\(x+2\sqrt{x}-3=\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\)

\(x^2+4y^2+9-4xy-6x+12y\)

\(=\left(x^2-4xy+4y^2\right)+\left(-6x+12y\right)+9\)

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

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

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: \(9x^2+6x-4y^2-4y\)

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

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

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

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

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

= ( x^2 - 4y^2 ) + ( 9 - 6x)

= [ x^2 - (2y)^2 ] + 3( 3 - 2x )

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

x² - 4y² + 9 - 6x 

= (x² - 6x + 9) - (2y)²

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

a 8xy²-12x²y+20xy=4xy(2y-3x+5)

a,=\(4xy\left(2y-3x+5\right)\)

b,=\(2\left(x^2-25\right)=2\left(x+5\right)\left(x-5\right)\)

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

a) = (x - 2)² - y²

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

b) = (x^2 + 6x + 9) - (2y)^2

= (x + 3)² - (2y)²

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

c) = (x - 3y)² - 6²

= (x - 3y - 6)(x - 3y + 6)