Lời giải :

1,

`3a^3-27a`

`=3a(a^2-9)`

`=3a(a-3)(a+3)`

2,

`3xy+6y+2z+xz`

`=3y(x+2)+z(2+x)`

`=(3y+z)(x+2)`

3,

`2x^2-4yz+2xz-4xy`

`=(2x^2+2xz)-(4yz+4xy)`

`=2x(x+z)-4y(z+x)`

`=(2x-4y)(x+z)`

`=2(x-2y)(x+z)`

4,

`x^2+y-xy-x`

`=(x^2-xy)+(y-x)`

`=x(x-y)-(x-y)`

`=(x-1)(x-y)`

5,

`16x^2+4xy-x^2-4y^2`

`=16x^2-(x^2-4xy+4y^2)`

`=(4x)^2-(x-2y)^2`

`=[4x-(x-2y)][4x+(x-2y)]`

`=(4x-x+2y)(4x+x-2y)`

`=(3x+2y)(5x-2y)`

6,

`a^4-4a^3-a^2+4a`

`=a^3(a-4)-a(a-4)`

`=(a^3-a)(a-4)`

`=a(a^2-1)(a-4)`

`=a(a-1)(a+1)(a-4)`

7,

`16x^4-x^2+4xy-4y^2`

`=16x^4-(x^2-4xy+4y^2)`

`=(4x^2)^2-(x-2y)^2`

`=[4x^2+(x-2y)][4x^2-(x-2y)]`

`=(4x^2+x-2y)(4x^2-x+2y)`

8,

`x^2-y^2+2x+1`

`=(x^2+2x+1)-y^2`

`=(x+1)^2-y^2`

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

9,

`x^3-5x^2-x+5`

`=x^2(x-5)-(x-5)`

`=(x^2-1)(x-5)`

`=(x-1)(x+1)(x-5)`

10,

`(x+1)(x+2)(x+3)(x+4)-3`         (1)

`=[(x+1)(x+4)][(x+2)(x+3)]-3`

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

`=(x^2+5x+4)(x^2+5x+6)-3`

Đặt `x^2+5x+4=t`

Phương trình (1) trở thành :

`t(t+2)-3`

`=t^2+2t-3`

`=t^2-t+3t-3`

`=t(t-1)+3(t-1)`

`=(t+3)(t-1)`

Trả lại biến cũ ta được :

`(x^2+5x+4+3)(x^2+5x+4-1)`

`=(x^2+5x+7)(x^2+5x+3)`

Hướng dẫn trả lời:

Bài 1:

1) `3a^3 - 27a`

`= 3acdota^2 + 3acdot(- 9)`

`= 3acdot(a^2 - 9)`

`= 3acdot(a^2 - 3^2)`

`= 3acdot(a + 3)cdot(a - 3)`

Giải thích:

Áp dụng HĐT `A^2 - B^2 = (A + B)cdot(A - B)`

2) `3xy + 6y + 2z + xz`

`= (3xy + 6y) + (xz + 2z)`

`= (3ycdotx + 3ycdot2) + (zcdotx + zcdot2)`

`= 3ycdot(x + 2) + zcdot(x + 2)`

`= (3y + z)cdot(x + 2)`

3) `2x^2 - 4yz + 2xz - 4xy`

`= (2xcdotx - 2xcdot2y) + (2zcdotx - 2zcdot2y)`

`= 2xcdot(x - 2y) + 2zcdot(x - 2y)`

`= (2x + 2z)cdot(x - 2y)`

`= 2cdot(x + z)cdot(x - 2y)`

4) `x^2 + y - xy - x`

`= (x^2 - xy) - (x - y)`

`= xcdot(x - y) - 1cdot(x - y)`

`= (x - 1)cdot(x - y)`

5) `16x^2 + 4xy - x^2 - 4y^2`

`= 16x^2 - (x^2 - 4xy + 4y^2)`

`= 16x^2 - [x^2 - 2cdotxcdot2y + (2y)^2]`

`= (4x)^2 - (x - 2y)^2`

`= [4x + (x - 2y)]cdot[4x - (x - 2y)]`

`= (4x + x - 2y)cdot(4x - x + 2y)`

`= (5x - 2y)cdot(3x + 2y)`

Giải thích:

Áp dụng HĐT `(A - B)^2 = A^2 - 2AB + B^2` và `A^2 - B^2 = (A + B)cdot(A - B)`

6) `a^4 - 4a^3 - a^2 + 4a`

`= (a^4 - a^2) - (4a^3 - 4a)`

`= a^2cdot(a^2 - 1) - 4acdot(a^2 - 1)`

`= (a^2 - 4a)cdot(a^2 - 1)`

`= acdot(a - 4)cdot(a^2 - 1^2)`

`= acdot(a - 4)cdot(a + 1)cdot(a - 1)`

Giải thích:

Áp dụng HĐT `A^2 - B^2 = (A + B)cdot(A - B)`

7) `16x^4 - x^2 + 4xy - 4y^2`

`= 16x^4 - (x^2 - 4xy + 4y^2)`

`= 16x^4 - [x^2 - 2cdotxcdot2y + (2y)^2]`

`= (4x^2)^2 - (x - 2y)^2`

`= [4x + (x - 2y)]cdot[4x - (x - 2y)]`

`= (4x^2 + x - 2y)cdot(4x^2 - x + 2y)`

Giải thích:

Áp dụng HĐT `(A - B)^2 = A^2 - 2AB + B^2` và `A^2 - B^2 = (A + B)cdot(A - B)`

8) `x^2 - y^2 + 2x + 1`

`= (x^2 + 2x + 1) - y^2`

`= (x^2 + 2cdotxcdot1 + 1^2) - y^2`

`= (x + 1)^2 - y^2`

`= [(x + 1) + y]cdot[(x + 1) - y]`

`= (x + y + 1)cdot(x - y + 1)`

Giải thích:

Áp dụng HĐT `(A + B)^2 = A^2 + 2AB + B^2` và `A^2 - B^2 = (A + B)cdot(A - B)`

9) `x^3 - 5x^2 - x + 5`

`= (x^3 - 5x^2) - (x - 5)`

`= x^2cdot(x - 5) - 1cdot(x - 5)`

`= (x^2 - 1)cdot(x - 5)`

`= (x^2 - 1^2)cdot(x - 5)`

`= (x + 1)cdot(x - 1)cdot(x - 5)`

Giải thích:

Áp dụng HĐT `A^2 - B^2 = (A + B)cdot(A - B)`

10) `(x + 1)cdot(x + 2)cdot(x + 3)cdot(x + 4) - 3`

`= [(x + 1)cdot(x + 4)]cdot[(x + 2)cdot(x + 3)] - 3`

`= [xcdot(x + 4) + 1cdot(x + 4)]cdot[xcdot(x + 3) + 2cdot(x + 3)] - 3`

`= (x^2 + 4x + x + 4)cdot(x^2 + 3x + 2x + 6) - 3`

`= (x^2 + 5x + 4)cdot(x^2 + 5x + 6) - 3` (1)

Đặt a = `x^2 + 5x + 5`. Từ (1) suy ra: `(a - 1)cdot(a + 1) - 3`

`= a^2 - 1^2 - 3`

`= a^2 - 1 - 3`

`= a^2 - 4`

`= a^2 - 2^2`

`= (a + 2)cdot(a - 2)`

`→ [(x^2 + 5x + 5) + 2]cdot[(x^2 + 5x + 5) - 2]`

`= (x^2 + 5x + 5 + 2)cdot(x^2 + 5x + 5 - 2)`

`= (x^2 + 5x + 7)cdot(x^2 + 5x + 3)`

Giải thích:

Áp dụng HĐT `A^2 - B^2 = (A + B)cdot(A - B)`