\(x^{4}+2018x^{2}+2017x+2018\)
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\(x^4+2018x^2+2017x+2018\)
\(\Rightarrow x^4+2018x^2+2018x-x+2018\)
\(\Rightarrow\left(x^4-x\right)+\left(2018x^2+2018x+2018\right)\)
\(\Rightarrow x\left(x^3-1\right)+2018\left(x^2+x+1\right)\)
\(\Rightarrow x\left(x-1\right)\left(x^2+x+1\right)+2018\left(x^2+x+1\right)\)
\(\Rightarrow\left(x^2+x+1\right)\left(x^2-x+2018\right)\)
\(\text{a) }4x^{16}+81=4x^4+36x^2+81-36x^8\)
\(=\left(4x^{16}+36x^8+81\right)-36x^8\)
\(=\left[\left(2x^8\right)^2+2.2x^8.9+9^2\right]+\left(6x^4\right)^2\)
\(=\left(2x^8+9\right)^2-\left(6x^4\right)^2\)
\(=\left(2x^8+9-6x^4\right)\left(2x^8+9+6x^4\right)\)
\(\text{b) }x^4+2018x^2+2017x+2018\)
\(=x^4+2018x^2+2018x-x+2018\)
\(=\left(x^4-x\right)+\left(2018x^2+2018x+2018\right)\)
\(=x\left(x^3-1\right)-2018\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2018\left(x^2+x+1\right)\)
\(=\left(x^2-x\right)\left(x^2+x+1\right)+2018\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2018\right)\)
x4+2018x2+2017x+2018=x4+2018x2+2018x-x+2018
=x(x3-1)+2018(x2+x+1)=(x2+x+1)(x2-x+2018)
Ktra xem mk có nhầm chỗ nào ko nhé. Cảm ơn bạn
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)
a)x2+xy-2y2=x2-xy+2xy-2y2
=x(x+2y)-y(x+2y)=(x+2y)(x-y)
câu b sai đề nha
a) x2 + xy - 2y2
=x2 + xy - y2 - y2
=(x2 - y2)+(xy-y2)
=(x-y)(x+y)+y(x-y)
=(x-y)(2y+x)
a) \(3x^2+8x-11\)
\(=3x^2-3+11x-11\)
\(=\left(3x^2-3x\right)+\left(11x-11\right)\)
\(=3x\left(x-1\right)+11\left(x-1\right)\)
\(=\left(x-1\right)\left(3x+11\right)\)
b) \(x^4+2018x^2-2017x+2018\)
\(=\left(x^4+x\right)+\left(2018x^2-2018x+2018\right)\)
\(=x\left(x^3+1\right)+2018\left(x^2-x+1\right)\)
\(=x\left(x+1\right)\left(x^2-x+1\right)+2018\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left[x\left(x+1\right)+2018\right]\)
\(=\left(x^2-x+1\right)\left(x^2+x+2018\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+2018\right)\)
a) 3x2 + 8x - 11
=3x2+11x-3x-11
=x(3x+11)-(3x+11)
= (x-1)(3x+11)
TH1: x<-2018 PT (1) <=>
-(x+2017)+2017x-(x+2018)+2018x=4040x
<=> -x-x+2017x+2018x-4040x=2017+2018
<=>-7x=4035
<=>x=(-4035/7)(loại).
TH2: -2018≤ x <-2017 PT (1) <=>
-(x+2017)+2017x+(x+2018)+2018x=4040x
<=> -x+x+2017x+2018x-4040x=2017-2018
<=>-5x=-1
<=>x=(1/5)(loại).
TH3: 2017≤x PT (1) <=>
(x+2017)+2017x+(x+2018)+2018x=4040x
<=> x+x+2017x+2018x-4040x=-2017-2018
<=>-3x=-4035
<=>x=1345
vậy PT (1) có một nghiệm duy nhất x=1345
x^4+2018x^2−2017x+2018
=(x^4+x)+(2018x^2−2018x+2018)
=x(x^3+1)+2018(x^2−x+1)
=x(x+1)(x^2−x+1)+2018(x^2−x+1)
=(x^2−x+1)[x(x+1)+2018]
=(x^2−x+1)(x^2+x+2018)
=(x^2−x+1)(x^2+x+2018)