\(\left(x^2-3x+2\right)\left(x^2-9x+20\right)-40=\left(x-1\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)-40\)

\(=\left(x^2-6x+5\right)\left(x^2-6x+8\right)-40\)
Đặt \(t=x^2-6x+5\) thì ta có \(t\left(t+3\right)-40=t^2+3t-40=\left(t+8\right)\left(t-5\right)\)

Suy ra \(\left(x^2-6x+5\right)\left(x^2-6x+8\right)-40=\left(x^2-6x+13\right)\left(x^2-6x\right)=x\left(x-6\right)\left(x^2-6x+13\right)\) 

x^2-2x+7x-14=x(x-2)+7(x-2)=(x+7).(x-2)

a, x² + 3x - 40

= x² - 8x + 5x - 40

= x ( x - 8 ) + 5 ( x - 8 )

= ( x - 8 ) ( x + 5 )

b, x² + 5x - 14

= x² - 2x + 7x - 14

= x ( x - 2 ) + 7 ( x - 2 )

= ( x - 2 ) ( x + 7 )

x^2-5x+8x-40=x(x-5)+8(x-5)=(x+8).(x-5).....xong rồi đấy

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

`#040911`

`x^2 - 3x - 10`

`= x^2 + 2x - 5x - 10`

`= (x^2 + 2x) - (5x + 10)`

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

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

Kq = ( x - 5 ) ( x + 2 )

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

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

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

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

b. \(=\left(1-x\right)^3\)

    x³ - x² + 3x - 27.

= (x³ - 27) - (x² - 3x).

= (x - 3) (x² + 3x + 9) - x (x - 3).

= (x - 3) (x² + 3x + 9 - x).

= (x - 3) (x² + 2x + 9).

x³ - x² + 3x - 27 = ( x³ - 27 ) - ( x² - 3x )

= ( x - 3 ) ( x² + 3x + 9 ) - x ( x - 3 ) 

= ( x - 3 ) ( x² + 3x + 9 - x )

= ( x - 3 ) ( x² - 2x + 9 )

* Chúc bạn học tốt

= (x³+3x²+3x+1)-(4y)³

=(x+1)³-(4y)³

=(x+1-4y)[(x+1)²+(x+1).4y+16y² ]

=(x+1-4y)[(x²+2x+1)+(4xy+4y)+16y²]

x²+3x+3y+xy

=x(x+3)+y(3+x)

=(x+3)(x+y)

     x² + 3x + 3y + xy

=   ( x² + xy) + ( 3x + 3y)

= x( x + y) + 3 ( x + y)

= ( x + y) ( x + 3) 

=(x+y)(x+3)

2(x+3)-x³-3x

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

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

Đa thức này ko phân tích được nha bạn