6   \(n^5+5n=n^5-n+6n=n\left(n^4-1\right)+6n=n\left(n^2-1\right)\left(n^2+1\right)+6n\)

\(=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)+6n\)

vì n,n-1 là 2 số nguyên lien tiếp  \(\Rightarrow n\left(n-1\right)⋮2\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮2\)

  n,n-1,n+1 là 3 sô nguyên liên tiếp \(\Rightarrow n\left(n-1\right)\left(n+1\right)⋮3\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮3\)

\(\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮2\cdot3=6\)

\(6⋮6\Rightarrow6n⋮6\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)-6n⋮6\Rightarrow n^5+5n⋮6\)(đpcm)

7   \(n\left(2n+7\right)\left(7n+1\right)=n\left(2n+7\right)\left(7n+7-6\right)=7n\left(n+1\right)\left(2n+7\right)-6n\left(2n+7\right)\)

\(=7n\left(n+1\right)\left(2n+4+3\right)-6n\left(2n+7\right)\)

\(=7n\left(n+1\right)\left(2n+4\right)+21n\left(n+1\right)-6n\left(2n+7\right)\)

\(=14n\left(n+1\right)\left(n+2\right)+21n\left(n+1\right)-6n\left(2n+7\right)\)

n,n+1,n+2 là 3 sô nguyên liên tiếp dựa vào bài 6 \(\Rightarrow n\left(n+1\right)\left(n+2\right)⋮6\Rightarrow14n\left(n+1\right)\left(n+2\right)⋮6\)

\(21⋮3;n\left(n+1\right)⋮2\Rightarrow21n\left(n+1\right)⋮3\cdot2=6\)

\(6⋮6\Rightarrow6n\left(2n+7\right)⋮6\)

\(\Rightarrow14n\left(n+1\right)\left(n+2\right)+21n\left(n+1\right)-6n\left(2n+7\right)⋮6\)

\(\Rightarrow n\left(2n+7\right)\left(7n+1\right)⋮6\)(đpcm)

......................?

mik ko biết

mong bn thông cảm 

nha ................

2x+1 chia hết cho x-5

=>2x-10+11 chia hết cho x-5

=>11 chia hết cho x-5

=>x-5 thuộc Ư(11)={-1;1;-11;11}

=>x thuộc{4;6;-6;16}

<=>2(x-5)+6 chia hết x-5

=>6 chia hết x-5

=>x-5\(\in\){-1,-2,-3,-6,1,2,3,6}

=>x\(\in\){4,3,2,-1,6,7,8,11}

\(3x+2⋮x-1\)

\(\Leftrightarrow3\left(x-1\right)+5⋮x-1\)

\(\Leftrightarrow5⋮x-1\)

\(\Leftrightarrow\left(x-1\right)\inƯ\left(5\right)\)

\(\Leftrightarrow\left(x-1\right)\in\left\{\pm1;\pm5\right\}\)

\(\Leftrightarrow x\in\left\{-4;0;2;6\right\}\)

Vậy để \(3x+2⋮x-1\) thì \(x\in\left\{-4;0;2;6\right\}\)

b) \(x^2+2x-7⋮x+2\)

\(\Leftrightarrow x\left(x+2\right)-7⋮x+2\)

\(\Leftrightarrow7⋮x+2\)

\(\Leftrightarrow\left(x+2\right)\inƯ\left(7\right)\)

\(\Leftrightarrow\left(x+2\right)\in\left\{\pm1;\pm7\right\}\)

\(\Leftrightarrow x\in\left\{-9;-3;-1;5\right\}\)

Vậy để \(x^2+2x-7⋮x+2\) thì \(x\in\left\{-9;-3;-1;5\right\}\)

Có \(\left(2x+5\right)⋮\left(x-1\right)\)

\(\Rightarrow2\left(x-1\right)+7⋮\left(x-1\right)\)

Mà \(2\left(x-1\right)⋮\left(x-1\right)\Rightarrow7⋮\left(x-1\right)\)

\(\Rightarrow x-1\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)

\(\Rightarrow x\in\left\{2;0;8;-6\right\}\)

Vậy x \(\in\left\{2;0;8;-6\right\}\)

\(2x+5⋮x-1\)

Ta có: \(2x+5=2\left(x-1\right)+7\)

Để \(2x+5⋮x-1\Leftrightarrow7⋮x-1\)

\(\Rightarrow x-1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Vậy ....

P/s: Hoq chắc ((:

2x + 5 chia hết cho x - 1

Mà x - 1 chia hết cho x - 1

=> 2 ( x - 1 ) chia hết cho x - 1

=> 2x - 2 chia hết cho x - 1

=> ( 2x + 5 ) - ( 2x - 2 ) chia hết cho x - 1

=>  2x + 5 - 2x + 2 chia hết cho x - 1

=> 7 chia hết cho x - 1

=> x - 1 thuộc Ư( 7 )

=> x - 1 thuộc { 1 ; - 1 ; 7 ; - 7 }

=> x thuộc { 2 ; 0 ; 8 ; - 6 }

dễ mà nhưng ngại làm