>\)\(3⋮\left(2n-1\right)=>2n-1\inƯ\left(3\right)\)

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

=>\(2n\in\left\{-2;0;2;4\right\}\)

=> n thuộc { -1;0;1;2}

a) \(25⋮n+2\left(n\in Z\right)\)

\(\Rightarrow n+2\in\left\{-1;1;-5;5;-25;25\right\}\)

\(\Rightarrow n\in\left\{-3;-1;-7;3;-27;23\right\}\)

b) \(2n+4⋮n-1\)

\(\Rightarrow2n+4-2\left(n-1\right)⋮n-1\)

\(\Rightarrow2n+4-2n+2⋮n-1\)

\(\Rightarrow6⋮n-1\)

\(\Rightarrow n-1\in\left\{-1;1;-2;2;-3;3;-6;6\right\}\)

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

c) \(1-4n⋮n+3\)

\(\Rightarrow1-4n+4\left(n+3\right)⋮n+3\)

\(\Rightarrow1-4n+4n+12⋮n+3\)

\(\Rightarrow13⋮n+3\)

\(\Rightarrow n+3\in\left\{-1;1;-13;13\right\}\)

\(\Rightarrow n\in\left\{-4;-2;-15;10\right\}\)

a) n ϵ{−3;−1;−7;3;−27;23}

b) n ∈{0;2;−1;3;−2;4;−5;7}

c) n ϵ {−4;−2;−15;10}

a. 

Đề bài sai, ví dụ \(n=1\) lẻ nhưng  \(1^2+4.1+8=13\) ko chia hết cho 8

b.

n lẻ \(\Rightarrow n=2k+1\)

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

\(=\left(2k+1-1\right)\left(2k+1+1\right)\left(2k+1+3\right)\)

\(=8k\left(k+1\right)\left(k+2\right)\)

Do \(k\left(k+1\right)\left(k+2\right)\) là tích 3 số tự nhiên liên tiếp nên chia hết cho 6

\(\Rightarrow8k\left(k+1\right)\left(k+2\right)\) chia hết cho 48

\(\dfrac{4n+7}{n+1}=\dfrac{n+1+n+1+n+1+n+1+3}{n+1}=1+1+1+1+\dfrac{3}{n+1}\)

Để nguyên thì \(\dfrac{3}{n+1}\in Z\) \(\Rightarrow n+1\in U\left(3\right)=\left\{\pm1;\pm3\right\}\)

- n+1=1 => n=0

- n+1=-1 => n=-2

- n+1=3 => n=2

- n+1=-3 => n=-4

Vậy \(n=\left\{0;-2;2;-4\right\}\)

2 n 3   –   3 n 2 + 3n – 1 = (2 n 2 – n + 2)(n – 1) + 1

Để 2 n 3   –   3 n 2 + 3n – 1 chia hết cho n – 1 thì 1 chia hết cho n – 1

=> (n – 1) Є {1;-1}

          n – 1             1        -1

             n                2        0

             P               9        1

                             TM       TM

Vậy n Є {0; 2} để P Є Z

Đáp án cần chọn là: A

A nguyên thì 3n^2-12+21 chia hết cho n-2

=>\(n-2\in\left\{1;-1;3;-3;7;-7;21;-21\right\}\)

=>\(n\in\left\{3;1;5;-1;9;-5;23;-19\right\}\)