7ⁿ⁺⁴-7ⁿ=7ⁿ(7⁴-1)=7ⁿ.2400

Do 2400 chia hết cho 30=>7ⁿ.2400 chia hết cho 30

Vậy 7ⁿ⁺⁴-7ⁿ chia hết cho 30 với mọi n thộc N

\(7^{n+4}-7^n=7^n.7^4-7^n=7^n.\left(7^4-1\right)=7^n.2400\)

=\(7^n.30.80\) chia hết cho 30

**** bạn

\(7^{n+4}-7^n=7^n\left(7^4-1\right)=7^n\left(2401-1\right)=7^n.2400=7^n.80.30\)

Chia hết cho 30 

Vậy 7^ n+4 - 7^n chia hết cho 30 

Nếu n = 3k (k \(\in N\)) thì 2ⁿ - 1 = 2^3k - 1 = 8^k - 1 = 7d \(⋮7\)

Nếu n = 3k+1 (k \(\in N\)) thì 2ⁿ - 1 = 2^3k+1 - 1 = 2^3k.2 - 2 + 1

= 2(2^3k - 1 ) +1

= BS7 + 1 ko chia hết cho 7

Nếu n = 3k+2 (k \(\in N\)) thì 2ⁿ - 1 = 2^3k+2 - 1 = 2^3k.4 - 4 + 3

= 4(2^3k - 1) + 3

= BS7 + 3 ko chia hết cho 7

Do đó: 2ⁿ - 1 chia hết cho 7 khi n = 3k (k \(\in N\))

\(7^{n+4}-7^n\)

\(\Rightarrow7^n\cdot7^4-7^n\)

\(\Rightarrow7^n\cdot\left(7^4-1\right)\)

\(\Rightarrow7^n\cdot\left(2401-1\right)\)

\(\Rightarrow7^n\cdot2400\)

\(\Rightarrow7^n\cdot30\cdot80⋮30\left(đpcm\right)\)

\(3^{n+2}+3^n\)

\(\Rightarrow3^n\cdot3^2+3^n\)

\(\Rightarrow3^n\cdot\left(3^2+1\right)\)

\(\Rightarrow3^n\cdot\left(9+1\right)\)

\(\Rightarrow3^n\cdot10⋮10\left(đpcm\right)\)

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 ................

hỏi là chứng minh cơ mà

Ta có :

\(n^4+7\left(7+2n^2\right)\)

\(=n^4+49+14n^2\)

\(=\left(n^2+7\right)^2\)

Vì n là số nguyên lẻ nên n có dạng 2k + 1 với k là số nguyên 

 \(\Rightarrow\left(n^2+7\right)^2=\left[\left(2k+1\right)^2+7\right]^2\)

\(=\left[\left(4k^2+4k+1\right)+7\right]^2\)

\(=\left(4k^2+4k+8\right)^2\)

\(=\left[4k\left(k+1\right)+8\right]^2\)

Vì \(\hept{\begin{cases}k\left(k+1\right)⋮2\forall k\in Z\\4⋮4\end{cases}}\) nên \(4k\left(k+1\right)⋮8\forall k\in Z\)

\(\Rightarrow4k\left(k+1\right)+8⋮8\forall k\in Z\)

\(\Rightarrow\left[4k\left(k+1\right)+8\right]^2⋮8^2\forall k\in Z\)

\(\Rightarrow\left[4k\left(k+1\right)+8\right]⋮64\forall k\in Z\)

=> đpcm 

n⁴ + 7( 7 + 2n² )

= n⁴ + 14n² + 49

= ( n² + 7 )²

Vì n lẻ và n ∈ Z => n = 2k + 1 ( k ∈ Z )

Thế vô ta được :

[ ( 2k + 1 )² + 7 ]²

= ( 4k² + 4k + 1 + 7 )²

= ( 4k² + 4k + 8 )²

= [ 4( k² + k + 2 ) ]²

= { 4[ k( k + 1 ) + 2 ] }²

Ta có : k( k + 1 ) chia hết cho 2

            2 chia hết cho 2

=> k( k + 1 ) + 2 chia hết cho 2

=> 4[ k( k + 1 ) + 2 ] chia hết cho 8

=>  { 4[ k( k + 1 ) + 2 ] }² chia hết cho 64

=> đpcm

24^54.54^24.2^10=(2^3.3)^54.(3^3.2)^24... 

=(2^3)^54.3^54.(3^3)^24.2^24.2^10 

= 2^162.2^24.2^10.3^54.3^72 

=2^196.3^126 

72^63=(2^3.3^2)^63 

=(2^3)^63(.3^2)^63=2^189.3^126 

vì 2^196.3^126 chia hết 2^189.3^126 

=>24^54.54^24.2^10 chia hết 72^63