BT phân tích thành nhân tử

1, x^4 +3x^3 +x^2 -12x -20
2, x^3 -10x -12
3, 4x^12 -12x +1
4, x^8 +x +1
5, x^8 +x^7 +1
6, 4.x^4.y^4 +1
7, (x^2+x)^2 + 4x(x+1) -12

 

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

 

1)
$x^4 +3x^3 +x^2 -12x -20$
$= x^4 - 4x^2 + 3x^3 - 12x + 5x^2 - 20$
$= x^2(x^2 - 4) + 3x(x^2 - 4) + 5(x^2 - 4)$
$= (x^2 + 3x + 5)(x^2 - 4)$
$= (x^2 + 3x + 5)(x - 2)(x + 2)$
2)
$x^3 - 10x - 12$
$= x^3 + 2x^2 - 2x^2 - 4x - 6x - 12$
$= x^2(x + 2) - 2x(x + 2) - 6(x + 2)$
$= (x^2 - 2x - 6)(x + 2)$
3)
Đề sai
4)
$x^8 + x + 1$
$= x^8 + x^7 + x^6 - x^7 - x^6 - x^5 + x^5 + x^4 + x^3 - x^4 - x^3 - x^2 + x^2 + x + 1$
$= x^6(x^2 + x + 1) - x^5(x^2 + x + 1) + x^3(x^2 + x + 1) - x^2(x^2 + x + 1) + x^2 + x + 1$
$= (x^6 - x^5 + x^3 - x^2 + 1)(x^2 + x + 1)$
5)
$x^8 + x^7 + 1$
$= x^8 + x^7 + x^6 - x^6 - x^5 - x^4 + x^5 + x^4 + x^3 - x^3 - x^2 - x + x^2 + x + 1$
$= x^6(x^2 + x + 1) - x^4(x^2 + x + 1) + x^3(x^2 + x + 1) - x(x^2 + x + 1) + (x^2 + x + 1)$
$= (x^6 - x^4 + x^3 - x + 1)(x^2 + x + 1)$
6)
$4x^4y^4 + 1$
$= 4x^4y^4 + 4x^2y^2 + 1 - 4x^2y^2$
$= (2x^2y^2)^2 + 2.(2x^2y^2) + 1 - 4x^2y^2$
$= (2x^2y^2 + 1)^2 - (2xy)^2$
$= (2x^2y^2 - 2xy + 1)(2x^2y^2 + 2xy + 1)$
7)
Đặt $t = x^2 + x$, ta có:
$(x^2 + x)^2 + 4x(x + 1) - 12$
$= t^2 + 4t - 12$
$= t^2 - 2t + 6t - 12$
$= (t - 2)t + (t - 2)6$
$= (t - 2)(t + 6)$
$= (x^2 + x - 2)(x^2 + x + 6)$
$= (x^2 -x + 2x - 2)(x^2 + x + 6)$
$= (x(x - 1) + 2(x - 1))(x^2 + x + 6)$
$= (x + 2)(x - 1)(x^2 + x + 6)$

 

1)
x^4+3x^3+x^2−12x−20
=x^4−4x^2+3x^3−12x+5x^2−20
=x^2(x^2−4)+3x(x^2−4)+5(x^2−4)
=(x^2+3x+5)(x^2−4)
=(x^2+3x+5)(x−2)(x+2)
2)
x^3−10x−12
=x^3+2x^2−2x^2−4x−6x−12
=x^2(x+2)−2x(x+2)−6(x+2)
=(x^2−2x−6)(x+2)

4)
x^8+x+1
=x^8+x^7+x^6−x^7−x^6−x^5+x^5+x^4+x^3−x^4 − x^3−x^2+x^2+x+1
=x^6(x^2+x+1)−x^5(x^2+x+1)+x^3(x^2+x+1)−x^2(x^2+x+1)+x^2+x+1
=(x^6−x^5+x^3−x^2+1)(x^2+x+1)
5)
x^8+x^7+1
=x^8+x^7+x^6−x^6−x^5−x^4+x^5+x^4 + x^3−x^3−x^2−x+x^2+x+1
=x^6(x^2+x+1)−x^4(x^2+x+1)+x^3(x^2+x+1)−x(x^2+x+1)+(x^2+x+1)
=(x^6−x^4+x^3−x+1)(x^2+x+1)
6)
4x4y4+1
=4x^4y^4+4x^2y^2+1−4x^2y^2
=(2x2y2)2+2.(2x^2y^2)+1−4x^2y^2
=(2x^2y^2+1)2−(2xy)2
=(2x^2y^2−2xy+1)(2x2y2+2xy+1)
7)
Đặt t=x^2+x, ta có:
(x^2+x)2+4x(x+1)−12
=t2+4t−12
=t2−2t+6t−12
=(t−2)t+(t−2)6
=(t−2)(t+6)
=(x^2+x−2)(x^2+x+6)
=(x^2−x+2x−2)(x^2+x+6)
=[x(x−1)+2(x−1)](x^2+x+6)
=(x+2)(x−1)(x^2+x+6)

 

1. $x^4 +3x^3 +x^2 -12x -20 = (x - 2)(x + 2)(x^2 + 3x + 5)$

2. $x^3 -10x -12 = (x + 2)(x^2 - 2x - 6)$

3. Đề lỗi rồi bạn ơi

4. $x^8 +x +1 = (x^2 + x + 1)(x^6 - x^5 + x^3 - x^2 +1)$

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

6. $4.x^4.y^4 +1 = (2x^2y^2 - 2xy + 1)(2x^2y^2 + 2xy + 1)$

7. $(x^2+x)^2 + 4x(x+1) -12 = (x - 1)(x + 2)(x^2 + x + 6)$