Phân tích đa thức thành nhân tử :
a, \(x^8+x^4+1\)
b, \(x^4+2008x^2+2007x+2008\)
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M
0
H
2
7 tháng 2 2018
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D
7 tháng 2 2018
x^4 + 2008x^2 + 2007x + 2008
\(=x^4+x^2+2007x^2+2007x+2007+1\)
\(=x^4+x^2+1+2007\left(x^2+x+1\right)\)
\(=\left(x^2+1\right)^2-x^2+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
NT
1
18 tháng 6 2018
x4+2008x2+2007x+2008
<=> x4-x+2008x2+2008x+2008
<=> x(x3-1)+2008(x2+x+1)
<=> x(x-1)(x2+x+1)+2008(x2+x+1)
<=> (x2+x+1)(x2-x+2008)
VA
1
14 tháng 2 2016
x^4+2008x^2+2007x+2008
=x^4+2008x^2+2008x-x+2008
=(x^4-x)+(2008x^2+2008x+2008)
=x(x^3-1)+2008(x^2+x+1)
=x(x-1)(x^2+x+1)+2008(x^2+x+1)
=(x^2+x+1)(x^2-x+2008)
VT
2
14 tháng 3 2015
\(\left(x^4+x^2+1\right)+\left(2007x^2+2007x+2007\right)\)
=\(\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
PN
2
MA
7 tháng 10 2016
a) \(x^8+x+1\)
\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
b) \(x^4+2008x^2+2007x+2008\)
\(=x^4+x^3+x^2-x^3-x^2-x+2008x^2+2008x+2008\)
\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
VM
1
NT
1
BT
5 tháng 12 2017
=x4+2008x2+2008x-x+2008
=(x4-x)+(2008x2+2008x+2008)
=x(x3-1)+2008(x2+x+1)
=x(x-1)(x2+x+1)+2008(x2+x+1)
=(x2+x++1)(x2-x+2008)
a)\(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-\left(x^2\right)^2\)
\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x^4-x^2+1\right)\)
\(=\left(x^4+x^3+x^2\right)-\left(x^3-2007x^2-2007x-2008\right)\)
\(=x^2\left(x^2+x+1\right)-\left[x\left(x^2+x+1\right)-2008\left(x^2-x-1\right)\right]\)
\(=x^2\left(x^2+x+1\right)-\left(x^2+x+1\right)\left(x-2008\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)