Đa thức này không phân tích được thành nhân tử bạn nhé. 

\(x^4+4\)

= \(\left(x^2+2\right)^2-4x^2\)

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

`x^4+x^2 y^2+y^4`

`=x^4+2x^2 y^2 +y^4-x^2 y^2`

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

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

Bài 1 yêu cầu gì em?

Bài 2:

\(a,x\left(x-1\right)+5\left(x-1\right)=\left(x+5\right)\left(x-1\right)\\ b,3x\left(x+1\right)+3\left(x+1\right)=\left(3x+3\right)\left(x+1\right)=3\left(x+1\right)\left(x+1\right)=3\left(x+1\right)^2\\ c,x\left(x-3\right)+xy\left(x-3\right)=\left(x+xy\right)\left(x-3\right)=x\left(y+1\right)\left(x-3\right)\\ d,2x\left(x-2\right)-6\left(x-2\right)=\left(2x-6\right)\left(x-2\right)=2\left(x-3\right)\left(x-2\right)\)

Bài 1:

a) \(3xy+6y\)

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

b) \(3x^2+9x\)

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

c) \(6x-9y^2\)

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

d) \(10xy^2-6x^2y\)

\(=2xy\left(5y-3x\right)\)

Bài 2:

a) \(x\left(x-1\right)+5\left(x-1\right)\)

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

b) \(3x\left(x+1\right)+3\left(x+1\right)\)

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

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

\(=3\left(x+1\right)^2\) 

c) \(x\left(x-3\right)+xy\left(x-3\right)\)

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

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

d) \(2x\left(x-2\right)-6\left(x-2\right)\)

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

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

Bài `1`

`a,3xy +6y`

`= 3y(x+2)`

`b,3x^2+9x`

`= 3x(x+3)`

`c,6x-9y^2`

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

`d,10xy^2-6x^2y`

`= 2xy(5y-3x)`

x⁴ - 2x³-2x² -2x -3

=(x⁴+x³)-(3x³+3x²)+(x²+x)-(3x+3)

=x³(x+1)-3x²(x+1)+x(x+1)-3(x+1)

= (x³-3x²+x-3)(x+1)

= ((x³-3x²)+(x-3))(x+1)

= (x²(x-3)+(x-3))(x+1)

=(x²+1)(x-3)(x+1)

3 x y 2 – 6 x 2 y = 3 x y (   y   –   2 x )

Lời giải:

Ta có:

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

x 3   –   6 x 2 y   +   12 x y 2   –   8 y 3   =   x 3   –   3 . x 2 . ( 2 y )   +   3 . x . ( 2 y ) 2   –   ( 2 y ) 3     =   ( x   –   2 y ) 3

đáp án cần chọn là: D

a) \(4x^2-9y^2+6x-9y\)

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

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

b) \(1-2x+2yz+x^2-y^2-z^2\)

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

\(=\left(x-1\right)^2-\left(y-z\right)^2\)

\(=\left(x-y+z-1\right)\left(x+y-z-1\right)\)

Tick hộ mình nha 😘