4x² - 4y²

= (2x)² - (2y)²

= (2x - 2y)(2x + 2y)

(2x - 2y)(2x + 2y)

4x²-4y²=4.(x²-y²)=4.(x-y).(x+y)

-(y-2*x-1)*(y+2*x+1)

Phân tích thành nhân tử

\(4x^2-y^2+4x+1\)

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

\(=\left(2x+1-y\right).\left(2x+1+y\right)\)

= (4x² + 4x + 1) - y²

= (2x +1)² - y²

= (2x + 1 - y)(2x + 1 + y)

\(=4x\left(x+2y\right)+3\left(x+2y\right)=\left(x+2y\right)\left(4x+3\right)\)

\(4x^2+8xy+3x+6y\)

\(=4\left(x+2y\right)+3\left(x+2y\right)\)

\(=7\left(x+2y\right)\)

=4x(x+2y)+3(x+2y)=(x+2y)(4x+3)

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

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

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

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

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

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

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

d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)

a) Ta có: \(x-2y+x^2-4y^2\)

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

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

b) Ta có: \(x^2+2xy+y^2-4x^2y^2\)

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

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

c) Ta có: \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

d) Ta có: \(x^3+3x^2+3x+1-8y^3\)

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

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

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

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

b, \(x^2-4x^2y^2+y^2+2xy=\left(x+y\right)^2-\left(2xy\right)^2\)

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

c, \(x^6-x^4+2x^3+2x^2=x^6+2x^3+1-x^4+2x^2-1\)

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

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

d, \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-\left(2y\right)^3=\left(x+1-2y\right)\left(x+1+2y\right)\)

a) Ta có: \(x-2y+x^2-4y^2\)

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

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

b: Ta có: \(x^2-4x^2y^2+y^2+2xy\)

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

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

ghytyyj

Trả lời:

7, 49y² - x² + 6x - 9

= 49y² - ( x² - 6x + 9 )

 = ( 7y )² - ( x - 3 )²

= ( 7y - x + 3 ) ( 7y - x - 3 )

8, sửa đề: 25x² - 4y² - 4y - 1

= 25x² - ( 4y² + 4y + 1 )

= ( 5x )² - ( 2y + 1 )

= ( 5x - 2y - 1 ) ( 5x + 2y + 1 )

9, 4x² - y² + 8y - 16

= 4x² - ( y² - 8y + 16 )

= ( 2x )² - ( y - 4 )²

= ( 2x - y + 4 ) ( 2x + y - 4 )

a, \(49y^2-x^2+6x-9=49y^2-\left(x-3\right)^2=\left(7y-x+3\right)\left(7y+x-3\right)\)

b, đề sai rồi bạn 

c, \(4x^2-y^2+8y-16=4x^2-\left(y-4\right)^2=\left(2x-y+4\right)\left(2x+y-4\right)\)

(2x +y-4)