-Đặt \(t=\left(x^2-x+1\right)\)

\(\left(x^2-x+1\right)^2-5x\left(x^2-x+1\right)+4x^2\)

\(=t^2-5xt+4x^2\)

\(=t^2-4xt-xt+4x^2\)

\(=t\left(t-4x\right)-x\left(t-4x\right)\)

\(=\left(t-4x\right)\left(t-x\right)\)

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

\(=\left(x^2-5x+1\right)\left(x^2-2x +1\right)\)

\(=\left(x^2-5x+1\right)\left(x-1\right)^2\)

\(\left(x^2+x+1\right)\left(x^2+x+5\right)-21=x^4+x^3+5x^2+x^3+x^2+5x+x^2+x+5-21=x^4+2x^3+7x^2+6x-16=\left(x-1\right)\left(x+2\right)\left(x^2+x+8\right)\)

\(=\left(x^2+x+1\right)\left(x^2+x+1+4\right)-21\)

\(=\left(x^2+x+1\right)^2+4\left(x^2+x+1\right)-21\)

\(=\left(x^2+x+1\right)^2-3\left(x^2+x+1\right)+7\left(x^2+x+1\right)-21\)

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

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

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

\(\left(x^2+x+1\right)\left(x^2+x+5\right)-21\)

\(=\left(x^2+x\right)^2+6\left(x^2+x\right)+5-21\)

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

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

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

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

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

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

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

Là nhân tử rồi bn ơi

\(1,\\ 12x^6y^3:4x^3y=3x^3y^2\\ \left(x+1\right)\left(x^2-x+1\right)=x^3+1\\ 2x^2y\left(x^2+3xy\right)=3x^4y+6x^3y^2\\ 2,\\ a,=2xy\left(2x+3y-4\right)\\ b,=\left(x-3\right)\left(x+y\right)\\ c,=\left(x-2\right)\left(x+2\right)+y\left(x-2\right)=\left(x+y+2\right)\left(x-2\right)\\ d,=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\\ 3,\\ a,\Leftrightarrow x^2-x^2+2x=2\\ \Leftrightarrow2x=2\Leftrightarrow x=1\\ b,\Leftrightarrow\left(x-2\right)\left(x-2+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) Áp dụng HĐT 1 thu được ( 2 x   +   y ) 2 .

b) Áp dụng HĐT 3 với A = 2x + l; B = x - l thu được

[(2x +1) + (x -1)] [(2x +1) - (x -1)] rút gọn thành 3x(x + 2).

c) Ta có: 9 - 6x +  x 2  -  y 2 = ( 3   -   x ) 2  -  y 2  = (3 - x - y)(3 -x + y).

d) Ta có: -(x + 2) + 3( x 2  - 4) = -{x + 2) + 3(x + 2)(x - 2)

= (x + 2) [-1 + 3(x - 2)] = (x + 2)(3x - 7).

\(\Leftrightarrow x^3-2x^2+x^2-2x+x-2\)

\(\Leftrightarrow x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)

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

`x^3-x^2-x-2`

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

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

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

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

`@` `\text {Ans}`

`\downarrow`

`x^3 - x^2 - x - 2`

`= x^3 - 2x^2 + x^2 - 2x + x - 2`

`= (x^3 - 2x^2) + (x^2 - 2x) + (x-2)`

`= x^2(x - 2) + x(x - 2) + (x-2)`

`= (x^2 + x + 1)(x-2)`