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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,\(x^5-x^4-x^4+x^3+2x^3-2x^2-2x^2+2\)2x-2x+2\(x^4\left(x-1\right)-x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)+2\left(x-1\right)\)
=\(\left(x^4-x^3+2x^2-2x+2\right)\left(x-1\right)\)
b,
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
x6+3x4y2-8x3y3+3x2y4+y6= x6+3x4y2+3x2y4+y6-8x3y3=(x2+y2)3-(2xy)3
= (x2+y2-2xy)[(x2+y2)2+2xy(x2+y2)+(2xy)2]= (x-y)2(x4+6x2y2+y4+2x3y+2xy3)
(x2+y2-5)2-4x2y2-16xy-16=(x2+y2-5)2-(4x2y2+16xy+16)=(x2+y2-5)2-(2xy+4)2
=(x2+y2-5+2xy+4)(x2+y2-5-2xy-4)=(x2+2xy+y2-1)(x2-2xy+y2-9)=[(x+y)2-1][(x-y)2-32]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)
x4+324=x4+36x2+324-36x2=(x2+18)2-(6x)2=(x2+18-6x)(x2+18+6x)
tớ lm câu a thui nha , tại khó quá ^^
a/ \(=3x^6+3x^5+6x^4+3x^3+3x^2-7x^5-7x^4-14x^3-7x^2-7x+3x^4+3x^3+6x^2+3x+1\)
\(=3x^2\left(x^4+x^3+2x^2+x+1\right)-7x\left(x^4+x^3+2x^2+x+1\right)+3\left(x^4+x^3+2x^2+x+1\right)\)
\(=\left(3x^2-7x+3\right)\left(x^4+x^3+x^2+x^2+x+1\right)\)
\(=\left(3x^2-7x+3\right)\left[x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]\)
\(=\left(3x^2-7x+3\right)\left(x^2+1\right)\left(x^2+x+1\right)\)
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)