3x³ + 6x²y+ 3xy²

=3x³ + 3x²y + 3x²y + 3xy²

= 3x²(x +y) + 3xy(x +y)

=(x+y)(3x² + 3xy)

= 3x(x+y)(x+y)

= 3x(x+y)²

a) 9x⁴+16y⁶-24x²y³

=(3x²)²-2.3x².4y³+(4y³)²

=(3x²-4y³)²

b) 16x²-24xy+9y²

=(4x)²-2.4x.3y+(3y)²

=(4x-3y)²

c) 36x²-(3x-2)²

=(36x-3x+2)(36x+3x-2)

=(33x+2)(39x-2)

d) 27x³+54x²y+36xy²+8y³

=(3x)³+3.(3x)².2y+3.3x.(2y)²+(2y)³

=(3x+2y)³

e) y⁹-9x²y⁶+27x⁴y³-27x⁶

=(y³)³-3.(y³)².3x²+3.y³.(3x²)²-(3x²)³

=(y³-3x²)³

f) 64x³+1

= (4x)³+1³

=(4x+1)[(4x)²-4x.1+1²]

=(4x+1)(16x²-4x+1)

e) 27x⁶-8x³  *sửa đề*

=(3x²)³-(2x)³

=(3x²-2x)[(3x)²+3x².2x+(2x)²]

=(3x²-2x)(9x²+6x³+4x²)

~~~

a) \(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)

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

c) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

b) 2xy(1+z)

d) 33x^2

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

b, 2xy+2xyz
=2xy(1+z)

d,27x^2+6x^2
=3x²(9+2)
=3x².11

f, (3x-6y)x+y(x-2y)
=3(x-2y)+y(x-2y)
=(3+y)(x-2y)

b: \(2xy+2xyz=2xy\left(z+1\right)\)

d: \(27x^2+6x^2=33x^2\)

f: \(\left(3x-6y\right)x+y\left(x-2y\right)=\left(x-2y\right)\left(3x+y\right)\)

\(x^3+y^3-3x^2+3x-1\\=(x^3-3x^2+3x-1)+y^3\\=(x-1)^3+y^3\\=(x-1+y)[(x-1)^2-(x-1)y+y^2]\\=(x+y-1)(x^2-2x+1-xy+y+y^2)\)

\(x^3-3x^2y+x+3xy^2-y-y^3\\=(x^3-3x^2y+3xy^2-y^3)+(x-y)\\=(x-y)^3+(x-y)\\=(x-y)[(x-y)^2+1]\\=(x-y)(x^2-2xy+y^2+1)\)