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`@` `\text {Ans}`

`\downarrow`

`a,`

`3x^2 + 6xy + 3y^2 - 3z`

`= 3*x^2 + 3*2xy + 3y^2 - 3z`

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

`b,`

`x^3 + x^2y - x^2z - xyz`

`= x(x + y)(x-z)`

14 tháng 11 2021

a) \(3x^4y-12x^2y^3=3x^2y\left(x^2-\left(2y\right)^2\right)=3x^2y\left(x+2y\right)\left(x-2y\right)\)

b) Sửa đề: \(x^2-y^2-8x+16=\left(x-4\right)^2-y^2=\left(x-4-y\right)\left(x-4+y\right)\)

c) \(x^3+3x^2+4x+12=x^2\left(x+3\right)+4\left(x+3\right)=\left(x^2+4\right)\left(x+3\right)\)

d) \(3x^2-6xy+3y^2-27=3\left(x^2-2xy+y^2-9\right)=3\left(\left(x-y^2\right)-3^2\right)=3\left(x-y-3\right)\left(x-y+3\right)\) 

NV
14 tháng 11 2021

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

\(x^2-y^2-8y-16=x^2-\left(y^2+8y+16\right)=x^2-\left(y+4\right)^2=\left(x+y+4\right)\left(x-y-4\right)\)

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

\(3x^2-6xy+3y^2-27=3\left[\left(x-y\right)^2-9\right]=3\left(x-y-3\right)\left(x-y+3\right)\)

26 tháng 10 2021

a: \(=x\left(x-3\right)-4y\left(x-3\right)\)

=(x-3)(x-4y)

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(x+2)

26 tháng 10 2021

\(a,=x\left(x-3\right)-4y\left(x-3\right)=\left(x-4y\right)\left(x-3\right)\\ b,=\left(x-1\right)\left(x^2+x+1\right)-4x\left(x-1\right)=\left(x-1\right)\left(x^2-3x+1\right)\\ c,=\left(x-y\right)\left(1-a\right)\\ d,=\left(x-2\right)\left(x-2+x+2\right)=2x\left(x-2\right)\\ e,=x^2\left(x+y\right)-xz\left(x+y\right)=x\left(x-z\right)\left(x+y\right)\\ f,=\left(x-y-2\right)\left(x+y\right)\)

13 tháng 10 2021

a/ x+ 4x - 21= x2 - 3x +4x - 21

                       = (x2+4x)-(3x+21)

                       = x(x+4)- 3(x+7)

                       = (x-3).(x+7)

 b/ 3x2-6xy+3y2-3z= 3(x2- 2xy+y2- z2)

                                  = 3[(x2 + 2xy + y2) – z2]

                                  = 3[(x + y)2 – z2]

                                  = 3(x + y – z)(x + y + z)

  c/ 2x2y + 12xy + 18y = 2y(x2+6x+9)

26 tháng 11 2023

a: \(70a+84b-20ab-24b^2\)

\(=\left(70a+84b\right)-\left(20ab+24b^2\right)\)

\(=14\left(5a+6b\right)-4b\left(5a+6b\right)\)

\(=\left(5a+6b\right)\left(14-4b\right)\)

\(=2\left(7-2b\right)\left(5a+6b\right)\)

b: \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)

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

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

\(=x^2\left(y+z\right)+x\left(y^2+z^2\right)+yz\left(y+z\right)+2xyz+xyz\)

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

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

\(=\left(y+z\right)\cdot x\left(x+y+z\right)+yz\left(y+z+x\right)\)

\(=\left(y+z+x\right)\cdot\left(xy+xz+yz\right)\)

c: \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)

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

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

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

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

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

9 tháng 9 2019

3x2 + 6xy + 3y2 – 3z2

= 3.(x2 + 2xy + y2 – z2)

(Nhận thấy xuất hiện x2 + 2xy + y2 là hằng đẳng thức nên ta nhóm với nhau)

= 3[(x2 + 2xy + y2) – z2]

= 3[(x + y)2 – z2]

= 3(x + y – z)(x + y + z)

19 tháng 8 2021

a) \(3x^2-6xy+3y^2-12x^2=3\left(x^2-2xy+y^2\right)-12x^2=3\left(x-y\right)^2-12x^2=3\left[\left(x-y\right)^2-4x^2\right]=3\left(x-y-2x\right)\left(x-y+2x\right)=3\left(-x-y\right)\left(3x-y\right)\)

b)\(3x^2y^2-6x^2y^3+12x^2y^2=3x^2y^2\left(1-2y+4\right)=3x^2y^2\left(5-2y\right)\)

c) \(3x^2-3y^2+12x-12y=3\left(x^2-y^2\right)+12\left(x-y\right)=3\left(x-y\right)\left(x+y+4\right)\)

a: \(3x^2-6xy+3y^2-12x^2\)

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

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

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

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

b: \(3x^2y^2-6x^2y^3+12x^2y^2\)

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

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

c: Ta có: \(3x^2-3y^2+12x-12y\)

\(=3\left(x-y\right)\left(x+y\right)+12\left(x-y\right)\)

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