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

b) \(\left(x+2\right)^2-9=\left(x-1\right)\left(x+5\right)\)

c) \(\left(a+b\right)^2-\left(a-2b\right)^2\)

\(=\left(a+b-a+2b\right)\left(a+b+a-2b\right)\)

\(=3b\left(2a-b\right)\)

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

`b, (x+2)^2-9 = (x+2-3)(x+2+3) = (x-1)(x+5)`

`c, (a+b)^2-(a-2b)^2 = (a+b+a-2b)(a+b-a+2b) = (2a-b)(3b)`

`a, 4a^2 + 4a + 1 = (2a+1)^2`

`b, -3x^2 + 6xy - 3y^2`

` = -3(x-y)^2`

`c, (x+y)^2 - 2(x+y)z + z^2`

`= (x+y-z)^2`

a: \(x^2-8x+16x=x^2+8x=x\left(x+8\right)\)

b: \(4x^2-8xyz+4y^2=4\left(x^2-2xyz+y^2\right)\)

c: \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)

\(4a^2b^2-\left(a^2+b^2-1\right)^2\)

\(=\left[2ab-\left(a^2+b^2-1\right)\right].\left[2ab+\left(a^2+b^2-1\right)\right]\)

\(=\left(2ab-a^2-b^2+1\right)\left(2ab+a^2+b^2+-1\right)\)

\(=\left[1-\left(a-b\right)^2\right]\left[\left(a+b\right)^2-1\right]\)

\(=\left(1-a+b\right)\left(1+a-b\right)\left(a+b+1\right)\left(a+b-1\right)\)

\(4a^2b^2-\left(a^2+b^2-1\right)^2=\left(2ab+a^2+b^2-1\right)\left(2ab-a^2-b^2+1\right)\)

\(=\left[\left(a+b\right)^2-1\right]\left[1-\left(a-b\right)^2\right]\)

\(=\left(a+b-1\right)\left(a+b+1\right)\left(1+a-b\right)\left(1-a+b\right)\)

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

4a²b²-(a²+b²-c²)²

= (4ab-a²-b²+c²)(4ab+a²+b²-c²)

= -[(a-b)²-c²][(a+b)²-c²]

=-(a-b+c)(a-b-c)(a+b-c)(a+b+c)

=(b-a-c)(b+c-a)(a+b-c)(a+b+c)

\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)

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

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

a: \(x^2-2xy+y^2+3x-3y-4\)

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

\(=\left(x-y+4\right)\left(x-y-1\right)\)

Bài 1: 

a: \(4a^2-6b=2\left(2a^2-3b\right)\)

b: \(m^3n-2m^2n^2-mn\)

\(=mn\left(m^2-2mn-1\right)\)

Bài 1:

a) \(4a^2-6b=2\left(a^2-3b\right)\)

b) \(=mn\left(m^2-2mn-1\right)\)

Bài 2:

a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)

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

a: Ta có: \(x^5-x^3+x^2-1\)

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

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

b: Ta có: \(5x^3-45x\)

\(=5x\left(x^2-9\right)\)

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

c: Ta có: \(16x^4y^2+2xy^5\)

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

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

d: Ta có: \(a^3-8+6a^2-12a\)

\(=\left(a-2\right)\left(a^2+2a+4\right)+6a\left(a-2\right)\)

\(=\left(a-2\right)\left(a^2+8a+4\right)\)

e: Ta có: \(x^4+x^3+x+1\)

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

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

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

`b, 6ab - 9ab^2 = 3ab(2-b)`

`c, 2a(x-1) + 3b(1-x)`

`= (2a-3b)(x-1)`

`d, (x-y)^2 - x(y-x)`

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

`= (2x-y)(x-y)`