a) 2x^2 -  2xy - 5x +5y

= (2x^2 - 2xy ) - ( 5x- 5 y)

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

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

b)8x²  +4xy-2ax-ay

=(8x² +4xy) -(2ax+ay)

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

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

c)=x(x² -4x +4)

=x(x-2)²

d)=16- (x² -2xy +y^2)

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

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

các câu còn lại tg tự

chúc bn hok tốt

a) x³ - 4x² + 4x

= x(x² - 4x + 4)

= x(x - 2)²

b) 2xy - x² - y² + 16

= 16 -x² + 2xy - y²

= 16 - (x² - 2xy + y²)

= 4² - (x - y)²

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

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

c) x² - y² - 2yz - z²

= x² - (y² + 2yz + z²)

= x² - (y + z)²

= [x -(y + z)].(x + y +z)

=(x - y - z)(x + y + z)

d) 3a² - 6ab + 3b² - 12c²

= 3(a² - 2ab + b² - 4c²)

= 3[(a2 - 2ab + b²) - (2c)²]

= 3[(a - b)² - (2c)²]

= 3(a - b - c)(a - b + c)

con D bạn chép sai đề bài rồi, phải là +3b2 chứ. tích cho mik nha, ko thì lần sau mik ko giúp đâu ihihihi.....!!!!!!!!!

a) \(2x^2-2xy-5x+5y\)

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

\(=\left(5-2x\right)\left(y-x\right).\)

b) \(8x^2+4xy-2ax-ay\)

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

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

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

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

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

d) \(2xy-x^2-y^2+16\)

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

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

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

e) \(x^2-y^2-2yz-z^2\)

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

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

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

g) \(3a^2-6ab+3b^2-12c^2\)

\(=3\left(a^2-2ab+b^2-4c^2\right)\)

\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)

\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)

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

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

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

b)\(3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)\)\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)

\(=3\left(a-b+2c\right)\left(a-b-2c\right)\)

i. Ta có: (x+y)^3 - (x-y)^3=[(x+y)-(x-y)][(x+y)^2 + (x+y)(x-y) + (x-y)^2]

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

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

có hằng đẳng thức (x+y)^3 vs (x-y)^3 thì sao bạn phải làm như vậy nữa

bạn xem mình làm như này có đúng không:

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

= 2 . 3x^2y = 6x^2y

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

\(c,3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)=3.\left(\left(a-b\right)^2-\left(2c\right)^2\right)\)

                                                     \(=3\left(a-b-2c\right).\left(a-b+2c\right)\)

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

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

\(e,a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b\right)\left(a+b-c\right)\)

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

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

\(h,x^2\left(x-1\right)+16\left(1-x\right)=x^2\left(x-1\right)-16\left(x-1\right)=\left(x-1\right)\left(x^2-16\right)=\)

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

\(A=3a^2c^2+bd+3abc+acd=\left(3a^2c^2+3abc\right)+\left(bd+acd\right)=3ac\left(ac+b\right)+d\left(b+ac\right)\\ =\left(3ac+d\right)\left(ac+b\right)\)

\(B=a^2c-a^2d-b^2d+b^2c=a^2\left(c-d\right)-b^2\left(c-d\right)=\left(a^2-b^2\right)\left(c-d\right)\\=\left(a-b\right)\left(a+b\right)\left(c-d\right)\)

\(C=8x^2+4xy-2ax-ay=\left(8x^2+4xy\right)-\left(2ax+ay\right)=4x\left(2x+y\right)-a\left(2x+y\right)\\ =\left(4x-a\right)\left(2x+y\right)\)

\(E=3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2\right)-12c^2=3\left(a-b\right)^2-12c^2\\ =3\left[\left(a-b\right)^2-4c^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)