K
Khách

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

14 tháng 8 2019

a)\(x^2y-x^3-9y+9y\)

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

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

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

\(b,9x^2-1=\left(3x+1\right)\left(3x-1\right)\)

\(c,\left(x-y\right)4-4=\left(x-y-1\right)4\)

14 tháng 8 2019

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

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

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

\(3,\left(x-y\right)4-4=4\left(x-y-1\right)\)

\(4,\)\(9\left(x-y\right)^2=3^2\left(x-y\right)^2=\left(3x-3y\right)^2\)

\(5,3x^2-6ab+3b^2-12c^2???\)

\(6,x^2-25+y^2+2xy=\left(x+y\right)^2-25\)

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

15 tháng 8 2019

\(x^3+1-x^2-x\)

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

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

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

15 tháng 8 2019

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

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

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

25 tháng 6 2018

a ) \(3a^2-6ab+3b^2-12c^2\)

\(=3\left(a^2-2ab+b^2-4c^2\right)\)

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

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

b ) \(x^2-25+y^2+2xy\)

\(=\left(x^2+2xy+y^2\right)-25\)

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

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

c )

\(x^2y-x^3-9y+9x\)

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

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

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

d )\(x^2\left(x-1\right)+16\left(1-x\right)\)

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

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

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

14 tháng 8 2019

 TL:

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

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

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

14 tháng 8 2019

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

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

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

28 tháng 9 2019

a) (3x-1)(9x2+3x+1)=27x3-1

27x3-1=27x3-1

27x3-1-(27x3-1)=0

27x3-1-27x3+1=0

⇒x=0

b)(x2-5x+25)(x+5)=x3+125

(x+5)(x2-x.5+52)=x3+125

x3+125-(x3+125)=0

x3+125-x3-125=0

⇒x=0

c)(x-3)(x2-6x+9)=(x-3)3

x3-33-(x-3)3=0

x3-27-x3+27=0

⇒x=0

28 tháng 9 2019

d) Đề phải là thế này chứ \(\left(x-y+4\right).\left(x-y-4\right)\)

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

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

\(=x^2-2.x.y+y^2-16\)

\(=x^2-2xy+y^2-16.\)

Chúc bạn học tốt!

14 tháng 10 2017

a. 6x3y2 ( 2-x) + 9x2y2 (x-2)

= -6x3y2 (x-2) + 9x2y( x-2)

= (x-2) 3x2y2 ( -2x + 3)

b. x2 - 4x + 4y - y

= x- y2 - (4x - 4y )

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

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

c. 81x+ 6yz -9y2-z2 

= 81x - (9y2 - 6yz + z2 )

= (9x)2 - ( 3y - z )2

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

14 tháng 10 2017

\(a,=6x^3y^2\left(2-x\right)-9x^2y^2\left(2-x\right)\)

\(=\left(2-x\right)\left(6x^3y^2-9x^2y^2\right)=\left(2-x\right)3x^2y^2\left(2x-3\right)\)

\(f,=\left(x-3-x-2\right)\left(x-3+x+2\right)\)

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

\(g,=\left(x-3\right)\left(x+3\right)+2\left(x+3\right)\)

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

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

25 tháng 5 2017

a)\(81x^2-6yz-9y^2-z^2\)

\(=81x^2-\left(z-3y\right)^2\)

\(=\left(9x-z+3y\right)\left(9x+z-3y\right)\)

b)\(x^2y-x^3-9y+9x\)

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

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

c)\(3a^2-6ab+3b^2-12c^2\)

\(=3\left(a^2-2ab+b^2-4z^2\right)\)

\(=3\left[\left(a-b\right)^2-4z^2\right]\)

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

26 tháng 5 2017

a)\(81x^2-6yz-9y^2-z^2=\left(9x\right)^2-\left(9y^2+6yz+z^2\right)=\left(9x\right)^2-\left(3y+z\right)^2=\left(9x-3y-z\right)\left(9x+3y+z\right)\)b)\(x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(x^2-9\right)\left(y-x\right)=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)

c)\(3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)

9 tháng 12 2017

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

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

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

b) \(\left(6x^5y^2-9x^4y^3+15x^3y^4\right):3x^3y^2\)

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

những câu khác tương tự

14 tháng 8 2019

\(x^4-y^4\)

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

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

14 tháng 8 2019

\(125x^3-1\)

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

\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)

4 tháng 12 2018

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

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

c. \(\left(x^2+9\right)^2-36x^2=\left(x^2+6x+9\right)\left(x^2-6x+9\right)=\left(x+3\right)^2\left(x-3\right)^2\)

d. \(25-x^2+2xy-y^2=25-\left(x-y\right)^2=\left(5+x-y\right)\left(5-x+y\right)\)

còn lại làm tương tự

4 tháng 12 2018

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

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

c) \(\left(x^2+9\right)^2-36x^2=\left(x^2+9\right)^2-\left(6x\right)^2=\left(x^2-6x+9\right)\left(x^2+6x+9\right)\)

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

d) \(25-x^2+2xy-y^2=5^2-\left(x-y\right)^2=\left(5-x+y\right)\left(5+x-y\right)\)

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

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

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

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

g) \(2x^2-9x+10=2x^2-4x-5x+10=2x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(2x-5\right)\)

h) \(x^2-5x-14=x^2-7x+2x-14=x\left(x-7\right)+2\left(x-7\right)=\left(x-7\right)\left(x+2\right)\)

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

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

k) \(x^4+4=\left(x^2\right)^2+2\cdot x^2\cdot2+2^2-2\cdot x^2\cdot2\)

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