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23 tháng 9 2021

\(a,x+y=1\Leftrightarrow\left(x+y\right)^3=1\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\\ \Leftrightarrow x^3+y^3+3xy\cdot1=1\Leftrightarrow x^3+y^3+3xy=1\)

\(b,x^3-y^3-3xy\\ =x^3-3x^2y+3xy^2-y^3-3xy+3x^2y-3xy^2\\ =\left(x-y\right)^3-3xy\left(x-y-1\right)\\ =1^3-3xy\left(1-1\right)=1-0=1\)

\(c,x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\\ =x^2-xy+y^2+3xy-6x^2y^2+6x^2y^2\\ =x^2+2xy+y^2=\left(x+y\right)^2=1\)

`#3107.101107`

`D = x^3 - y^3 - 3xy` biết `x - y - 1 = 0`

Ta có:

`x - y - 1 = 0`

`=> x - y = 1`

`D = x^3 - y^3 - 3xy`

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

`= 1 * (x^2 + xy + y^2) - 3xy`

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

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

`= x^2 - 2*x*y + y^2`

`= (x - y)^2`

`= 1^2 = 1`

Vậy, với `x - y = 1` thì `D = 1`

________

`E = x^3 + y^3` với `x + y = 5; x^2 + y^2 = 17`

`x + y = 5`

`=> (x + y)^2 = 25`

`=> x^2 + 2xy + y^2 = 25`

`=> 2xy = 25 - (x^2 + y^2)`

`=> 2xy = 25 - 17`

`=> 2xy = 8`

`=> xy = 4`

Ta có:

`E = x^3 + y^3`

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

`= 5 * [ (x^2 + y^2) - xy]`

`= 5 * (17 - 4)`

`= 5 * 13`

`= 65`

Vậy, với `x + y = 5; x^2 + y^2 = 17` thì `E = 65`

________

`F = x^3 - y^3` với `x - y = 4; x^2 + y^2 = 26`

Ta có:

`x - y = 4`

`=> (x - y)^2 = 16`

`=> x^2 - 2xy + y^2 = 16`

`=> (x^2 + y^2) - 2xy = 16`

`=> 2xy = (x^2 + y^2) - 16`

`=> 2xy = 26 - 16`

`=> 2xy = 10`

`=> xy = 5`

Ta có:

`F = x^3 - y^3`

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

`= 4 * [ (x^2 + y^2) + xy]`

`= 4 * (26 + 5)`

`= 4*31`

`= 124`

Vậy, với `x - y = 4; x^2 + y^2 = 26` thì `F = 124.`

5 tháng 6 2021

`a)(x-1)(x^2+x+1)`

`=x^3+x^2+x-x^2-x-1`

`=x^3-1`

`b)(x^3+x^2y+xy^2+y^3)(x-y)`

`=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4`

`=x^4-y^4`

5 tháng 6 2021

a) VT`=(x-1)(x^2+x+1)`

`=x^3 +x^2 +x -x^2-x-1 `

`=x^3-1=` VP.

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

`=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4`

`=x^4-y^4=` VP.

1) 

Ta có: x+y=2

nên \(\left(x+y\right)^2=4\)

\(\Leftrightarrow x^2+y^2+2xy=4\)

\(\Leftrightarrow2xy=2\)

hay xy=1

Ta có: \(x^3+y^3\)

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

\(=2^3-3\cdot1\cdot2\)

=2

2)\(x^2+y^2=\left(x+y\right)^2-2xy=8^2-2\cdot\left(-20\right)=104\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=8^3-3\cdot\left(-20\right)\cdot8=512+480=992\)

\(x^2+y^2+xy=\left(x+y\right)^2-xy=8^2-\left(-20\right)=64+20=84\)

10: \(x\left(x-y\right)+x^2-y^2\)

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

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

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

11: \(x^2-y^2+10x-10y\)

\(=\left(x^2-y^2\right)+\left(10x-10y\right)\)
\(=\left(x-y\right)\left(x+y\right)+10\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y+10\right)\)

12: \(x^2-y^2+20x+20y\)

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

\(=\left(x-y\right)\left(x+y\right)+20\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y+20\right)\)

13: \(4x^2-9y^2-4x-6y\)

\(=\left(4x^2-9y^2\right)-\left(4x+6y\right)\)

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

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

14: \(x^3-y^3+7x^2-7y^2\)

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

\(=\left(x-y\right)\left(x^2+xy+y^2\right)+7\cdot\left(x^2-y^2\right)\)

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

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

15: \(x^3+4x-\left(y^3+4y\right)\)

\(=x^3-y^3+4x-4y\)

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

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

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

16: \(x^3+y^3+2x+2y\)

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

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

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

17: \(x^3-y^3-2x^2y+2xy^2\)

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

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

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

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

18: \(x^3-4x^2+4x-xy^2\)

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

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

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

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

8 tháng 12 2023

Phân tích đa thức thành nhân tử nha

5 tháng 5 2017

a) A = -1;                        b) B = ( x   +   y ) 3  =1.

`a, x^3 + y^3 + x + y`

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

`= (x+y)(x^2-xy+y^2+1)`

`b, x^3 - y^3 + x -y`

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

`= (x-y)(x^2+xy+y^2+1)`

`c, (x-y)^3 + (x+y)^3`

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

`= (2x)(x^2 + 3y^2)`

`d, x^3 - 3x^2y + 3xy^2 - y^3 + y^2 - x^2`

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

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

a: =(x+y)(x^2-xy+y^2)+(x+y)

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

b: =(x-y)(x^2+xy+y^2)+(x-y)

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

c: =x^3-3x^2y+3xy^2-y^3+x^3+3x^2y+3xy^2-y^3

=2x^3+6xy^2

d: =(x-y)^3+(y-x)(y+x)

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

a: \(\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(y-x\right)^2}\)

\(=\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(x-y\right)^2}\)

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

b: \(\dfrac{\left(x-2y\right)^3}{x^2-4xy+4y^2}\)

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

=x-2y

c: \(\dfrac{x^3+y^3}{x+y}\)

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

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