\(x^2-x-y^2-y\)

b)

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2 tháng 8 2017

a, \(x^2-x-y^2-y\)

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

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

b, \(a^3-a^2x-ay+xy\)

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

c, sai đề?

d, \(x^3-x+3x^2y+3xy^2+y^3-y\)

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

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

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

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

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

\(=5\left(x^2-2xy+y^2\right)-20z^2\)

\(=5\left(x-y\right)^2-20z^2\)

\(=5\left[\left(x-y\right)^2-4z^2\right]\)

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

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

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

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

Câu 2 nha

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

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

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

\(c,x^2-x+3x^2y+3xy^2+y^3-y\)

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

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

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

16 tháng 7 2017

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

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

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

c) \(5x^2-10xy+5y^2-20z^2=\left(\sqrt{5}x-\sqrt{5}y\right)^2-20z^2\)

Câu b :

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

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

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

Câu c :

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

\(=5\left(x^2-2xy+y^2\right)-20z^2\)

\(=5\left(x-y\right)^2-20z^2\)

\(=5\left[\left(x-y\right)^2-4z^2\right]\)

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

27 tháng 10 2021

helpppppp

23 tháng 8 2019

k) \(x^3-x+3x^2+3xt^2+y^3-y\)

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

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

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

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

23 tháng 8 2019

h) \(a^3-a^2x-ay+xy\)

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

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

16 tháng 7 2018

a)   \(x^3-2x^2-6x+12\)

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

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

\(=\left(x-2\right)\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\)

b)  \(x^4-7x^2+12\)

\(=x^4-3x^2-4x^2+12\)

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

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

\(=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x-2\right)\left(x+2\right)\)

c)  \(x^2-5x+4\)

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

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

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

d)  \(3x^2+5x+2\)

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

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

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

e)  \(x^3-x+3x^2y+3xy^2+y^3-y\)

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

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

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

1 tháng 11 2016

Đây, bản full đây thím, tớ thực sự đã kiên nhẫn lắm đấy ...

a)\(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)=4\left(x^2-y^2-2x+2ay-a^2+1\right)\)

\(=4\left[\left(x^2-2x+1\right)-\left(a^2-2ay+y^2\right)\right]\)

\(=4\left[\left(x-1\right)^2-\left(a-y\right)^2\right]\)

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

b)\(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)

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

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

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

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

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

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

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

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

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

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

d)\(a^5+a^4+a^3+a^2+a+1=a^4\left(a+1\right)+a^2\left(a+1\right)+\left(a+1\right)\)

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

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

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

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

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

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

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

f)\(5x^3-3x^2y-45xy^2+27y^3=5x\left(x^2-9y^2\right)-3y\left(x^2-9y^2\right)\)

\(=\left(x^2-9y^2\right)\left(5x-3y\right)\)

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

g)\(3x^2\left(a-b+c\right)+36xy\left(a-b+c\right)+108y^2\left(a-b+c\right)\)

\(=\left(a-b+c\right)\left(3x^2+36xy+108y^2\right)\)

\(=3\left(a-b+c\right)\left(x^2+12xy+36y^2\right)\)

\(=3\left(a-b+c\right)\left(x+6y\right)^2\)

1 tháng 11 2016

a/ \(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)\)

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

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

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

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

Giải giúp bạn 2 bài tiêu biểu thôi nha

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

xog a

tốt

Bài làm

a) 3x2 - 6x2 + 3x

= -3x2 + 3x

= 3x( 1 - x )

b) 3x2 + 5x - 3xy - 5y

= ( 3x2 - 3xy ) + ( 5x - 5y )

= 3x( x - y ) + 5( x - y )

= ( x - y )( 3x + 5 )

c) x3 + 2x2 + x

= x( x2 + 2x + 1 )

= x( x2 + 2.x.1 + 12 )

= x( x + 1 )2

d) xy + y2 - x - y

= ( xy - x ) + ( y2 - y )

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

= ( y - 1 )( x +  y )

# Học tốt #

29 tháng 6 2018

\(1\hept{\begin{cases}6x^2-8x+3x-4\\2x\left(3x-4\right)+\left(3x-4\right)\\\left(3x-4\right)\left(2x+1\right)\end{cases}}\)

\(2\hept{\begin{cases}7x^2-7xy-5x+5y+6xy\\7x\left(x-y\right)-5\left(x-y\right)+\frac{6xy\left(x-y\right)}{\left(x-y\right)}\\\left(x-y\right)\left(7x-5+\frac{6xy}{\left(x-y\right)}\right)\end{cases}}\)

\(3\hept{\begin{cases}5x\left(x-y\right)-15\left(x-y\right)\\\left(x-y\right)\left(5x-15\right)\end{cases}}\)

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

\(5\hept{\begin{cases}x^3-4x-3x^2+12\\x\left(x^2-4\right)-3\left(x^2-4\right)\\\left(x+2\right)\left(x-2\right)\left(x-3\right)\end{cases}}\)

\(6\hept{\begin{cases}2x+2y+x^2-y^2\\2\left(x+y\right)+\left(x+y\right)\left(x-y\right)\\\left(x+y\right)\left(2+x-y\right)\end{cases}}\)

\(7\hept{\begin{cases}\left(x^2y-2xy\right)-\left(xy-2y\right)+\left(xy-y\right)\\xy\left(x-2\right)-y\left(x-2\right)+y\left(x-1\right)\\y\left(X-2\right)\left(x-1\right)+y\left(x-1\right)\end{cases}}\Leftrightarrow y\left(x-1\right)\left(x-2+1\right)\)

\(8\hept{\begin{cases}x\left(2-y\right)+z\left(2-y\right)\\\left(2-y\right)\left(x+1\right)\end{cases}}\)

16 tháng 9 2018

\(2x^2+x\)

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

.

hk 

tốt

12 tháng 8 2020

a) 3( x - y ) - 5x( y - x )

= 3( x - y ) - 5x[ -( x - y ) ]

= 3( x - y ) + 5x( x - y )

= ( 3 + 5x )( x - y )

b) x3 + 2x2y + xy2 - 9x

= x( x2 + 2xy + y2 - 9 )

= x[ ( x + y )2 - 32 ]

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

c) 14x2y - 21xy2 + 28x2y2

= 7xy( 2x - 3y + 4xy )

12 tháng 8 2020

                                              Bài giải

\(a,\text{ }3\left(x-y\right)-5x\left(y-x\right)\)

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

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

\(b,\text{ }x^3+2x^2y+xy^2-9x\)

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

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

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

\(c,\text{ }14x^2y-21xy^2+28x^2y\)

\(=7xy\left(2x-3y+4x\right)\)

\(=7xy\left(6x-3y\right)\)