\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)

\(x^3-6x^2y+12xy^2-8y^3\)

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

a

\(8x^3-\dfrac{1}{125}y^3\\ =\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\\ =\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\\ =\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)

b

\(-x^3+6x^2y-12xy^2+8y^3\\ =-\left(x^3-6x^2y+12xy^2-8y^3\right)\\ =-\left(x^3-3.2y.x^2+3.\left(2y\right)^2.x-\left(2y\right)^3\right)\\ =-\left(x-2y\right)^3\\ =-\left(x-2y\right)\left(x-2y\right)\left(x-2y\right)\)

a: 8x^3-1/125y^3

=(2x)^3-(1/5y)^3

=(2x-1/5y)(4x^2+2/5xy+1/25y^2)

b: =(2y-x)^3

a) \(8x^3-\dfrac{1}{125}y^3\)

\(=\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\)

\(=\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x\cdot\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\)

\(=\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{24}y^2\right)\)

b) \(-x^3+6x^2y-12xy^2+8y^3\)

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

\(=-\left(x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\right)\)

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

\(3,x\left(x-1\right)-y\left(1-x\right)=\left(x+y\right)\left(x-1\right)\\ 4,x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\\ 5,x^2-2xy+y^2-xz+yz=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y-z\right)\left(x-y\right)\\ 6,x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\\ 9,x^3+x^2-xy+xy+y^2+y^3\\ =x^2\left(x+1\right)+y^2\left(x+1\right)=\left(x^2+y^2\right)\left(x+1\right)\\ 10,x^2-6\left(x+3\right)-9\\ =x^2-6x-18-9\\ =x^2-6x-27=\left(x-9\right)\left(x+3\right)\)

10: \(x^2-6\left(x+3\right)-9\)

\(=x^2-6x-18-9\)

\(=x^2-6x-27\)

\(=\left(x-9\right)\left(x+3\right)\)

3)(x-1).(x-y)

x=67e

a,3x³y³-15x²y²=3x²y²(xy-5)

b,2x(x-5y)+8y(5y-x)=2x(x-5y)-8y(x-5y)=(x-5y).(2x-8y)

c,(3x-1)²-16=(3x-1)²-4²=(3x-1+4)(3x-1-4)=(3x+3)(3x-5)

d,x³-3x²+3x-1=x³-1-(3x²+3x)=x³-1-3x(x+1)=(x³-1-3x)(x+1)

e,125x³+1=(5x)³+1³=(5x+1)(25x²-5x.1+1²)

f,x³+6x²y+12xy²+8y³=x³+3.x².2y+3.x.(2y)²+(2y)³=(x+2y)³

Bài 1:

\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)

Bài 3:

\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)

Bài 1;

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

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

3) \(2y^3+8y^3+8y=10y^3+8y=2y\left(5y^2+4\right)\)

4) \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

1) \(x^3-64x=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\)

2) \(8x^2y-18y=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\)

3) \(24x^3-3=3\left(8x^3-1\right)=3\left(2x-1\right)\left(4x^2+2x+1\right)\)

Bài 3:

1) \(5x^2+10x+5-5y^2=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y\right]=5\left(x-y+1\right)\left(x+y+1\right)\)

2) \(3x^3-6x^2+3x-12xy^2=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=3x\left(x-2y-1\right)\left(x+2y-1\right)\)

3) \(a^3b-ab^3+a^2+2ab+b^2=ab\left(a^2-b^2\right)+\left(a+b\right)^2=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2=\left(a+b\right)\left(a^2b-ab^2+a+b\right)\)

4) \(2x^3-2xy^2-8x^2+8xy=2x\left(x^2-y^2-4x+4y\right)=2x\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]=2x\left(x-y\right)\left(x+y-4\right)\)

\(a,6x^2y-12xy^2+18x^2y^2\)

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