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12 tháng 2 2020

\(x^4+2019x^2+2018x+2019\)

\(=x^4-x^3+x^3+2019x^2-x^2+x^2+2019x-x+2019\)

\(=\left(x^4-x^3+2019x^2\right)+\left(x^3-x^2+2019x\right)+\left(x^2-x+2019\right)\)

\(=x^2\left(x^2-x+2019\right)+x\left(x^2-x+2019\right)+\left(x^2-x+2019\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)

3 tháng 1 2019

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)

20 tháng 2 2019

a) \(x^4+2019x^2+2018x+2019\)

\(=\left(x^4-x\right)+\left(2019x^2+2019x+2019\right)\)

\(=x\left(x^3-1\right)+2019\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2019\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+2019\right]\)

\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)

b) \(E=2x^2-8x+1=2x^2-8x+8-7\)

\(=2\left(x^2-4x+4\right)-7=2\left(x-2\right)^2-7\)

Vì \(2\left(x-2\right)^2\ge0\forall x\Rightarrow E\ge-7\)

Dấu "=" xảy ra <=> \(2\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy MinE = -7 <=> x = 2

20 tháng 2 2019

b) \(E=2x^2-8x+1\)

\(E=2\left(x^2-4x+\frac{1}{2}\right)\)

\(E=2\left(x^2-2\cdot x\cdot2+2^2+\frac{7}{2}\right)\)

\(E=2\left[\left(x-2\right)^2+\frac{7}{2}\right]\)

\(E=2\left(x-2\right)^2+7\ge7\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy....

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(x+ x + 1)

= (x2 + x + 1)(x2 - x + 2018)

23 tháng 3 2018

2 câu riêng hay chung ?? 

thử dùng hệ số bất định xem thanh niên

13 tháng 10 2019

hệ số bất định là cqq j?

NV
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

\(x^4+2019x^2+2018x+2019\)

\(=\left(x^4-x\right)+\left(2019x^2+2019x+2019\right)\)

\(=x\left(x^3-1\right)+2019\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2019\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+2019\right]\)

\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)

5 tháng 8 2019

\(\text{a) }4x^{16}+81=4x^4+36x^2+81-36x^8\)

                          \(=\left(4x^{16}+36x^8+81\right)-36x^8\)

                          \(=\left[\left(2x^8\right)^2+2.2x^8.9+9^2\right]+\left(6x^4\right)^2\)

                          \(=\left(2x^8+9\right)^2-\left(6x^4\right)^2\)

                         \(=\left(2x^8+9-6x^4\right)\left(2x^8+9+6x^4\right)\)                    

5 tháng 8 2019

\(\text{b) }x^4+2018x^2+2017x+2018\)

\(=x^4+2018x^2+2018x-x+2018\)

\(=\left(x^4-x\right)+\left(2018x^2+2018x+2018\right)\)

\(=x\left(x^3-1\right)-2018\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2018\left(x^2+x+1\right)\)

\(=\left(x^2-x\right)\left(x^2+x+1\right)+2018\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2018\right)\)

12 tháng 10 2021

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

28 tháng 11 2021
Lol .ngudoots