Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, =x4-x + 2019x2+2019x+2019
=x(x3-1)+2019(x2+x+1)
=x(x-1)(x2+x+1)+2019(x2+x+1)
=(x2-x+2019)(x2+x+1)
b, =(x-y+y-z)[(x-y)2-(x-y)(y-z)+(y-z)2 ] + (z-x)3
=(x-z)(x2-2xy+y2-xy+xz+y2-yz+y2-2yz+z2) - (x-z)3
=(x-z)(x2-2xy+y2-xy+xz+y2-yz+y2-2yz+z2-x2+2xz-z2)
=(x-z)(-3xy+3y2+3xz-3yz)
=3(x-z)(-xy+y2+xz-yz)
=3(x-z)[(-xy+xz)+(y2-yz)]
=3(x-z)[-x(y-z)+y(y-z)]
=3(y-x)(x-z)(y-z)
\(x^4+2019x^2+2018x+2019\)
\(=x^4-x^3+x^3+2019x^2-x^2+x^2+2019x-x+2019\)
\(=\left(x^4-x^3+2019x^2\right)+\left(x^3-x^2+2019x\right)+\left(x^2-x+2019\right)\)
\(=x^2\left(x^2-x+2019\right)+x\left(x^2-x+2019\right)+\left(x^2-x+2019\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)
\(x^4+2018x^2+2017x+2018\)
\(=\left(x^4-x\right)+\left(2018x^2+2018x+2018\right)\)
\(=x.\left(x^3-1\right)+2018.\left(x^2+x+1\right)\)
\(=x.\left(x-1\right)\left(x^2+x+1\right)+2018.\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2018\right)\)
\(x^5+x^4+2\)
\(=x^5+x^4+x^2-x^2+1+1\)
\(=\left(x^5-x^2\right)+\left(x^4+x^2+1\right)\)
\(=\left(x^5-x^2\right)+\left(x^4+2x^2-x^2+1\right)+1\)
\(=x^2\left(x^3-1\right)+\left(x^4+2x^2-x^2+1\right)+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(\left(x^2+1\right)^2-x^2\right)+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+1+x\right)\cdot\left(x^2+1-x\right)+1\)
\(=\left(x^3-x^2\right)\left(x^2+x+1\right)+\left(x^2+1+x\right)\cdot\left(x^2+1-x\right)+1\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+x^2+1-x\right)+1\)
\(=\left(x^2+x+1\right)\left(x^3+1-x\right)+1\)
<=>x4-x+x2 +x+1= x (x-1) (x2+x+1) + (x2+x+1) = (x2+x+1)(x2-x+1)
chắc có lẽ đúng đó
Nghịch xíu :v
a, \(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)-2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x+2\right)\)
b, \(x^2+4x+3\)
\(=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
Chúc bạn học tốt!!!
\(x^4+3x^2-4\)
\(=x^4+4x^2-x^2-4\)
\(=x^2\left(x^2+4\right)-\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x^2-1\right)\)
\(=\left(x^2+4\right)\left(x-1\right)\left(x+1\right)\)
Chúc bạn học tốt.
Ta có: x3 - x2 - 4
= (x3 - 1) - (x2 - 2x + 1) - 2(x + 1)
= (x - 1)(x2 + x + 1) - (x - 1)2 - 2(x + 1)
= (x - 1)(x2 + x + 1 - x + 1 - 2)
= x2(x - 1)
x3 - x2 - 4
= x3 + x2 - 2x2 - 4
= (x3 - 2x2) + (x2 - 4)
= x2 (x - 2) - (x2 - 22)
= x2 (x - 2) - (x + 2) (x - 2)
= (x - 2) [x2 + (x + 2)]
= (x - 2) (x2 + x + 2)
#Học tốt!!!
~NTTH~