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6 tháng 10 2019

8x2+4xy-2ax-2y

= 2( x2+2xy-ax-y)

2xy-x2-y2+16

= - (x2-2xy+y2-16)

= - [(x-y)2 - 16 ]

=-(x-y-4)(x-y+4)

3a2-6ab+3b2-12c2

=3(a2-2ab+b2-4c2)

=3[(a-b)2-4c2]

=3(a-b-2c)(a-b+2c)

(nếu đề bài là phân tích đa thức thành nhân tử bn nhé)

chúc bn hc tốt

1 tháng 6 2021

a,x2-y2-2x+2y
= (x+y)(x-y) - 2(x-y)
= (x-y)(x+y-2)
b,2x+2y-x2-xy
= 2(x+y) - x(x+y)
= (x+y)(2-x)
c,3a2-6ab+3b2-12c2
= 3(a2 - 2ab + b2 - 4c2)
= 3[(a-b)2 - 4c2)
= 3(a-b-2c)(a-b+2c)
d,x2-25+y2+2xy
= (x+y)2 - 25
= (x+y+5)(x+y-5)

e) a2+2ab+b2-ac-bc

= (a+b)2-c(a+b)

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

f) x2-2x-4x2-4y

= -3x2-2x-4y

= -(3x2+2x+4y)

g)x2y-x3-9y+9x

= x2(y-x)-9(y-x)

= (y-x)(x2-9)

h) x2(x-1)+16(1-x)

= x2(x-1)-16(x-1)

= (x-1)(x2-16)

= (x-1)(x-4)(x+4)

n) 81x2-6yz-9y2-z2

= (9x)2-[(3y)2+6yz+z2]

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

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

m) xz- yz-x2+2xy-y2

= z(x-y)-(x2-2xy+y2)

= z(x-y)-(x-y)2

= (x-y)(z-x+y)

 p) x2 + 8x + 15

= x2 + 3x + 5x + 15

= x(x+3) + 5(x+3)

= (x+3)(x+5)

k) x2 - x - 12

= x2 + 3x - 4x - 12

= x(x+3) - 4(x+3)

= (x+3)(x-4)

26 tháng 9 2018

1 : 8x2+4xy-2ax -ay=4x(2x+y)-a(2x+y)=(2x+y)(4x-a)

2,3 tương tự 

\(\left(2x-y\right)\left(4x^2-4xy+y^2\right)-8x^2\left(x-y\right)\)

\(=8x^3-y^3-8x^3+8x^2y\)

\(=8x^2y-y^3\)

7 tháng 10 2021

cảm ơn nha yêuuuuu

 

5 tháng 10 2018

Bạn Hoa giải đúng

\(\left(2x-y\right)\left(4x^2-4xy+y^2\right)-8x^2\left(x-y\right)\)

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

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

\(=-4x^2y-6xy^2-y^3\)

\(=-y\left(4x^2+6xy+y^2\right)\)

1 tháng 10 2021

`a)x^3-8x^2+16x`

`=x(x^2-8x+16)`

`=x(x-4)^2`

`b)x^2+4y^2+2x-4y-4xy-24`

`=(x-2y)^2+2(x-2y)-24`

`=(x-2y)^2-4(x-2y)+6(x-2y)-24`

`=(x-2y-4)(x-2y+6)`

`c)x^4+x^3-x^2-2x-2`

`=x^4-2x^2+x^3-2x+x^2-2`

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

`=(x^2-2)(x^2+x+1)`

4 tháng 7 2023

\(6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)+5y^2\left(x^2+xy\right)\\ =6x^2y^2-6xy^3-8x^3+8x^2y^2+5x^2y^2+5xy^3\\ =\left(6x^2y^2+8x^2y^2+5x^2y^2\right)+\left(-6xy^3+5xy^3\right)-8x^3\\ =19x^2y^2-xy^3-8x^3\)

Với `x=1/2;y=2` ta có :

 \(19x^2y^2-xy^3-8x^3\\ =19.\left(\dfrac{1}{2}\right)^2.2^2-\dfrac{1}{2}.2^3-8.2^3\\ =19.\dfrac{1}{4}.4-\dfrac{1}{2}.8-8.8\\ =19-4-64\\ =-49\)

14 tháng 8 2021

a) \(=\left(x+6y\right)\left(x-6y\right)-\left(x-6y\right)\)

\(=\left(x-6y\right)\left(x-6y-1\right)\)

b) \(=x\left(x^2-8x+16\right)\)

\(=x\left(x-4\right)^2\)

c) \(=2\left(x-y\right)^2-18\)

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

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

a: \(x^2-36y^2-x+6y\)

\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)

\(=\left(x-6y\right)\left(x+6y-1\right)\)

b: \(x^3-8x^2+16x\)

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

\(=x\left(x-4\right)^2\)

c: \(2x^2-4xy+2y^2-18\)

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

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

d: \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-10\right)\)

a: Ta có: \(x^2-36y^2-x+6y\)

\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)

\(=\left(x-6y\right)\left(x+6y-1\right)\)

b: Ta có: \(16x-8x^2+x^3\)

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

\(=x\left(x-4\right)^2\)

c: Ta có: \(2x^2-4xy+2y^2-18\)

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

\(=2\cdot\left[\left(x-y\right)^2-9\right]\)

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

d: Ta có: \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-10\right)\)

e: Ta có: \(x^4-x^2-30\)

\(=x^4-6x^2+5x^2-30\)

\(=x^2\left(x^2-6\right)+5\left(x^2-6\right)\)

\(=\left(x^2-6\right)\left(x^2+5\right)\)

f: Ta có: \(x^2-xy-2y^2\)

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

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

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

g: Ta có: \(x^4-13x^2y^2+4y^4\)

\(=x^4-4x^2y^2+4y^4-9x^2y^2\)

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

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

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

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

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

h: Ta có: \(\left(x^2-2x\right)^2-2\left(x^2-2x\right)-3\)

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

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

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

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