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17 tháng 3 2020

a) \(x^8+3x^4+4=x^8+4x^4-x^4+4\)

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

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

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

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

11) Ta có: \(a^6+a^4+a^2b^2+b^4-b^6\)

\(=a^6-b^6+a^4+a^2b^2+b^4\)

\(=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)+\left(a^4+a^2b^2+b^4\right)\)

\(=\left(a^4+a^2b^2+b^4\right)\left(a^2-b^2+1\right)\)

12) Ta có: \(x^3+3xy+y^3-1\)

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

\(=\left[\left(x+y\right)^3-1\right]-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left[x^2+2xy+y^2+x+y+1\right]-3xy\left(x+y-1\right)\)

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

14) Ta có: \(x^8+x+1\)

\(=x^8+x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4+x^3-x^3+x^2-x^2+x+1\)

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

15) Ta có: \(x^8+3x^4+4\)

\(=x^8+4x^4+4-x^4\)

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

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

6 tháng 6 2017

1)  \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)

2)  \(x^3-9x^2+6x+16\)

\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)

\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)

\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)

3)   \(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2-x-1\right)\)

4)  \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)

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

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

5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)

\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)

\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)

\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)

gửi phần này trước còn lại làm sau !!! tk mk nka !!!

5 tháng 6 2017

nhiều thế

15 tháng 6 2017

9) \(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)

=\(3x^4+3x^2+3-\left(x^4+x^2+1+2x^3+2x+2x^2\right)\)

= \(3x^4+3x^2+3-x^4-x^2-1-2x^3-2x-2x^2\)

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

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

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

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

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

= \(\left(x-1\right)^2.2.\left(x^2+x+1\right)\)

10) \(64x^4+y^4\)

= \(\left(8x^2\right)^2+2.8x^2.y^2+\left(y^2\right)^2-16x^2y^2\)

= \(\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)

= \(\left(8x^2+y^2-4xy\right).\left(8x^2+y^2+4xy\right)\)

6 tháng 6 2017

a,\(x^3-7x+6\)

\(=x^3-2x^2+2x^2-4x-3x+6\)

\(=\left(x^3-2x^2\right)+\left(2x^2-4x\right)-\left(3x-6\right)\)

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

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

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

\(=\left(x-2\right).\left[\left(x^2-x\right)+\left(3x-3\right)\right]\)

\(=\left(x-2\right).\left[x.\left(x-1\right)+3.\left(x-1\right)\right]\)

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

b,\(x^3-9x^2+6x+16\)

\(=x^3-8x^2-x^2+8x-2x+16\)

\(=\left(x^3-8x^2\right)-\left(x^2-8x\right)-\left(2x-16\right)\)

\(=x^2.\left(x-8\right)-x.\left(x-8\right)-2.\left(x-8\right)\)

\(=\left(x-8\right).\left(x^2-x-2\right)\)

\(=\left(x-8\right).\left(x^2+x-2x-2\right)\)

\(=\left(x-8\right).\left[\left(x^2+x\right)-\left(2x+2\right)\right]\)

\(=\left(x-8\right).\left[x.\left(x+1\right)-2.\left(x+1\right)\right]\)

\(=\left(x-8\right).\left(x+1\right).\left(x-2\right)\)

c,\(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

\(=\left(x^3-5x^2\right)-\left(x^2-5x\right)-\left(6x-30\right)\)

\(=x^2.\left(x-5\right)-x.\left(x-5\right)-6.\left(x-5\right)\)

\(=\left(x-5\right).\left(x^2-x-6\right)\)

\(=\left(x-5\right).\left(x^2+2x-3x-6\right)\)

\(=\left(x-5\right).\left[\left(x^2+2x\right)-\left(3x+6\right)\right]\)

\(=\left(x-5\right).\left[x.\left(x+2\right)-3.\left(x+2\right)\right]\)

\(=\left(x-5\right).\left(x+2\right).\left(x-3\right)\)

Chúc bạn học tốt!!!

6 tháng 6 2017

d,\(2x^3-x^2+5x+3\)

\(=2x^3+x^2-2x^2-x+6x+3\)

\(=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)

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

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

e, \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=\left(27x^2-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)

\(=9x^2.\left(3x-1\right)-6x.\left(3x-1\right)+4.\left(3x-1\right)\)

\(=\left(3x-1\right).\left(9x^2-6x+4\right)\)

Chúc bạn học tốt!!!

7 tháng 9 2017

a)\(x^3+3xy+y^3-1\)

\(=x^3+3x^2y+3xy^2+y^3-1-3x^2y-3xy^2+3xy\)

\(=\left(x+y\right)^3-1^3-3xy\left(x+y-1\right)\)

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

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

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

b) Đặt \(B=3x^2+22xy+11x+37y+7y^2+10\)

Giả sử \(B=\left(ax+by+c\right)\left(mx+ny+p\right)\)

\(=amx^2+anxy+apx+bmxy+bny^2+bpy+cmx+cny+cp\)

\(=amx^2+\left(an+bm\right)xy+\left(ap+cm\right)x+bny^2+\left(bp+cn\right)y+cp\)

Ta được hệ: \(\left\{{}\begin{matrix}am=3;an+bm=22\\ap+cm=11;bn=7\\bp+cn=37;cp=10\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=3;b=1\\c=5;m=1\\n=7;p=2\end{matrix}\right.\)

Vậy B phân tích được thành \(\left(3x+y+5\right)\left(x+7y+2\right)\).

8 tháng 9 2017

a/ =(x+y)3-1-3xy(x+y-1)

=(x+y-1)(x2+2xy+y2+xy+1)-3xy(x+y-1)

=(x+y-1)(x2+y2+1)

mơn nha

13 tháng 6 2016

a) \(4x^4+4x^3+5x^2+2x+1=\left[\left(2x^2\right)^2+4x^3+x^2\right]+2\left(2x^2+x\right)+1=\left(2x^2+x\right)^2+2\left(2x^2+x\right)+1=\left(2x^2+x+1\right)^2\)

b) \(3x^2+22xy+11x+37y+7y^2+10=\left(3x^2+21xy+6x\right)+\left(7y^2+xy+2y\right)+\left(5x+35y+10\right)\)

\(=3x\left(x+7y+2\right)+y\left(x+7y+2\right)+5\left(x+7y+2\right)\)

\(=\left(3x+y+5\right)\left(x+7y+2\right)\)

c) Không phân tích được.

d) \(x^4-8x+63=\left(x^4+4x^3+9x^2\right)-\left(4x^3+16x^2+36x\right)+\left(7x^2+28x+63\right)\)

\(=x^2\left(x^2+4x+9\right)-4x\left(x^2+4x+9\right)+7\left(x^2+4x+9\right)\)

\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)

13 tháng 6 2016

c) \(x^4-7x^3+14x^2-7x+1=\left(x^4-3x^3+x^2\right)-\left(4x^3-12x^2+4x\right)+\left(x^2-3x+1\right)\)

\(=x^2\left(x^2-3x+1\right)-4x\left(x^2-3x+1\right)+\left(x^2-3x+1\right)\)

\(=\left(x^2-3x+1\right)\left(x^2-4x+1\right)\)

16 tháng 11 2019

a, x^2+2xy+y^2-x-y-12

=(x+y)^2-(x+y+12)

=(x+y)(x+y-1+12 )

=(x+y)(x+y+11)

c, x^4-8x+63

=x(x^3-8)+63

=x[(x-2)(x^2+2x+4)]+63

=x(x-2)(x^2+2x+67)

P/S: câu b dài quá, nhác làm. nhưng k cho mk với :)))

16 tháng 11 2019

Bạn lm sai r nha bạn .nhg cx cảm ơn đã TL câu hỏi của mk