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14 tháng 12 2021

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

14 tháng 12 2021

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

\(=\left(3x+1+2x+1\right)\left(3x+1-2x-1\right)\\ =x\left(5x+2\right)\)

\(c,-5x^2+10xy-5y^2+20z^2\\ =-5\left(x^2-2xy+y^2-4z^2\right)\\ =-5\left[\left(x^2-2xy+y^2\right)-\left(2z\right)^2\right]\\ =-5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\\ =-5\left(x-y+2z\right)\left(x-y-2z\right)\)

23 tháng 7 2016

1/a ) = (x+y)3 -(x+y)

= (x+y)[(x+y)2+1]

c) = 5(x2-xy+y2)-20z2

=5(x-y)2-20z2

= 5 [ (x-y)2- 4z2 ]

=5(x-y-4z)(x-y+4z)
 

23 tháng 7 2016

Bài 1:

a) x3-x+3x2y+3xy2+y3-y

=x3+2x2y-x2+xy2-xy+x2y+2xy2-xy+y3-y2+x2+2xy-x+y2-y

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

=(x2+2xy-x+y2-y)(x+y+1)

=[x(x+y-1)+y(x+y-1)](x+y+1)

=(x+y-1)(x+y)(x+y+1) 

c) 5x2-10xy+5y2-20z2

=-5(2xy-y2+4z2-2)

Bài 2:

5x(x-1)=x-1   

=>5x2-6x+1=0

=>5x2-x-5x+1

=>x(5x-1)-(5x-1)

=>(x-1)(5x-1)=0

=>x=1 hoặc x=1/5

b) 2(x+5)-x2-5x=0

=>2(x+5)-x(x+5)=0

=>(2-x)(x+5)=0

=>x=2 hoặc x=-5

23 tháng 7 2016

1) 

a) (x+y)3-(x+y)= (x+y)(x+y-1)

b) xem lại đề câu B nha bạn

2)

a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc=0

(a+b)3+c3-3ab(a+b+c)=0

(a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c)=0

(a+b+c)(a2+b2+c2-xy-yz-xz)=0

Suy ra: a3+b3+c3=3abc

 

7 tháng 10 2016

1. a) = (x+y)3 -(x+y) =(x+y)((x+y)2 -1)

     = (x+y)(x+y+1)(x+y-1)

b) = 5(( x-y)2 - 4z2)

     = 5( x-y +2z)(x-y-2z)

2. áp dụng ( a+b+c)3 = .....rồi biến đổi

     

10 tháng 8 2016

a.\(x^2\left(x^2+2x+1\right)\)

   \(x^2\left(x+1\right)^2\)

17 tháng 8 2019

Bài 1

a) x4 + 2x3 + x2 = x2(x2 + 2x + 1) = x2.(x + 1)2

b) 5x2 - 10xy + 5y2 - 20z2

= 5(x2 - 2xy + y2 - 4z2)

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

= 5.(x - y - 2z).(x - y + 2z)

c) 25x2 - y2 + 4y - 4

= 25x2 - (y2 - 4y + 4 )

= (5x)2 - (y - 2)2

= (5x - y + 2)(5x + 2 -y)

17 tháng 8 2019

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11 tháng 12 2016

a, \(4y^2+1-4y=\left(2y\right)^2-2.2y.1+1^2=\left(2y-1\right)^2\)

b, \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\)

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

 

 

16 tháng 8 2018

a) Sửa đề

\(x^4+2x^3+x^2\)

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

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

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

\(=\left(x+1\right).x^2\left(x+1\right)\)

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

b) \(x^3-x+3x^2y+3xy^2+y^3-y\)

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

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

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

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

c) \(5x^2-10xy+5y^2-20z^2\)

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

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

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

a) Ta có: \(5x^2-10xy+5y^2-20z^2\)

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

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

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

b) Ta có: \(x^2-8x+15\)

\(=x^2-3x-5x+15\)

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

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

c) Ta có: \(2x^2-5xy+3y^2\)

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

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

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

d) Ta có: \(16y^3-2x^3-6x\left(x+1\right)-2\)

\(=16y^3-2x^3-6x^2-6x-2\)

\(=2\left[8y^3-x^3-3x^2-3x-1\right]\)

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

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

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

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

10 tháng 10 2017

Câu a trước đi ạ ^^

a) 7x - 6x- 2

= - 6x2 + 7x - 2

= (- 6x2 + 3x) + (4x - 2)

= 3x (- 2x + 1) + 2 (2x-1)

= - 3x ( 2x -1) + 2 (2x - 1)

= ( 2x -1 ) ( - 3x +2 )