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

6) \(\left(x+y\right)^2-\left(x+y\right)-12\)\(=\left(x+y\right)^2-2\cdot\frac{1}{2}\left(x+y\right)+\frac{1}{4}-\frac{49}{4}\)

\(=\left(x+y-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)\(=\left(x+y-\frac{1}{2}-\frac{7}{2}\right)\left(x+y-\frac{1}{2}+\frac{7}{2}\right)\)

\(=\left(x-4\right)\left(x+3\right)\)

7)   \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)          (NHÂN x + 2 vs x +  5  và  x + 3 vs x + 4 )

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

ĐẶT   \(x^2+7x+11=y\)   ta được :  

\(\left(y+1\right)\left(y-1\right)-24=y^2-1-24\)

\(=y^2-25=\left(y-5\right)\left(y+5\right)\)

8)  \(4x^4-32x^2+1=4x^4+4x^2+1-36x^2\)

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

9) sai đề rùi bạn ơi ! đề đúng nè 

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

Ta thấy :  

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

Thay vào biểu thức bài cho ta được : 

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

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

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

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

bài ở trên câu 3 : kết luận là  \(\left(x-3\right)\left(x^2-x-6\right)\)bạn sửa lại giúp mk nka !!! Th@nk !!! Tk Mk vs  

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!!!

7, \(\left(x+2\right).\left(x+3\right).\left(x+4\right).\left(x+5\right)-24\)

\(=\left[\left(x+2\right).\left(x+5\right)\right].\left[\left(x+3\right).\left(x+4\right)\right]-24\)

\(=\left(x^2+5x+2x+10\right).\left(x^2+4x+3x+12\right)-24\)

\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)(1)

Đặt \(t=x^2+7x+10\Rightarrow t+2=x^2+7x+12\)

\(\Rightarrow\left(1\right)=t.\left(t+2\right)-24\)

\(=t^2+2t-24=t^2-4t+6t-24\)

\(=\left(t^2-4t\right)+\left(6t-24\right)=t.\left(t-4\right)+6.\left(t-4\right)\)

\(=\left(t-4\right).\left(t+6\right)\) (2)

Vì \(t=x^2+7x+10\) nên:

(2) \(=\left(x^2+7x+10-4\right).\left(x^2+7x+10+6\right)\)

\(=\left(x^2+7x+6\right).\left(x^2+7x+16\right)\)

\(=\left(x^2+x+6x+6\right).\left(x^2+7x+16\right)\)

\(=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)

\(=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)

\(=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)

Chúc bạn học tốt!!!

`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: \(x^3-6x^2-x+30\)

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

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

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

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

4) Ta có: \(2x^3-x^2+5x+3\)

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

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

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

5) Ta có: \(27x^3-27x^2+18x-4\)

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

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

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

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

b: \(27x^3+1=\left(3x+1\right)\left(9x^2-3x+1\right)\)

c: \(64x^3-27y^3=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)

a: x^3+8=(x+2)(x^2-2x+4)

b: =(3x+1)(9x^2-3x+1)

c: =(x+3)(x^2-3x+9)

d: =(4x-3y)(16x^2+24xy+9y^2)

\(a.x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\) 

\(b.27x^3+1=\left(3x+1\right)\left(9x-3x+1\right)\)

\(c.x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\) 

\(d.64x^3-27y^3=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)  

\(a,=\left(x-1\right)^3\\ b,=\left(1-2x\right)\left(1+2x\right)\\ c,=x^3-8\\ d,=\left(3x-1\right)\left(9x^2+3x+1\right)\\ e,=\left(x+2\right)\left(x^2-2x+4\right)\\ g,=\left(x-2\right)^2\\ h,=x^2-4y^2\\ j,=\left(x-4\right)^2\)

1, x³-9x²y+27xy²-27y³=(x-3y)³

2, 27x³-9x²y+xy²-\(\dfrac{1}{27}\)y³=(3x-\(\dfrac{1}{3}\)y)³

3)x⁶-3x⁴y+3xy²-y³=(x²-y)³

1) \(x^3-9x^2y+27xy^2-27y^3=\left(x-3y\right)^3\)

2) \(27x^3-9x^2y+xy^2-\dfrac{1}{27}y^3=\left(3x-\dfrac{1}{3}y\right)^3\)

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

\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)

a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)

2)
a)
\(=x^3+125\)
\(\)b)\(=1-x^3\)
c)
=\(y^3+27t^3\)