Phân tích đa thức thành nhân tử
\(x^4+2011x^2+2010x+2011\)
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MT
2
AM
8 tháng 6 2015
x4+2011x2+2010x+2011
=(x4+x3+x2)+(2011x2+2011x+2011)-(x3+x2+x)
=x2(x2+x+1)+2011(x2+x+1)-x(x2+x+1)
=(x2+x+1)(x2+2011-x)
MT
8 tháng 6 2015
x4+2011x2+2010x+2011=x4-x+2011x2+2011x+2011
=x(x3-1)+2011(x2+x+1)
=x(x- 1)(x2+x+1)+2011(x2+x+1)
=(x2+x+1)[x(x-1)+2011]
=(x2+x+1)(x2-x+2011)
NT
3
22 tháng 7 2017
a) \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-xz-yz+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
b) \(x^4+2011x^2+2010x+2011\)
\(=x^4+2010x^2+x^2+2010x+2010+1\)
\(=\left(x^4+x^2+1\right)+\left(2010x^2+2010x+2010\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2010\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2011\right)\)
5 tháng 9 2018
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
TD
1
2 tháng 9 2016
a ) \(3x^3-7x^2+17x-5\)
\(=\left(3x^2-x^2\right)-\left(6x^2-2x\right)+\left(15x-5\right)\)
\(=x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)
\(=\left(x^2-2x+5\right)\left(3x-1\right)\)
b \(x^4+2011x^2+2010x+2011\)
\(=x^4-x+2011x^2+2011x+2011\)
\(=x\left(x^3-1\right)+2011\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2011\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2011\right)\)
KJ
2
R
18 tháng 8 2019
x4+2012x2+2011x+2012
=(x4-x)+(2012x2+2012x+2012)
=x(x3-1)+2012(x2+x+1)
=x(x-1) (x2+x+1) + 2012 (x2+x+1)
=(x2+x+1) [x(x-1)+2012]
=(x2+x+1) (x2-x+2012)
NP
1
PN
26 tháng 11 2015
\(x^4+2010x^2+2009x+2010\)
\(=x^4-x+\left(2010x^2+2010x+2010\right)\)
\(=x\left(x^3-1\right)+2010\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2010\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+2010\right]=\left(x^2+x+1\right)\left(x^2-x+2010\right)\)
TT
1
20 tháng 8 2017
1) \(\left(x^2+3x+1\right)^2-1=\left(x^2+3x\right)\left(x^2+3x+2\right)=x\left(x+3\right)\left[\left(x^2+2x\right)+\left(x+2\right)\right]\)
\(=x\left(x+3\right)\left[x\left(x+2\right)+\left(x+2\right)\right]=x\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
2) \(x^4+2012x^2+2011x+2012\)
\(=\left(x^4-x\right)+\left(2012x^2+2012x+2012\right)\)
\(=x\left(x^3-1\right)+2012\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2012\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+2012\right]\)
\(=\left(x^2+x+1\right)\left(x^2-x+2012\right)\)
NT
Phân tích các đa thức sau thành nhân tử
a) (x+y+z)^3 - x^3 - y^3 - z^3
b) x^4 + 2012x^2 + 2011x + 2012
3
4 tháng 12 2014
= x3 + y3 + z3 + 3x2yz + 3xy2z + 3xyz2 - x3 -y3 - z3
=3x2yz + 3xy2z + 3xyz2
= 3xyz( x + y + z)
LT
4 tháng 12 2014
b.
x^4+2012x^2+2012x-x+2012=
(x^4-x)+2012(x^2+x+1)=
x(x-1)(x^2+x+1)+2012(x^2+x+1)=
(x+2012)(x^2+x+1)
TT
1 tháng 9 2020
a) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)^3+z^3+3.\left(x+y\right).z.\left(x+y+z\right)\right]-x^3-y^3-z^3\)
\(=\left[x^3+y^3+3xy.\left(x+y\right)+z^3+3\left(x+y\right).z.\left(x+y+z\right)\right]-x^3-y^3-z^3\)
\(=3xy\left(x+y\right)+3\left(x+y\right)z.\left(x+y+z\right)\)
\(=3.\left(x+y\right)\left(xy+zx+zy+z^2\right)\)
\(=3.\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
b) \(x^4+2012x^2+2011x+2012\)
\(=x^4-x+2012x^2+2012x+2012\)
\(=x.\left(x^3-1\right)+2012.\left(x^2+x+1\right)\)
\(=x.\left(x-1\right)\left(x^2+x+1\right)+2012.\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2012\right)\)
=(x4−x3+2011x2)+
(x3−x2+2011x)+(x2−x+2011)
=x2(x2−x+2011)+x(x2−x+2011)+(x2−x+2011)
=(x2+x+1)(x2−x+2011)
=(x4−x3+2011x2)+(x3−x2+2011x)+(x2−x+2011)
=x2(x2−x+2011)+x(x2−x+2011)+(x2−x+2011)
=(x2+x+1)(x2−x+2011)
x3−x2+2011x)+(x2−x+2011)
=x2(x2−x+2011)+x(x2−x+2011)+(x2−x+2011)=(x2+x+1)(x2−x+2011)
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