1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)

2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)

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

3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)

1) Ta có: \(x^3+2x^2-6x-27\)

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

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

2: Ta có: \(9x^2+6x-4y^2-4y\)

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

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

a) \(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)

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

c) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)

\(x^3+9x^2+6x-16\)

\(=x^3+x^2-2x+8x^2+8x-16\)

\(=x\left(x^2+x-2\right)+8\left(x^2+x-2\right)\)

\(=\left(x^2+x-2\right)\left(x+8\right)\)

\(=\left(x^2-x+2x-2\right)\left(x+8\right)\)

\(=\left[x\left(x-1\right)+2\left(x-1\right)\right]\left(x+8\right)\)

\(=\left(x-1\right)\left(x+2\right)\left(x+8\right)\)

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

\(x^3+6x^2+9x\)

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

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

1: \(6x^2y-9xy^2+3xy\)

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

2: \(\left(4-x\right)^2-16\)

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

\(=-x\cdot\left(8-x\right)\)

3: \(x^3+9x^2-4x-36\)

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

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

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

2) \(\left(4-x\right)^2-16=\left(4-x\right)^2-4^2=\left(4-x-4\right)\left(4-x+4\right)=-x\left(8-x\right)\)

3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)

\(9x^2-6x^2-3\)

\(=3x^2-3\)

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

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

\(9x^2-6x^2-3\)

\(=3x^2-3\)

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

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

\(9x^2-6x^2-3\)

\(=3x^2-3\)

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

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

Nguồn: kudo shinichi

Nguyễn Ngọc Tuấn : hám fame vc

\(x^3-9x^2+6x+16=\left(x^3+x^2\right)-\left(10x^2+10x\right)+\left(16x+16\right)\)

\(=x^2.\left(x+1\right)-10x\left(x+1\right)+16\left(x+1\right)\)

\(=\left(x+1\right).\left(x^2-10x+16\right)\)

\(=\left(x+1\right).\left[\left(x^2-8x\right)-\left(2x-16\right)\right]\)

\(=\left(x+1\right)\left[x\left(x-8\right)-2\left(x-8\right)\right]\)

\(=\left(x+1\right)\left(x-2\right)\left(x-8\right)\)

x^3-9x^2+6x+16

=x^3+x^2-10x^2-10x+16x+16

=(x^3+x^2)-(10x^2+10x)+(16x+16)

=x^2(x+1)-10x(x+1)+16(x+1)

=(x+1)(x^2-10x+16)

=(x+1)(x^2-2x-8x+16)

=(x+1)[(x^2-2x)-(8x-16)]

=(x+1)[x(x-2)-8(x-2)]

=(x+1)(x-2)(x-8)