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

Check lại đề hộ mình nhé:vv

\(=12x^2-6x=6x\left(2x-1\right)\)

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

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

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

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

b. \(=\left(1-x\right)^3\)

a) \(A=x^2-6x+9-9y^2\)

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

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

b) \(B=x^3-3x^2+3x-1+2\left(x^2-1\right)\)

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

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

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

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

b, \(B=\left(x-1\right)^3+2\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left[\left(x-1\right)^2+2\left(x+1\right)\right]\)

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

a) 3xy² - 3x³ - 6xy + 3x 

=3x (y² - x² - 2y +1)

= 3x [ (y-1)² -x² ]

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

b) 3x² +11x+6

= 3 x² +9x +2x +6

=3x (x+3)+2(x+3)

= (x+3)(3x+2)

c) -x³ - 4xy² + 4x²y +16x

= -x (x² + 4y² - 4xy -16 )

= -x [(x -2y)² - 4² ]

= -x(x-2y-4)(x-2y+4) 

a)2x^2+xy-y^2-x+2y-1

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

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

=2a^2+ab-b^2         với a=x,b=y-1

=2a^2+2ab-ab-b^2

=(2a-b)(a+b)

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

3x3 – 6x2 + 3x = 3x(x2 - 2x + 1) = 3x(x - 1)2

ý D nhé