Question

Phân tích đa thức thành nhân tử :

a 

Answer 1

7a/

$x^3y+x-y-1=(x^3y-y)+(x-1)=y(x^3-1)+(x-1)$

$=y(x-1)(x^2+x+1)+(x-1)=(x-1)[y(x^2+x+1)+1]$

$=(x-1)(x^2y+xy+y+1)$

7b/

$x^2(x-2)+4(2-x)=x^2(x-2)-4(x-2)=(x-2)(x^2-4)$

$=(x-2)(x-2)(x+2)=(x-2)^2(x+2)$
7c/

$x^3-x^2-20x=x(x^2-x-20)=x[(x^2+4x)-(5x+20)]$

$x[x(x+4)-5(x+4)]=x(x+4)(x-5)$

 

Answer 2

7d/

$(x^2+1)^2-(x+1)^2=[(x^2+1)-(x+1)][(x^2+1)+(x+1)]$

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

$=x(x-1)(x^2+x+2)$

7e/

$6x^2-7x+2=(6x^2-3x)-(4x-2)=3x(2x-1)-2(2x-1)=(2x-1)(3x-2)$

7f/

$x^4+8x^2+12=(x^4+6x^2)+(2x^2+12)=x^2(x^2+6)+2(x^2+6)$

$=(x^2+6)(x^2+2)$

7g/

$(x^3+x+1)(x^3+x)-2=(t+1)t-2$ (đặt $x^3+x=t$)

$=t^2+t-2=(t^2+2t)-(t+2)=t(t+2)-(t+2)$

$=(t+2)(t-1)=(x^3+x+2)(x^3+x-1)$

$=[(x^3+x^2)-(x^2+x)+(2x+2)](x^3+x-1)$

$=[x^2(x+1)-x(x+1)+2(x+1)](x^3+x-1)$

$=(x+1)(x^2-x+2)(x^3+x-1)$

Answer 3

7h.

$(x+1)(x+2)(x+3)(x+4)+1$

$=[(x+1)(x+4)][(x+2)(x+3)]+1=(x^2+5x+4)(x^2+5x+6)+1$

$=t(t+2)+1$ (đặt $x^2+5x+4=t$)

$=t^2+2t+1=(t+1)^2=(x^2+5x+5)^2$
7i/

Đặt $x^2+2=a$ thì:

$-(x^2+2)^2+4x(x^2+2)-3x^2=-a^2+4xa-3x^2$

$=-(a^2-ax)+(3ax-3x^2)$

$=-a(a-x)+3x(a-x)=(a-x)(3x-a)$

$=(x^2+2-x)(3x-x^2-2)$

$=-(x^2-x+2)(x^2-3x+2)=-(x^2-x+2)(x-1)(x-2)$

 

Answer 4

7k/

$81x^4+4y^4=(9x^2)^2+(2y^2)^2$

$=(9x^2)^2+(2y^2)^2+2.9x^2.2y^2-36x^2y^2$

$=(9x^2+2y^2)^2-(6xy)^2=(9x^2+2y^2-6xy)(9x^2+2y^2+6xy)$