a) \(3x^2-6xy+3y^2-12x^2=3\left(x^2-2xy+y^2\right)-12x^2=3\left(x-y\right)^2-12x^2=3\left[\left(x-y\right)^2-4x^2\right]=3\left(x-y-2x\right)\left(x-y+2x\right)=3\left(-x-y\right)\left(3x-y\right)\)

b)\(3x^2y^2-6x^2y^3+12x^2y^2=3x^2y^2\left(1-2y+4\right)=3x^2y^2\left(5-2y\right)\)

c) \(3x^2-3y^2+12x-12y=3\left(x^2-y^2\right)+12\left(x-y\right)=3\left(x-y\right)\left(x+y+4\right)\)

a: \(3x^2-6xy+3y^2-12x^2\)

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

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

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

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

b: \(3x^2y^2-6x^2y^3+12x^2y^2\)

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

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

c: Ta có: \(3x^2-3y^2+12x-12y\)

\(=3\left(x-y\right)\left(x+y\right)+12\left(x-y\right)\)

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

a) \(5x+10y=5\left(x+2y\right)\)

b) \(3x^3-12x=3x\left(x^2-4\right)=3x\left(x-2\right)\left(x+2\right)\)

c) \(4x^2+9x-4xy-9y=4x\left(x-y\right)+9\left(x-y\right)=\left(x-y\right)\left(4x+9\right)\)

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

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

3x²-3y²-12x+12y

=3(x²-y²)-12(x-y)

=3(x-y)(x+y)-4.3(x-y)

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

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

\(3x^2-3y^2-12x+12y\)

\(=3\left(x^2-y^2\right)-12\left(x-y\right)\)

\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)

\(=\left(x-y\right)\left[3\left(x+y\right)-12\right]\)

\(=\left(x-y\right)\left(3x+3y-12\right)\)

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

a) 10x(x-y) - 6y(y-x)
= 10x(x-y) +6y ( x-y)
=(10x+6y) (x-y)
b) 3x² + 5y - 3xy -5x
= 3x(x-y) + 5(y-x)
= 3x(x-y) -5(x-y)
= (3x-5) ( x-y)
c) 3y^₂ - 3z² +3x² + 6xy
=3(y² - z² + x² + 2xy)
=3[(x² +2xy+y²)-z² ]
=3[(x+y)² - z² ]
=3(x+y-z) (x+y+z)
d) 16x³ + 54y3
=2(8x³ + 27y³ )
=2[(2x)³ + (3y)³ ]
=2(2x+3y) (4x² - 6xy + 9y² )
e) x² - 25 -2xy+y²
=(x²-2xy+y²)-25
=(x-y)² -5²
=(x-y-5) (x-y+5)
f) (mình chưa làm ra )
{mong m.n bổ sung thêm..}

mấy câu trên bạn kia đã trả lời rồi nên mk k làm lại nx

f, x⁵ - 3x⁴ + 3x³ - x²

= x² (x³ - 3x² + 3x -1)

= x² (x - 1)³

Chúc bạn học tốt!

nhìu giữ cha !!!!

a.

$12x^3y-24x^2y^2+12xy^3=12xy(x^2-2xy+y^2)=12xy(x-y)^2$
b.

$x^2-6x+xy-6y=(x^2+xy)-(6x+6y)=x(x+y)-6(x+y)=(x-6)(x+y)$
c.

$2x^2+2xy-x-y=2x(x+y)-(x+y)=(x+y)(2x-1)$

d.

$x^3-3x^2+3x-1=(x-1)^3$

e.

$3x^2-3y^2-12x-12y=(3x^2-3y^2)-(12x+12y)$

$=3(x-y)(x+y)-12(x+y)=(x+y)[3(x-y)-12]=3(x-y)(x-y-4)$

f.

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

g.

$x^2+2x+1=(x+1)^2$

h. Không phân tích được thành nhân tử

i.

$x^2-2x-3=(x^2-3x)+(x-3)=x(x-3)+(x-3)=(x+1)(x-3)$

j.

$x^2+4x-12=(x^2-2x)+(6x-12)=x(x-2)+6(x-2)=(x-2)(x+6)$

k.

$x^2-8x-9=(x^2+x)-(9x+9)=x(x+1)-9(x+1)=(x+1)(x-9)$

l.

$x^2+x-6=(x^2+3x)-(2x+6)=x(x+3)-2(x+3)=(x-2)(x+3)$

a: \(x^2+3y^2-4x+6y+7=0\)

\(\Leftrightarrow x^2-4x+4+3y^2+6y+3=0\)

\(\Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\)

\(\Leftrightarrow\left(x,y\right)=\left(-2;1\right)\)

A

A. (x-2)³ = x³ - 6x² +12x - 8 (hằng đẳng thức)

A

\(12x^3-12x^2+3x\)

\(=12x^3-9x+12x-12x^2\)

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

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

\(=3x.\left(-1\right)=-3x\)

p/s: ko chắc =]

sorry ;<

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

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

bn sửa lại cái dòng thứ ba nha