`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: \(27x^3-27x^2+18x-4\)

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

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

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

đề có sai ko ạ?

x^3-2x-4

=x^3-2x-8+4 (Ta thấy - 8 + 4 là bằng -4 nên ta thêm vào thì cũng giống nhau)

=(x^3-8)-(2x-4) (Nhóm hạng tử)

=(x-2)(x^2+2x+4)-2(x-2) \([\)(Hằng đẳng thức 6) và ta thấy -2 là nhân tử chung\(]\)

=(x-2)(x^2+2x+4-x+2)  (Rút gọn)

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

x³ - 2x - 4

= x³ - 2x² + 2x² - 4x + 2x - 4

= (x³ - 2x²) + (2x² - 4x) + (2x - 4)

= x²(x - 2) + 2x(x - 2) + 2(x - 2)

= (x - 2)(x² + 2x + 2)

\(27x^3-a^3b^3\)

\(=\left(3x\right)^3-\left(ab\right)^3\)

\(=\left(3x-ab\right)\left[\left(3x\right)^2+3x\cdot ab+\left(ab\right)^2\right]\)

\(=\left(3x-ab\right)\left(9x^2+3xab+a^2b^2\right)\)

27x³ - a³b³

= (3x)³ - (ab)³

= (3x - ab)(9x² + 3xab + a²b²)

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 =(((

Câu 2: 

a: \(\Leftrightarrow3x^2+2x-1=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

b: \(\Leftrightarrow x^3-4x-x^3-8=4\)

hay x=-3

a) ( 3 x   -   2 y ) 3 .        b) ( x   -   1 ) ( x   +   3 ) 2 .

\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)

\(=\left(x^2+2x+4\right)\left(x^2+2x+5\right)\)

1: \(-x^2+2x+8\)

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

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

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

\(18x^2-36xy+18x^2-72z^2\)

\(=36x^2-36xy-72z^2\)

\(=36\left(x^2-xy-2z^2\right)\)