Theo bài ra , ta có : 

4x² - 4xy + 8y² 

= (2x)² - 2.2xy + y² + 7y² 

= (2x + y)² + 7y² 

Chúc bạn hôc tốt =))

x²-2x+1-y²+2y-1

=(x-1)²-(y²-2y+1)

=(x-1)²-(y-1)²

a.)x^2-y^2-2x+2y

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

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

b)

4.(x-xy+2y^2)

\(a,=\left(3-x+1\right)\left(9+3x-3+x^2-2x+1\right)\\ =\left(4-x\right)\left(x^2+x+7\right)\\ b,=4x^2-4xy-13xy+13y^2\\ =4x\left(x-y\right)-13y\left(x-y\right)\\ =\left(4x-13y\right)\left(x-y\right)\\ c,=4\left(x^2-xy-2y^2\right)\\ =4\left(x^2+xy-2xy-2y^2\right)\\ =4\left(x+y\right)\left(x-2y\right)\\ d,=x^3+4x^2+5x^2+20x+6x+24\\ =\left(x+4\right)\left(x^2+5x+6\right)\\ =\left(x+4\right)\left(x^2+2x+3x+6\right)\\ =\left(x+4\right)\left(x+2\right)\left(x+3\right)\\ f,=x\left(x+4y\right)-3\left(x+4y\right)=\left(x-3\right)\left(x+4y\right)\\ g,=4x^3+4x^2-29x^2-29x-24x-24\\ =\left(x+1\right)\left(4x^2-29x-24\right)\\ =\left(x+1\right)\left(4x^2-32x+3x-24\right)\\ =\left(x+1\right)\left(x-8\right)\left(4x+3\right)\)

\(a,27-\left(x-1\right)^3=\left(3-x+1\right)\left[9+3\left(x-1\right)+\left(x+1\right)^2\right]=\left(4-x\right)\left(9+3x-3+x^2+2x+1\right)=\left(4-x\right)\left(x^2+5x+7\right)\)

\(b,4x^2-17xy+13y^2=\left(4x^2-4xy\right)-\left(13xy-13y^2\right)=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)

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

\(d,x^3+9x^2+26x+24=\left(x^3+2x^2\right)+\left(7x^2+14x\right)+\left(12x+24\right)=\left(x+2\right)\left(x^2+7x+12\right)=\left(x+2\right)\left[\left(x^2+3x\right)+\left(4x+12\right)\right]=\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(f,4xy+x^2-3x-12y=x\left(4y+x\right)-3\left(x+4y\right)=\left(x+4y\right)\left(x-3\right)\)

\(g,4x^3-25x^2-53x-24=\left(4x^3-32x^2\right)+\left(7x^2-56x\right)+\left(3x-24\right)=\left(4x^2+7x+3\right)\left(x-8\right)=\left[\left(4x^2+4x\right)+\left(3x+3\right)\right]=\left(4x+3\right)\left(x+1\right)\left(x-8\right)\)

a) 5x² - 10x = 5x( x - 2 )

b) x² - y² - 2x + 2y = (x² - y²) - (2x - 2y)

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

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

c) 4x² - 4xy - 8y² = (4x² - 4xy + 8y²) - 9y²

                           = (2x - 9y²) - 3y²

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

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

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

a) 5x² - 10x = 5x( x - 2 )

b) x² - y² - 2x + 2y = (x² - y²) - (2x - 2y)

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

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

c) 4x² - 4xy - 8y² = (4x² - 4xy + 8y²) - 9y²

                           = (2x - 9y²) - 3y²

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

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

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

a) 5x² - 10x = 5x( x - 2 )

b) x² - y² - 2x + 2y = (x² - y²) - (2x - 2y)

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

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

c) 4x² - 4xy - 8y² = (4x² - 4xy + 8y²) - 9y²

                           = (2x - 9y²) - 3y²

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

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

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

3: =4x^2+4x+1-2

=(2x+1)^2-2

\(=\left(2x+1-\sqrt{2}\right)\left(2x+1+\sqrt{2}\right)\)

4: =x^2+xy-5xy-5y^2

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

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

5x² + 10xy - 4x - 8y

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

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

=(5x² + 10xy) - (4x + 8y) = 5x.(x + 2y) - 4.(x + 2y) = (5x - 4).(x + 2y)

\(5x^2+10xy-4x-8y=\left(5x^2+10xy\right)-\left(4x+8y\right)=5x\left(x+2y\right)-4\left(x+2y\right)=\left(5x-4\right)\left(x+2y\right)\)

\(4x^2-y^2+8x-16\)

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

4x² - y² + 8y - 16

= 4x² - (y² - 8y + 16)

= (2x)² - (y - 4)²

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

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

Bạn thử xem lại đề câu d nhé.

a) Ta có: \(4x\left(2x-3y\right)-8y\left(3y-2x\right)\)

\(=4x\left(2x-3y\right)+8y\left(2x-3y\right)\)

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

b) Ta có: \(4x^2-4xy+y^2-9z^2\)

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

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

c) Ta có: \(x^2y+yz+xy^2+xz\)

\(=xy\left(x+y\right)+z\left(x+y\right)\)

\(=\left(x+y\right)\left(xy+z\right)\)

\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)