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

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

\(=\left(4x^2+6x\right)-\left(2x+3\right)\)

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

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

4x² - 4y²

= (2x)² - (2y)²

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

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

4x²-4y²=4.(x²-y²)=4.(x-y).(x+y)

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

\(4x^2+8xy+3x+6y\)

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

\(=7\left(x+2y\right)\)

=4x(x+2y)+3(x+2y)=(x+2y)(4x+3)

Ta có: \(4x^2+12x+9\)

\(=4x^2+6x+6x+9\)

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

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

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

\(4x-8y-x^2+4y^2\)

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

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

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

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

 = ( x - y)^2 - z^2 

= ( x - y -z)(x - y + z) 

= (4x² + 4x+ 1) - y²

=(4x² + 4x +1) -y²

=[(2x)²  + 2.2x.1 +1] - y²

=(2x +1)² - y²

=[(2x +1) -y].[(2x +1) +y]

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

4x^2-y^2+4x+1

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

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

Cưu là mình vs (x^2+x)^2-2(x^2+x)-15