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

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

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

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

a) 5x^2 + 6xy + y^2

=5x²+5xy+xy+y²

=5x.(x+y)+y.(x+y)

=(x+y)(5x+y)

b) x^2 + 2xy - 15y^2.

=x²-3xy+5xy-15y²

=x.(x-3y)+5y.(x-3y)

=(x-3y)(x+5y)

c) (x-y)^2 + 4(x-y) - 12

=(x-y)²+4(x-y)+4-16

=(x-y+2)²-16

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

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

d) x^3 - 2x - 4.

=x³+2x²+2x-2x²-4x-4

=x.(x²+2x+2)-2.(x²+2x+2)

=(x²+2x+2)(x-2)

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

2, \(x^2-25+y^2+2xy=\left(x+y\right)^2-5^2=\left(x+y-5\right)\left(x+y+5\right)\)

3, \(x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)

4, \(x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

5, \(x^4+8x=x\left(x^3+8\right)=x\left(x+8\right)\left(x^2-8x+64\right)\)

\(1,\)

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

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

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

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

\(2,\)

\(x^2-25+y^2+2xy\)

\(=\left(x^2+2xy+y^2\right)-25\)

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

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

\(3,\)

\(x^2y-x^3-9y+9x\)

\(=\left(x^2y-x^3\right)-\left(9y-9x\right)\)

\(=x^2\left(y-x\right)-9\left(y-x\right)\)

\(=\left(x^2-9\right)\left(y-x\right)\)

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

\(4,\)

\(x^4+2x^3+x^2\)

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

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

\(5,\)

\(x^4-8x\)

\(=x\left(x^3-8\right)\)

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

a) x² - y² - 4x + 4y

= (x² - 4x + 4) - (y² - 4y + 4)

= (x - 2)² - (y - 2)²

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

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

b) (xy + 4)² - 4(x + y)²

= (xy + 4)² - [2(x + y)]²

= (xy + 4)² - (2x + 2y)²

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

c) 25 - x² + 2xy - y²

= 25 - (x² - 2xy + y²)

= 5² - (x - y)²

=> (5 - x + y)(5 + x - y)

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

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

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

c) \(25-x^2+2xy-y^2=25-\left(x^2-2xy+y^2\right)=5^2-\left(x-y\right)^2=\left(5+x-y\right)\left(5-x+y\right)\)

x² - y² - 4x + 4y 

= ( x² - y² ) - ( 4x - 4y )

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

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

( xy + 4 )² - 4( x + y )²

= ( xy + 4 )² - 2²( x + y )²

= ( xy + 4 )² - [ 2( x + y ) ]²

= ( xy + 4 )² - ( 2x + 2y )²

= [ ( xy + 4 ) - ( 2x + 2y ) ][ ( xy + 4 ) + ( 2x + 2y )

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

25 - x² + 2xy - y²

= 25 - ( x² - 2xy + y² )

= 5² - ( x - y )²

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

a) \(x^4+2x^3-4x-4=\left[\left(x^2\right)^2-4\right]+\left(2x^3-4x\right)\)

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

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

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

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

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

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

\(=\left(1-y\right)\left[\left(x-1\right)x+y\left(x-1\right)\left(x+1\right)\right]=\left(1-y\right)\left(x-1\right)\left(x+xy+y\right)\)

c) Không phân tích được.

x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử

= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung

= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung