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DM
0
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 2 2019
\(=x^4-x+2019x^2+2019x+2019\)
\(=x\left(x^3-1\right)+2019\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2019\left(x^2+x+1\right)\)
\(=\left(x^2-x\right)\left(x^2+x+1\right)+2019\left(x^2+x+1\right)\)
\(=\left(x^2-x+2019\right)\left(x^2+x+1\right)\)
22 tháng 2 2019
\(x^4+2019x^2+2018x+2019\)
\(=\left(x^4-x\right)+\left(2019x^2+2019x+2019\right)\)
\(=x\left(x^3-1\right)+2019\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2019\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+2019\right]\)
\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)
HT
2
KT
20 tháng 10 2018
\(x^4+2019x^2+2018x+2019\)
\(=x^4+x^2+1+2018\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2018\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)
23 tháng 3 2018
Ta có : x4 + 2018x2 + 2017x + 2018
= x4 - x + 2018x2 + 2018x + 2018
= x(x3 - 1) + 2018(x2 + x + 1)
= x(x - 1)(x2 + x + 1) + 2018(x2 + x + 1)
= (x2 + x + 1)(x2 - x + 2018)
S
5
KB
0
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 3 2019
a/ \(x^2\left(y-z\right)-y^2\left(y-z+x-y\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left(y-z\right)-y^2\left(x-y\right)+z^2\left(x-y\right)\)
\(=\left(x^2-y^2\right)\left(y-z\right)-\left(x-y\right)\left(y^2-z^2\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x+y-y-z\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
b/ \(2a^3+7a^2b+7ab^2+2b^3\)
\(=2a^3+3a^2b+ab^2+4a^2b+6ab^2+2b^3\)
\(=a\left(2a^2+3ab+b^2\right)+2b\left(2a^2+3ab+b^2\right)\)
\(=\left(a+2b\right)\left(2a^2+3ab+b^2\right)=\left(a+2b\right)\left(2a^2+ab+2ab+b^2\right)\)
\(=\left(a+2b\right)\left(a\left(2a+b\right)+b\left(2a+b\right)\right)=\left(a+2b\right)\left(a+b\right)\left(2a+b\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 3 2019
c/ \(x^3-x^2-14x+24=x^3+x^2-12x-2x^2-2x+24\)
\(=x\left(x^2+x-12\right)-2\left(x^2+x-12\right)=\left(x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-2\right)\left(x^2-3x+4x-12\right)=\left(x-2\right)\left(x\left(x-3\right)+4\left(x-3\right)\right)\)
\(=\left(x-2\right)\left(x-3\right)\left(x+4\right)\)
d/ \(x^3+y^3+z^3-3xyz=x^3+3x^2y+3xy^2+y^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)-3xy\left(x+y+z\right)\)
\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
e/ \(x^4+2019x^2+2018x+2019=x^4-x+2019x^2+2019x+2019\)
\(=x\left(x^3-1\right)+2019\left(x^2+x+1\right)=x\left(x-1\right)\left(x^2+x+1\right)+2019\left(x^2+x+1\right)\)
\(=\left(x^2-x+2019\right)\left(x^2+x+1\right)\)
DN
1
\(x^4+2019x^2+2018x+2019\)
\(=x^4-x^3+x^3+2019x^2-x^2+x^2+2019x-x+2019\)
\(=\left(x^4-x^3+2019x^2\right)+\left(x^3-x^2+2019x\right)+\left(x^2-x+2019\right)\)
\(=x^2\left(x^2-x+2019\right)+x\left(x^2-x+2019\right)+\left(x^2-x+2019\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)