Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Cách 1:
Ta có:
\(A=n^4-6n^3+27n^2-54n+32=(n^4-n^3)-5n^3+5n^2+22n^2-22n-32n+32\)
\(=n^3(n-1)-5n^2(n-1)+22n(n-1)-32(n-1)\)
\(=(n-1)(n^3-5n^2+22n-32)\)
\(=(n-1)(n^3-2n^2-3n^2+6n+16n-32)\)
\(=(n-1)[n^2(n-2)-3n(n-2)+16(n-2)]\)
\(=(n-1)(n-2)(n^2-3n+16)\)
Ta thấy $(n-1)(n-2)$ là tích 2 số nguyên liên tiếp nên \((n-1)(n-2)\vdots 2\)
\(\Rightarrow A=(n-1)(n-2)(n^2-3n+16)\vdots 2\)
Ta có đpcm.
Cách 2:
\(A=n^4-6n^3+27n^2-54n+32\)
\(=(n^4+27n^2)-(6n^3+54n-32)\)
\(=n^2(n^2+27)-2(3n^3+27n-16)\)
Ta thấy \(n^2+27-n^2=27\) lẻ nên $n^2, n^2+27$ khác tính chẵn lẻ
Do đó trong 2 số $n^2$ và $n^2+27$ có 1 số chẵn, 1 số lẻ
\(\Rightarrow n^2(n^2+27)\vdots 2\)
Mà \(2(3n^3+27n-16)\vdots 2\)
Suy ra \(A=n^2(n^2+27)-2(3n^3+27n-16)\vdots 2\)
Ta có đpcm.
\(n^4-6n^3+27n^2-54n+32\)
\(=n^4-n^3-5n^3+5n^2+22n^2-22n+32n-32\)
\(=\left(n-1\right)\left(n^3-5n^2+22n+32\right)\)
\(=\left(n-1\right)\left(n^3-2n^2-3n^2+6n+16n+32\right)\)
\(=\left(n-1\right)\left(n-2\right)\left(n^2-3n+16\right)\) chia hếtcho 2
Bài 1:
\(M=x^4-x^3-x^3+x^2+2x^2-2x+2\)
\(=x^2\left(x^2-x\right)-x\left(x^2-x\right)+2\left(x^2-x\right)+2\)
\(=3x^2-3x+6+2\)
\(=3x^2-3x+8\)
\(=3\left(x^2-x\right)+8=3\cdot3+8=17\)
\(=x^3\left(x+2\right)-x\left(x+2\right)\)
\(=\left(x+2\right)\cdot x\cdot\left(x+1\right)\left(x-1\right)\)
Vì đây là tích của bốn số nguyên liên tiếp
nên \(\left(x+2\right)\cdot x\cdot\left(x+1\right)\cdot\left(x-1\right)⋮24\)
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)
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)
\(A=n^4-6n^3+27n^2-54n+32\)
\(=\left(n^4-3n^3+16n^2\right)-\left(3n^3-9n^2+48n\right)+\left(2n^2-6n+32\right)\)
\(=n^2\left(n^2-3n+16\right)-3n\left(n^2-3n+16\right)+2\left(n^2-3n+16\right)\)
\(=\left(n^2-3n+2\right)\left(n^2-3n+16\right)\)
\(=\left(n-2\right)\left(n-1\right)\left(n^2-3n+16\right)\)
Nhận thấy: \(\left(n-2\right)\left(n-1\right)\)là tích 2 số nguyên liên tiếp \(\left(n\in Z\right)\)
=> \( \left(n-2\right)\left(n-1\right)\)\(⋮\)\(2\)
=> A chia hết cho 2