K
Khách

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

14 tháng 4 2020

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

14 tháng 4 2020

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

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

31 tháng 8 2021

\(4\left(x^2y^2+z^2t^2+2xyzt\right)-\left(x^2+y^2-z^2-t^2\right)^2\)

\(=\left[2\left(xy+zt\right)\right]^2-\left(x^2+y^2-z^2-t^2\right)^2\)

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

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

Ta có: \(4\left(x^2y^2+2xyzt+z^2t^2\right)-\left(x^2+y^2-z^2-t^2\right)^2\)

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

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

\(=\left[-\left(x^2-2xy+y^2\right)+\left(z^2+2tz+t^2\right)\right]\left[\left(x^2+2xy+y^2\right)-\left(t^2-2tz+z^2\right)\right]\)

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

12 tháng 9 2021

\(4(x^2y^2+z^2t^2+2xyzt)-(x^2+y^2-z^2-t^2)^2\)

\(=[2(xy+zt]^2-(x^2+y^2-z^2-t^2)^2\)

\(=(2xy+2zt)^2-(x^2+y^2-z^2-t^2)^2\)

\(=(2xy+2zt-x^2-y^2+z^2+t^2)(2xy+2zt+x^2+y^2-z^2-t^2)^2\)

19 tháng 8 2015

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

b/ \(=y^4+\left(-2x^2-34\right)y^2+32xy+x^4-34x^2+225\)

câu này bn coi lại đc k , mk k lm ra 

4 tháng 9 2021

undefined

Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)

\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)

\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)

 

29 tháng 10 2019

a) \(x^2-5xy+6y^2\)

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

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

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

b) \(16\left(x-1\right)^2-36y^2\)

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

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

c) \(4\left(x+y\right)-12\left(x+y\right)^2\)

\(=\left(x+y\right)\left[4-12\left(x+y\right)\right]\)

\(=4\left(x+y\right)\left[1-3x-3y\right]\)

18 tháng 5 2016

đúng ,mình k 2 nhé

7 tháng 9 2020


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

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

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

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

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

\(c,x^3+y^3+z^3-3xyz\)

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

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

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

\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y\right)-3xyz\)

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

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

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

7 tháng 9 2020

a) 4(2 - x)2 + xy - 2y = 4(x - 2)2 + y(x - 2) = (4x - 8 + y)(x - 2)

b) 2(x - 1)3 - 5(x - 1)2 - (x - 1) = (x - 1)[2(x - 1)2 - 5(x - 1) - 1]

= (x - 1)(2x2 - 4x + 2 - 5x + 5 - 1) = (x - 1)(2x2 - 9x + 6)

c) x3 + y3 + z3 - 3xyz = (x + y)(x2 - xy + y2) + z3 - 3xyz

= (x + y)3 + z3 - 3xy(x + y) - 3xyz = (x + y + z)(x2 + 2xy + y2 - xz - yz + z2) - 3xy(x + y + z)

= (x + y + z)(x2 + y2 + z2 - xz - yz + 2xy - 3xy) = (x + y + z)(x2 + y2 + z2 - xy - yz - xz)

1 tháng 9 2021

\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\\ =\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)

1 tháng 9 2021

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

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

\(=(x+3)(x+5)(x-1)(x+1)\)