K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

4 tháng 11 2018

a) \(2x^2-2y^2\)

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

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

b) \(x^2-4x+4\)

\(=x^2-2\cdot x\cdot2+2^2\)

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

c) \(x^2+2x+1-y^2\)

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

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

d) \(x^2-4x\)

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

e) \(x^2+10x+25\)

\(=x^2+2\cdot x\cdot5+5^2\)

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

g) \(x^2-2xy+y^2-9\)

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

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

h) \(2x^2-2\)

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

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

i) \(5x^2-5xy+9x-9y\)

\(=5x\left(x-y\right)+9\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+9\right)\)

k) \(y^2-4y+4-x^2\)

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

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

l) \(x^2-16\)

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

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

m) \(3x^2-3xy+2x-2y\)

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

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

o) \(3x^4-6x^3+3x^2\)

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

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

4 tháng 11 2018

a) 2x2 - 2y2

 = (2x - 2y)(2x + 2y)

 = 4(x - y)(x + y)

b) x2 - 4x + 4

 = (x - 2)2

c) x+ 2x + 1 - y2

 = (x + 1)2 - y2

 = (x + 1 - y)(x + 1 + y)

d) x2 - 4x 

 = x(x - 4)

e) x+10x + 25

 = (x + 5)2

g) x2 - 2xy + y2 - 9

= (x - y)2 - 32

 = (x - y - 3)(x - y + 3)

h) 2x2 - 2

= 2(x2 - 1) 

 = 2(x - 1)(x + 1)

i) 5x- 5xy + 9x - 9y

  = 5x(x - y) + 9(x- y)

 = (5x + 9)(x - y)

k) y2 - 4y + 4 - x2

 = (y - 2)2 - x2

 = (y - 2 - x)(y - 2 + x)

l) x- 16

 = x- 42

 = (x - 4)(x + 4)

m) 3x2 - 3xy + 2x -2y

 = 3x(x - y) +2(x-y)

 = (3x + 2)(x - y)

o) 3x- 6x+ 3x2

 = 3x4 - 3x3 - 3x3 + 3x2

 = 3x3(x - 1) - 3x2(x - 1)

 = (3x- 3x2)(x - 1)

 = 3x2(x - 1)(x - 1)

 = 3x2.(x - 1)2

12 tháng 10 2021

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

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

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

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

18 tháng 10 2021

ỳtct7ct7c7c7t79tc9

 

giỏi vậy tui ngồi làm quài ko ra lun :^

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

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

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

c, \(x^6-x^4+2x^3+2x^2=x^6+2x^3+1-x^4+2x^2-1\)

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

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

d, \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-\left(2y\right)^3=\left(x+1-2y\right)\left(x+1+2y\right)\)

a) Ta có: \(x-2y+x^2-4y^2\)

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

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

b: Ta có: \(x^2-4x^2y^2+y^2+2xy\)

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

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

17 tháng 12 2023

Bài 1

a) 5x²y - 20xy²

= 5xy(x - 4y)

b) 1 - 8x + 16x² - y²

= (1 - 8x + 16x²) - y²

= (1 - 4x)² - y²

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

c) 4x - 4 - x²

= -(x² - 4x + 4)

= -(x - 2)²

d) x³ - 2x² + x - xy²

= x(x² - 2x + 1 - y²)

= x[(x² - 2x+ 1) - y²]

= x[(x - 1)² - y²]

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

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

e) 27 - 3x²

= 3(9 - x²)

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

f) 2x² + 4x + 2 - 2y²

= 2(x² + 2x + 1 - y²)

= 2[(x² + 2x + 1) - y²]

= 2[(x + 1)² - y²]

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

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

17 tháng 12 2023

Bài 2:

a: \(x^2\left(x-2023\right)+x-2023=0\)

=>\(\left(x-2023\right)\left(x^2+1\right)=0\)

mà \(x^2+1>=1>0\forall x\)

nên x-2023=0

=>x=2023

b: 

ĐKXĐ: x<>0

\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)

=>\(-x\left(x-4\right)+2x^2-4x-9=0\)

=>\(-x^2+4x+2x^2-4x-9=0\)

=>\(x^2-9=0\)

=>(x-3)(x+3)=0

=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

c: \(x^2+2x-3x-6=0\)

=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)

=>\(x\left(x+2\right)-3\left(x+2\right)=0\)

=>(x+2)(x-3)=0

=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

d: 3x(x-10)-2x+20=0

=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)

=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)

=>\(\left(x-10\right)\left(3x-2\right)=0\)

=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)

Câu 1:

a: \(5x^2y-20xy^2\)

\(=5xy\cdot x-5xy\cdot4y\)

\(=5xy\left(x-4y\right)\)

b: \(1-8x+16x^2-y^2\)

\(=\left(16x^2-8x+1\right)-y^2\)

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

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

c: \(4x-4-x^2\)

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

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

d: \(x^3-2x^2+x-xy^2\)

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

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

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

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

e: \(27-3x^2\)

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

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

f: \(2x^2+4x+2-2y^2\)

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

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

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

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

19 tháng 7 2018

a) 5xy ( x - y ) - 2x + 2y

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

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

b) 6x-2y-x(y-3x)

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

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

c)  x+ 4x - xy-4y

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

( x + 4 ) ( x - y )

d) 3xy + 2z - 6y - xz 

= ( 3xy - 6y ) + ( 2z - xz )

= 3y ( x - 2 ) + z ( x - 2 )

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

19 tháng 7 2018

a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)

b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)

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

d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)

11)

a,4-9x^2=0

(2-3x)(2+3x)=0

2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3

b,x^2 +x+1/4=0

(x+1/2)^2 =0

x+1/2=0

x=-1/2

c,2x(x-3)+(x-3)=0

(x-3)(2x+1)=0

x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2

d,3x(x-4)-x+4=0

3x(x-4)-(x-4)=0

(x-4)(3x-1)=0

x-4=0=>x=4 hoặc 3x-1=0=>x=1/3

e,x^3-1/9x=0

x(x^2-1/9)=0

x(x+1/3)(x-1/3)=0

x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3

f,(3x-y)^2-(x-y)^2 =0

(3x-y-x+y)(3x-y+x-y)=0

2x(4x-2y)=0

4x(2x-y)=0

x=0hoặc 2x-y=0=>x=y/2

22 tháng 10 2023

2:

a: \(x^2-12x+20\)

\(=x^2-2x-10x+20\)

=x(x-2)-10(x-2)

=(x-2)(x-10)

b: \(2x^2-x-15\)

=2x^2-6x+5x-15

=2x(x-3)+5(x-3)

=(x-3)(2x+5)

c: \(x^3-x^2+x-1\)

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

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

d: \(2x^3-5x-6\)

\(=2x^3-4x^2+4x^2-8x+3x-6\)

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

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

e: \(4y^4+1\)

\(=4y^4+4y^2+1-4y^2\)

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

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

f; \(x^7+x^5+x^3\)

\(=x^3\left(x^4+x^2+1\right)\)

\(=x^3\left(x^4+2x^2+1-x^2\right)\)

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

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

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

\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-3\left(x^2+x\right)+6\)

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

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

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

h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)

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

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

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

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

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

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

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

i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)

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

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

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

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

j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)

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

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

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

10 tháng 12 2021

\(a,=xy\left(x+2y+1\right)\\ b,=x^2\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x-2\right)\left(x+2\right)\\ c,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ d,=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2=\left(x-2\right)\left(x+2+x-2\right)=2x\left(x-2\right)\\ e,=\left(x+1\right)^2-y^2=\left(x+y+1\right)\left(x-y+1\right)\\ g,=\left(x+9-6x\right)\left(x+9+6x\right)=\left(9-5x\right)\left(7x+9\right)\\ h,=\left(x-y\right)^2-\left(z-t\right)^2=\left(x-y-z+t\right)\left(x-y+z-t\right)\\ i,=\left(x-1\right)^3-y^3=\left(x-y-1\right)\left(x^2-2x+1+xy+y+y^2\right)\)

10 tháng 12 2021

c: =(x-5)(x+3)

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

20 tháng 8 2021

a) \(3xy-6xy^2=3xy\left(1-2y\right)\)

b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)

c) \(x^3-x^2+2\)

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

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

f) \(x^2+2x+1-9y^2=\left(x+1\right)^2-\left(3y\right)^2=\left(x-3y+1\right)\left(x+3y+1\right)\)

g) \(6x^2-12x=6x\left(x-2\right)\)

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

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

20 tháng 8 2021

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

l) \(x^3-7x^2+9x+3x^2-21x+27=x\left(x^2-7x+9\right)+3\left(x^2-7x+9\right)=\left(x+3\right)\left(x^2-7x+9\right)\)

1 tháng 9 2023

a) \(3x^2-3xy-5x+5y\)

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

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

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

b) \(2x^3y-2xy^3-4xy^2-2xy\)

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

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

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

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

c) \(x^2+1+2x-y^2\)

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

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

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

d) \(x^2+4x-2xy-4y+y^2\)

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

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

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

e) \(x^3-2x^2+x\)

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

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

f) \(2x^2+4x+2-2y^2\)

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

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

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

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

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

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

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

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

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

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

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

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

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

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

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