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\(\left(4n+3\right)^2-25\)
\(=\left(4n+3\right)^2-5^2\)
\(=\left(4n+3-5\right)\left(4n+3+5\right)\)
\(=\left(4n-2\right)\left(4n+8\right)\)
![](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)
a: \(=\left(4n-7-5\right)\left(4n-7+5\right)\)
\(=\left(4n-12\right)\left(4n-2\right)\)
\(=8\left(n-3\right)\left(2n-1\right)⋮8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(4n-3\right)^2-5^2=\left(4n-3+5\right)\left(4n-3-5\right)\)
\(=\left(4n+2\right)\left(4n-8\right)\)
\(=8\left(2n+1\right)\left(n-2\right)\)
Vậy \(\left(4n-3\right)^2-25\)chia hết cho 8
Áp dụng hằng đẳng thứ số 3 ta dc :
\(\left(4n-3\right)^2-25\)
\(=\left(4n-3\right)^2-5^2\)
\(=\left(4n-3-5\right)\left(4n-3+5\right)\)
\(=\left(4n-8\right)\left(4n-2\right)\)
Thanks nha
(4n-7)2-25=(4n-7+5)(4n-7-5)
=(4n-2)(4n-12)
=2.(2n-1).4.(n-3)
=8.(2n-1)(n-3)
=> (4n-7)2-25 chia hết cho 8
cmr (4n-7)^2-25 chia hết cho 8