a, \(=\left(xy+1+x-y\right)\left(xy+1-x+y\right)\)

b, \(\left(x+y-x+y\right)[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2]\)

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

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

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

câu a, b áp dụng hằng đẳng thức rồi làm nha 

c) 3x⁴y² + 3x³y² + 3xy² + 3y²

= ( 3x⁴y² + 3x³y² ) + ( 3xy² + 3y² )

= 3x³y² ( x + 1) + 3y² ( x + 1 )

= ( 3x³y² + 3y² ) ( x + 1 )

= 3y² ( x³ + 1 ) ( x + 1 )

= 3y² ( x + 1 ) ( x² - x + 1 ) ( x + 1 )

= 3y² ( x + 1 )² ( x² - x + 1 )

a) (xy +1)²- (x-y)²

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

b) (x + y)³ - (x - y)³

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

= 2y(x²+2xy+y²+x²+xy-xy-y²+x²-2xy+y²)

=2y(3x²+y²)

c) 3x⁴y² + 3x³y² + 3xy² + 3y²

=3y²(x⁴+x³+x+1)

= 3y²(x³(x+1)+(x+1)

= 3y²(x+1)(x³+1)

ko bt đúng ko

`a, 8x^3 - 1 = (2x-1)(4x^2 + 2x - 1)`

`b, x^3 + 27y^3 = (x+3y)(x^3 - 3xy + 9y^2)`

`c, x^3 - y^6 = (x-y^2)(x+xy^2 + y^4)`

1A:

a: \(x^3+2x=x\left(x^2+2\right)\)

b: \(3x-6y=3\left(x-2y\right)\)

c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)

\(=5\left(x+3y\right)\left(1-3x\right)\)

d: \(3\left(x-y\right)-5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+3\right)\)

1A. a. x(x²+2) 

b. 3(x-2y)

c. 5(x+3y)(1-3x) 

d. (x-y) (3-5x)

1B. a. 2x(2x-3)

b.xy(x²-2xy+5)

c. 2x(x+1)(x+2)

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

1B:

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(x^3y-2x^2y^2+5xy\)

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

a) \(3\left(x-y\right)^2+9y\left(y-x\right)^2\)

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

\(=\left(x-y\right)^2\left(3-9y\right)\)

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

b) \(3\left(x-y\right)^2+9y\left(y-x\right)\)

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

\(=\left(y-x\right)\left[3\left(y-x\right)+9y\right]\)

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

\(=3\left(y-x\right)\left(4y-x\right)\)

a: =3(x-y)^2+9y(x-y)^2

=(x-y)^2(3+9y)

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

b: =3(x-y)^2-9y(x-y)

=3(x-y)(x-y-9y)

=3(x-y)(x-10y)

`a, P = 2x(3 - x^2)`

`b, Q = 5x^2(x-3y)`

`c, R = xy(3x^2y^2 - 6y^2z + 1)`

a) \(P=6x-2x^3\)

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

b) \(Q=5x^3-15x^2y\)

\(Q=5x^2\left(x-3y\right)\)

c) \(R=3x^3y^3-6xy^3z+xy\)

\(R=xy\left(3x^2y^2-6y^2z+1\right)\)

Do câu d mình ko biết làm bởi v mình không làm được

`a, 4x^3 - 16x = 4x(x^2-4) = 4x(x-2)(x+2)`

`b, x^4 - y^4 = (x^2-y^2)(x^2+y^2) = (x-y)(x+y)(x^2+y^2)`

`c, xy^2 + x^2y + 1/4y^3`

`= y(xy + x^2 + 1/4y^2)`

`d, x^2 + 2x - y^2 + 1 = (x+1)^2 - y^2`

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

\(a,=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\\ b,=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\\ c,=\left[3x-2y-2\left(x+y\right)\right]\left[3x-2y+2\left(x+y\right)\right]\\ =5x\left(x-4y\right)\\ d,=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\\ f,=\left(x+3\right)\left(x^2-3x+9\right)\\ g,=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\\ h,=\left(5x-1\right)\left(25x^2+5x+1\right)\)

\(a)x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)\\ b)x^2-3y^2=\\ c)(3x-2y)^2-4(x+y)^2=(3x-2y)^2-[2(x+y)]^2\\=(3x-2y+2x+2y)(3x-2y-2x-2y)=5x(x-4y)\\ d)9(x-y)^2-4(x+y)^2=[3(x-y)]^2-[2(x+y)]^2=(3x-3y+2x+2y)(3x-3y-2x-2y)\\=(5x-y)(x-5y)\\ f)x^3+27=(x+3)(x^2-3x+9)\\ g)27x^3-0,001=(3x-0,1)(9x+0,3x+0,01)\\ h)125x^3-1=(5x-1)(25x^2+5x+1)\)

a: Ta có: \(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)\)

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

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

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

b: Ta có: \(x\left(y-x\right)^3-y\left(x-y\right)^2+xy\left(x-y\right)\)

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

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

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

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

a) \(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)=\left(x-1\right)\left(2x^2-9x+6\right)\)

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

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

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

\(a.2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)\\ =\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]\\ =\left(x-1\right)\left(2x^2-9x+6\right)\)

a: 6x-2y=2(3x-y)

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

Lời giải:
a. Không phân tích được nữa

b. $x^2(x-y)+4(y-x)=x^2(x-y)-4(x-y)=(x-y)(x^2-4)=(x-y)(x-2)(x+2)$
c. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$

$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$

ko phân tích dc

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