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\(A=n^4-1=\left(n-1\right)\left(n+1\right)\left(n^2+1\right)\)
\(=\left(2k-1-1\right)\left(2k-1+1\right)\left(4k^2-4k+1+1\right)\)
\(=2k\left(2k-2\right)\left(4k^2-4k+2\right)\)
\(=8k\left(k-1\right)\left(2k^2-2k+1\right)⋮8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Với n=3 thì \(n^3+4n+3=3^3+4\cdot3+3=42⋮̸8\) nha bạn
b: Đặt \(A=n^3+3n^2-n-3\)
\(=\left(n^3+3n^2\right)-\left(n+3\right)\)
\(=n^2\left(n+3\right)-\left(n+3\right)\)
\(=\left(n+3\right)\left(n^2-1\right)\)
\(=\left(n-1\right)\left(n+1\right)\left(n+3\right)\)
n lẻ nên n=2k+1
=>\(A=\left(2k+1-1\right)\left(2k+1+1\right)\left(2k+1+3\right)\)
\(=2k\cdot\left(2k+2\right)\left(2k+4\right)\)
\(=8k\left(k+1\right)\left(k+2\right)\)
Vì k;k+1;k+2 là ba số nguyên liên tiếp
nên \(k\left(k+1\right)\left(k+2\right)⋮3!=6\)
=>\(A=8k\left(k+1\right)\left(k+2\right)⋮6\cdot8=48\)
c:
d: Đặt \(B=n^4-4n^3-4n^2+16n\)
\(=\left(n^4-4n^3\right)-\left(4n^2-16n\right)\)
\(=n^3\left(n-4\right)-4n\left(n-4\right)\)
\(=\left(n-4\right)\left(n^3-4n\right)\)
\(=n\left(n-4\right)\left(n^2-4\right)\)
\(=\left(n-4\right)\cdot\left(n-2\right)\cdot n\cdot\left(n+2\right)\)
n chẵn và n>=4 nên n=2k
B=n(n-4)(n-2)(n+2)
\(=2k\left(2k-2\right)\left(2k+2\right)\left(2k-4\right)\)
\(=2k\cdot2\left(k-1\right)\cdot2\left(k+1\right)\cdot2\left(k-2\right)\)
\(=16k\left(k-1\right)\left(k+1\right)\left(k-2\right)\)
Vì k-2;k-1;k;k+1 là bốn số nguyên liên tiếp
nên \(\left(k-2\right)\cdot\left(k-1\right)\cdot k\cdot\left(k+1\right)⋮4!=24\)
=>B chia hết cho \(16\cdot24=384\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 8:
a) Ta có: \(2^9-1=\left(2^3-1\right)\cdot\left(2^6+2^3+1\right)\)
\(=7\cdot\left(64+8+1\right)=7\cdot73⋮73\)(đpcm)
b) Ta có: \(5^6-10^4=5^4\cdot5^2-5^4\cdot2^4=5^4\left(5^2-2^4\right)\)
\(=5^4\left(25-16\right)=5^4\cdot9⋮9\)(đpcm)
c) Ta có: \(\left(n+3\right)^2-\left(n-1\right)^2\)
\(=\left(n+3-n+1\right)\left(n+3+n-1\right)\)
\(=4\cdot\left(2n+2\right)=4\cdot2\cdot\left(n+1\right)=8\left(n+1\right)⋮8\)(đpcm)
d) Ta có: \(\left(n+6\right)^2-\left(n-6\right)^2\)
\(=\left(n+6-n+6\right)\left(n+6+n-6\right)\)
\(=12\cdot2n=24n⋮24\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a) Ta có: \(\left(3n-1\right)^2-4\)
\(=\left(3n-1-2\right)\left(3n-1+2\right)\)
\(=\left(3n-3\right)\left(3n+1\right)\)
\(=3\cdot\left(n-1\right)\cdot\left(3n+1\right)⋮3\forall n\in N\)(đpcm)
b) Ta có: \(100-\left(7n+3\right)^2\)
\(=\left[10-\left(7n+3\right)\right]\left[10+\left(7n+3\right)\right]\)
\(=\left(10-7n-3\right)\left(10+7n+3\right)\)
\(=\left(7-7n\right)\left(13+7n\right)\)
\(=7\cdot\left(1-n\right)\cdot\left(13+7n\right)⋮7\forall n\in N\)(đpcm)
c) Ta có: \(\left(3n+1\right)^2-25\)
\(=\left(3n+1-5\right)\left(3n+1+5\right)\)
\(=\left(3n-4\right)\left(3n+6\right)\)
\(=3\cdot\left(3n-4\right)\cdot\left(n+2\right)⋮3\forall n\in N\)(đpcm)
d) Ta có: \(\left(4n+1\right)^2-9\)
\(=\left(4n+1-3\right)\left(4n+1+3\right)\)
\(=\left(4n-2\right)\left(4n+4\right)\)
\(=2\cdot\left(2n-1\right)\cdot4\cdot\left(n+1\right)\)
\(=8\cdot\left(2n-1\right)\cdot\left(n+1\right)⋮8\forall n\in N\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(a=5k+1,b=5n+4\left(k,n\in N\right)\)
\(\Rightarrow ab+1=\left(5k+1\right)\left(5n+4\right)+1=25kn+20k+5n+4+1=25kn+20k+5n+5=5\left(5kn+5k+n+1\right)⋮5\forall k,n\in N\)
Ta có: ab+1
\(=\left(5k+1\right)\left(5c+4\right)+1\)
\(=25kc+20k+5c+4+1\)
\(=25kc+20k+5c+5⋮5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
b) Ta có: \(\left(2n-3\right)\left(2n+3\right)-4n\left(n-9\right)\)
\(=4n^2-9-4n^2+36n\)
\(=36n-9⋮9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = n^2 ( n+ 3 ) - ( n+ 3 )
= ( n^2 - 1 )(n+ 3 )
= ( n+ 1 )(n- 1 )(n + 3)
Vì n lẻ => n = 2k+ 1 thay vào ta có :
A = ( 2k + 1 + 1 )(2k+1 - 1 )(2k + 1 + 3) = (2k+2).2k (2k+4) = 2(k+1).2k . 2(k+2) = 8k(k+1)(k+2)
Luôn luôn chia hết cho 8 mới mọi n lẻ
=> A chia hết cho 8
\(n^4-1\)
\(=\left(n^2\right)^2-1^2\)
\(=\left(n^2-1\right)\left(n^2+1\right)\)
\(=\left(n-1\right)\left(n+1\right)\left(n^2+1\right)\)
Vì n lẻ \(\Rightarrow\hept{\begin{cases}n-1\text{chẵn}\\n+1\text{chẵn}\\n^2+1\text{chẵn}\Rightarrow n^2+1⋮2\left(1\right)\end{cases}}\)
mặt khác n - 1 và n + 1 là 2 số chẵn liên tiếp \(\Rightarrow\left(n-1\right)\left(n+1\right)⋮4\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮8\left(đpcm\right)\)
Phân tích thành nhân tử:
\(n^4-1=\left(n^2-1\right)\left(n^2+1\right)=\left(n-1\right)\left(n+1\right)\left(n^2+1\right)\)
Vì n là số tự nhiên lẻ nên n = 2k + 1 với k là số tự nhiên
Khi đó:
\(n^4-1=\left(2k-1+1\right)\left(2k+1+1\right)\left(n^2+1\right)\)
\(=2k\left(2k+2\right)\left(n^2+1\right)\)
\(=2k.2.\left(k+1\right)\left(n^2+1\right)\)
\(=4k\left(k+1\right)\left(n^2+1\right)\)
Vì k(k+1) là tích hay số tự nhiên liên tiếp nên k(k+1) chia hết cho 2 \(\Rightarrow4k\left(k+1\right)⋮8\)
\(\Rightarrow4k\left(k+1\right)\left(n^2+1\right)⋮8\)
hay \(n^4-1⋮8\)(với n là số tự nhiên lẻ)
Ta có điều phải chứng minh.