-Đặt \(t=\left(x^2-x+1\right)\)

\(\left(x^2-x+1\right)^2-5x\left(x^2-x+1\right)+4x^2\)

\(=t^2-5xt+4x^2\)

\(=t^2-4xt-xt+4x^2\)

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

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

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

\(=\left(x^2-5x+1\right)\left(x^2-2x +1\right)\)

\(=\left(x^2-5x+1\right)\left(x-1\right)^2\)

b) 4x2 – 9 + (2x + 3)2

= (4x2 - 9) + (2x + 3)2

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

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

= 4x(2x + 3)

a/ \(4x^2-9\)

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

b/ \(3x\left(3x-2\right)+1\)

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

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

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

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

a) $4x^2+4x+1$

$=(2x)^2+2\cdot2x\cdot1+1^2$

$=(2x+1)^2$

b) $x^2+6x-y^2+9$

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

$=(x^2+2\cdot x\cdot3+3^2)-y^2$

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

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

$\text{#}Toru$

a: \(4x^2+4x+1\)

\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2\)

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

b: \(x^2+6x-y^2+9\)

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

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

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

4 x 2 –   4 x y = 4 x x − y

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

4x² + 20x + 24 

= 4 ( x² + 5x + 6 ) = 4 ( x + 3 ) ( x + 2 )

\(4x^2-7x-2\\ =4x^2-8x+x-2\\ =4x\left(x-2\right)+\left(x-2\right)\\ =\left(x-2\right)\left(4x+1\right)\)

a: \(=x\left(x^2+4x+4-z^2\right)\)

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