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

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

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

b)\(x^2+4y^2+4xy\)

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

c)\(x^2+8x+16\)

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

d)\(9x^2-12xy+4y^2\)

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

e)\(-25x^2y^2+10xy-1\)

\(=-\left(25x^2y^2-10xy+1\right)\)

\(=-\left(5xy-1\right)^2\)

f)\(4x^2-4x+1\)

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

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

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

h)\(9x^2-6x+1\)

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

#H

a, 6x - 9 - x² = - x² + 6x - 9 = - (x² - 6x + 9) = - (x - 3)²

b, x² + 4y² + 4xy = x² + 2. x . 2y + (2y)² = (x + 2y)²

c, x² + 8x + 16 = x² + 2 . x . 4 + 4² = (x + 4)²

d, 9x² - 12xy + 4y² = (3x)² - 2 . 3x . 2y + (2y)² = (3x - 2y)²

e, - 25x²y² + 10xy - 1 = - (25x²y² - 10xy + 1) = - [(5xy)² - 2 . 5xy + 1] = - (5xy - 1)²

f, 4x² - 4x + 1 = (2x)² - 2 . 2x + 1 = (2x - 1)²

j, x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

h, 9x² - 6x + 1 = (3x)² - 2 . 3x + 1 = (3x - 1)²

Biểu thức này không phân tích được thành nhân tử nhé. Bạn xem xem có viết sai đề không.

a) x^2 - 5xy +4y^2= x^2 -xy -4xy+4y^2= (x^2-xy) - (4xy - 4y^2)= x(x-y)-4y(x-y)=(x-y)*(x - 4y)

b) x^2 -y^4+9y -x(9+y-y^3= x^2-y^4 +9y-9x-xy+xy^3=  (x^2-xy)-(9x-9y)+(xy^3-y^4)=x(x-y)-9(x-y)+y^3(x-y)=(x-y)*(y^3+x-9)

d) 2u^2+2v^2-5uv=(2u^2-4uv)+(2v^2-uv)=2u(u-2v)+v(2v-u)= 2u(u-2v)-v(u-2v)=(u-2v)*(2u-v)

d,

\(a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)\)

\(=\left(a+b\right)\left(a+b-c\right)\)

Vậy..

e

\(x^2-2x-4y^2-4y\)

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

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

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

d)\(a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)\)\(=\left(a+b\right)\left(a+b-c\right)\)

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

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

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

a, \(6x^3y^2.\left(2-x\right)+9x^2y^2\left(x-2\right)\)
\(=6x^3y^2.\left(2-x\right)-9x^2y^2\left(2-x\right)\)
\(=y^2.\left(2-x\right)\left(6x^3-9x^2\right)\)
\(=3x^2y^2.\left(2-x\right)\left(2x-3\right)\)

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