Phân tích: \(x^4+2017x^2+2016x+2017\)
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NN
2
8 tháng 12 2016
Ta có : x^4+2017x^2+2016x+2017
=x^4+x^3-x^3+x^2-x^2+2017x^2+2017x-x+2017
=x^4+x^3+x^2-x^3-x^2-x+2017x^2+2017x+2017
=x^2(x^2+x+1)-x(x^2+x+1)+2017(x^2+x+1)
=(x^2+x+1)(x^2-x+2017)
Nhớ k mk nha
3 tháng 12 2017
Ta có : x^4+2017x^2+2016x+2017
=x^4+x^3-x^3+x^2-x^2+2017x^2+2017x-x+2017
=x^4+x^3+x^2-x^3-x^2-x+2017x^2+2017x+2017
=x^2(x^2+x+1)-x(x^2+x+1)+2017(x^2+x+1)
=(x^2+x+1)(x^2-x+2017)
chúc cậu hok tốt _@
NN
0
AD
3
KT
27 tháng 1 2018
\(x^4+2017x^2+2016x+2017\)
\(=\left(x^4+x^2+1\right)+2016\left(x^2+x+1\right)\)
\(=\left(x^4+2x^2+1-x^2\right)+2016\left(x^2+x+1\right)\)
\(=\left[\left(x^2+1\right)-x^2\right]+2016\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2016\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2017\right)\)
27 tháng 1 2018
\(x^4+2017x^2+2016x+2017\)
\(=\left(x^4-x\right)+\left(2007x^2+2007x+2007\right)\)
\(=x.\left(x^3-1\right)+2007.\left(x^2+x+1\right)\)
\(=x.\left(x-1\right)\left(x^2+x+1\right)+2007.\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2007\right)\)
DN
1
9 tháng 1 2018
a,\(x^5-x^4-x^4+x^3+2x^3-2x^2-2x^2+2\)2x-2x+2\(x^4\left(x-1\right)-x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)+2\left(x-1\right)\)
=\(\left(x^4-x^3+2x^2-2x+2\right)\left(x-1\right)\)
b,
KN
1
TT
1
KN
20 tháng 10 2019
\(x^4+2016x^2+2017x+2016\)
\(=x^4+2016x^2+2016x+x+2016\)
\(=\left(x^4+x\right)+\left(2016x^2+2016x+2016\right)\)
\(=x\left(x^3+1\right)+2016\left(x^2+x+1\right)\)
\(=x\left(x+1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2+x+2016\right)\)
\(x^4+2017x^2+2016x+2017\)
\(=x^4+2017x^2-x+2017x+2017\)
\(=\left(x^4-x\right)+\left(2017x^2+2017x+2017\right)\)
\(=x.\left(x^3-1\right)+2017.\left(x^2+x+1\right)\)
\(=x.\left(x^3-x^2+x^2-x+x-1\right)+2017.\left(x^2+x+1\right)\)
\(=x.\left[x^2.\left(x-1\right)+x.\left(x-1\right)+\left(x-1\right)\right]+2017.\left(x^2+x+1\right)\)
\(=x.\left(x-1\right)+\left(x^2+x+1\right)+2017.\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right).\left[x\left(x-1\right)+2017\right]\)
\(=\left(x^2+x+1\right).\left(x^2-x+2017\right)\)
Chúc bạn học tốt!!!