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 !!!

nhiều thế

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!!!

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!!!

Chia nhỏ ra cậu ơi :v

Cậu đặt câu hỏi free nên đặt nhỏ ra thì mới có người làm nha để như này dày cộp không ai dám làm đou =(((

cảm ơn nhé

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

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

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

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

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: \(x^8+3x^4+4\)

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

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

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

`1)x^3-7x+6`

`=x^3-x-6x+6`

`=x(x-1)(x+1)-6(x-1)`

`=(x-1)(x^2+x-6)`

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

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

`=(x-1)(x-2)(x+3)`

`2)x^3-9x^2+6x+16`

`=x^3-2x^2-7x^2+14x-8x+16`

`=x^2(x-2)-7x(x-2)-8(x-2)`

`=(x-2)(x^2-7x-8)`

`=(x-2)(x^2-8x+x-8)`

`=(x-2)[x(x-8)+x-8]`

`=(x-2)(x-8)(x+1)`

`3)x^3-6x^2-x+30`

`=x^3+2x^2-8x^2-16x+15x+30`

`=x^2(x+2)-8x(x+2)+15(x+2)`

`=(x+2)(x^2-8x+15)`

`=(x+2)(x^2-3x-5x+15)`

`=(x+2)[x(x-3)-5(x-3)]`

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

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

`=2x^3+x^2-2x^2-x+6x+3`

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

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

`5)27x^3-27x^2+18x-4`

`=27x^3-9x^2-18x^2+6x+12x-4`

`=9x^2(3x-1)-6x(3x-1)+4(3x-1)`

`=(3x-1)(9x^2-6x+4)`

1) Ta có: \(x^3-7x+6\)

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

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

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

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

2) Ta có: \(x^3-9x^2+6x+16\)

\(=x^3-2x^2-7x^2+14x-8x+16\)

\(=x^2\left(x-2\right)-7x\left(x-2\right)-8\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2-7x-8\right)\)

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

3) Ta có: \(x^3-6x^2-x+30\)

\(=x^3+2x^2-8x^2-16x+15x+30\)

\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-8x+15\right)\)

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

bạn hỏi từng câu 1 lần thôi cũng đc hỏi 1 lần 17 câu thì thánh nào vô kiên nhẫn trả lời hết đc ^^

hoa mắt quá

6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)

\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)

7: \(\dfrac{2}{3}x^2y\cdot\left(3xy-x^2+y\right)\)

\(=2x^3y^2-\dfrac{2}{3}x^4y+\dfrac{2}{3}x^2y^2\)

8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)

\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)

9: \(2x^2\left(x^2+3x+\dfrac{1}{2}\right)=2x^4+6x^3+x^2\)

10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)

\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)

11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)

12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)

13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)

Bài 2:

a) \(3x^2-7x-10=\left(x+1\right)\left(3x-10\right)\)

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

c) \(x^2-2xy+y^2-5x+5y=\left(x-y\right)^2-5\left(x-y\right)=\left(x-y\right)\left(x-y-5\right)\)

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

e) \(1-2a+2bc+a^2-b^2-c^2=\left(a-1\right)^2-\left(b-c\right)^2=\left(a-1-b+c\right)\left(a-1+b-c\right)\)

f) \(x^3-3x^2-4x+12=\left(x+2\right)\left(x-3\right)\left(x-2\right)\)

g) \(x^4+64=\left(x^2+8\right)^2-16x^2=\left(x^2+8-4x\right)\left(x^2+6+4x\right)\)h) \(x^4-5x^2+4=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x-2\right)\)

i) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+16=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+16=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+16=\left(x^2+8x+11\right)^2\)

a: \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=\left(x+1\right)\left(3x-10\right)\)

b: \(x^2+6x+9-4y^2\)

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

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

c: \(x^2-2xy+y^2-5x+5y\)

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

\(=\left(x-y\right)\left(x-y-5\right)\)