3x² + 6x + 12

= 3(x² + 2x + 4)

= 3(x + 2)²

sorry, mik sửa:

3x² + 6x + 12

= 3(x² + 2x + 4)

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

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

\(x^3-3x^2+6x-4\)

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

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

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

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

x^3 - 3x^2 + 6x - 4 

<=> x^3-3x^2+3x-1+3x-3

<=>(x-1)^3+3(x-1)

<=>(x-1)+((x-1)^2+3)

<=>(x-1)+(x^2-2x+4)

3(x²-2x+1-y²⁾ 

=3((x-1)²-y²)

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

tik nha 3:)

a) 3x^2 -6x + 3 - 3y^2

=3.(x²-2x+1-y²)

=3.[(x-1)²-y²]

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

3x³ + 6x² + 3x - 12xy² 

= 3x(x² + 2x + 1 - 4y²) 

= 3x[(x + 1)² - (2y)²]

= 3x(x + 1 + 2y)(x - 2y + 1)

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

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

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

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

a ) x=0; x = -(căn bậc hai(7)*i-3)/8;x = (căn bậc hai(7)*i+3)/8;

b ) -(y-x-3)*(y+x+3)

a) \(12x^3-9x^2+3x\)

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

b) \(x^2-y^2+6x+9\)

\(=\left(x^2+6x+9\right)-y^2\)

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

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

a: \(12x^3-6x^2+3x\)

\(=3x\cdot4x^2-3x\cdot2x+3x\cdot1\)

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

b: \(\dfrac{2}{5}x^2+5x^3+x^2y\)

\(=x^2\cdot\dfrac{2}{5}+x^2\cdot5x+x^2\cdot y\)

\(=x^2\left(\dfrac{2}{5}+5x+y\right)\)

c: \(14x^2y-21xy^2+28x^2y^2\)

\(=7xy\cdot2x-7xy\cdot3y+7xy\cdot4xy\)

\(=7xy\left(2x-3y+4xy\right)\)

\(3x^2-6x+9x^2=12x^2-6x=6x\left(2x-1\right)\)