làm câu

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

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

2)

a) 2n+5 chia het cho n-1 

=> 2(n-1) +7 chia het cho n-1 

=: n-1 thuoc uoc cua 7 den day ke bang la xong. 

may cau con lai lam tuong tu

dài quá ko mún làm

Ta có: \(2n^2-n-1=2n^2+3n-4n-6+5=n\left(2n+3\right)-2\left(2n+3\right)+5\)

Vì \(n\left(2n+3\right)\)và \(-2\left(2n+3\right)\)chia hết cho \(2n+3\) nên để \(2n^2-n-1\)chia hết cho \(2n+3\) thì \(5\)phải chia hết cho \(2n+3\), tức là \(2n+3\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

Với  \(2n+3=1\)thì \(n=-1\)

Với  \(2n+3=-1\) thì \(n=-2\)

Với  \(2n+3=5\)thì \(n=1\)

Với  \(2n+3=-5\) thì \(n=-4\)

Vậy, để đa thức \(2n^2-n-1\) chia hết cho đa thức \(2n+3\) thì \(n=\left\{-2;-1;1;-4\right\}\) và  \(n\in Z\)

Đặt \(Q=\frac{2n^2+7n-2}{2n-1}\)

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

\(Q\in Z\Leftrightarrow\frac{2n^2+7n-2}{2n-1}\in Z\Leftrightarrow\frac{2}{2n-1}\in Z\Leftrightarrow2n-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Sau đó tìm n

Có:n+1\(⋮\)n+1

=>2n+2\(⋮\)n+1

Mà 2n-1 \(⋮\)n+1

=>(2n+2)-(2n-1)\(⋮\)n+1

=>2n+2-2n+1\(⋮\)n+1

=>3\(⋮\)n+1

=>n+1\(\in\)Ư(3)={-1;1;3;-3}

Nếu n+1=1=>n=0

Nếu n+1=-1=>n=-2

Nếu n+1=3=>n=2

Nếu n+1=-3=>n=-4

2n-1 chia het cho n+1

=>2.(n+1)-3 chia het cho n+1

=>-3 chia het cho n+1

=>n+1 E Ư(-3)={-3;-1;1;3}

=> n E {-4;-2;0;2}

2n-1 chia hết n+1

=> 2(n+1)-2-1 chia hết n+1

=> 2(n+1)-3 chia hết n+1

=> 3 chia hết cho n+1

=> n+1  =Ư(3)={-1;1;-3;3}

=>n={-2;0;-4;2}

Để 2n - 1 chia hết cho n + 1 <=> n + n + 1 + 1 - 3 chia hết cho n + 1

<=> ( n + 1 ) + ( n + 1 ) - 3 chia hết cho n + 1 

Vì n + 1 chia hết co n + 1

Để ( n + 1 ) + ( n + 1 ) - 3 chia hết cho n + 1 <=> 3 chia hết cho n + 1

<=> n + 1 là ước của 3

        Ư(3) = { - 3; - 1; 1; 3 }

Ta có n + 1 = - 3 => n = - 4 (TM)

         n + 1 = - 1 => n = - 2 (TM)

         n + 1 = 1 => n = 0 (TM)

         n + 1 = 3 => n = 2 (TM)

Vậy n = { - 4; - 2; 0; 2 }

\(2n-1⋮n-1\)

\(\Rightarrow2\left(n-1\right)+1⋮n-1\)

\(\Rightarrow1⋮n-1\)

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

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

Vậy............................

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

\(\Rightarrow3⋮n+1\Rightarrow n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

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

Vậy ...................