\(A=9x^2+3yx+6x+y+1\)

\(\Rightarrow A=\left(9x^2+3x\right)+\left(3xy+y\right)+\left(3x+1\right)\)

\(\Rightarrow A=3x\left(3x+1\right)+y\left(3x+1\right)+\left(3x+1\right)\)

\(\Rightarrow A=\left(3x+y+1\right)\left(3x+1\right)\)

\(3x^2-6x+9x^2=12x^2-6x=6x\left(2x-1\right)\)

TL:

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

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

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

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

\(=\left(3x\right)^2-2.3x.1+1^2\)

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

9x² - 6x +1= (3x-1)²

9x^2 -6x+1

=3x^2-2.3x.1+1

=(3x-1)^2

=3x^2-1^2

=(3x+1).(3x-1)

. là dấu nhân nhak :3

5x^2+10xy+5y^2

=5.(x²+2xy+y²)

=5.(x+y)²

x^3-6x^2+9x

=x.(x²-6x+9)

=x.(x-3)²

xy+y^2-x-y

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

=(x+y)(y-1)

haha bạn ấy nhấn  đúng cho tui nè

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: \(9x^2+6x-4y^2-4y\)

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

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

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

\(=3x^2-3\)

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

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

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

\(=3x^2-3\)

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

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

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

\(=3x^2-3\)

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

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

Nguồn: kudo shinichi

Nguyễn Ngọc Tuấn : hám fame vc