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

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

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

Ok

Mình không giảnh để trả lời câu hỏi này 

Ta có \(\left(x-2y\right)^3=x^3-6x^2y+12xy^2-8y^3=-8\)

\(\Rightarrow x-2y=-2\)

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

\(=3.\left(-2\right)^2=12\)

\(x^3-6x^2y+12xy^2-8y^3=-8\)

\(\Leftrightarrow\left(x-2y\right)^3=-8\)

=>x-2y=-2

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

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

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

Ta có: 3x^2-12xy+12y^2=3(x-2y)^2. Lại có x^3-6x^2y+12xy^2-8y^3=-8\(\Rightarrow\)(x-2y)^3=-8\(\Rightarrow\)x-2y=-2. Thay vào biểu thức ta được biểu thức bằng 12.    Học tốt!

\(x^3-6x^2y+12xy^2-8y^3=-8\Leftrightarrow\left(x-2y\right)^3=-8\Leftrightarrow x-2y=-2\)

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

a: =(2x-3y)^2-4(2x-3y)

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

b: =3x^2+21x-x-7

=(x+7)(3x-1)

c: =(3x-1)^4+2(3x-1)^2+1

=[(3x-1)^2+1]^2

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

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

\(3x^2-12xy+12y^2-6x+12y\)

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

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

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