-(x²+3xy²+3x²y+y²)

=-(x+y)³

\(x^2-3yx^2-3x^2y-y^2\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Đây nè bạn.

a) Ta có: \(M=x^2-2xy+y^2-10x+10y\)

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

\(=9^2-10\cdot9=-9\)

Ta có

N   =   x 3   +   3 x 2 y   +   3 x y 2   +   y 3   +   x 2   +   2 x y   +   y 2     =   ( x 3   +   3 x 2 y   +   3 x y 2   +   y 3 )   +   ( x 2   +   2 x y   +   y 2 )     =   ( x   +   y ) 3   +   ( x   +   y ) 2   =   ( x   +   y ) 2 ( x   +   y   +   1 )

Từ đề bài x = 10 – y ó x + y = 10. Thay x + y = 10 vào N = ( x   +   y ) 2 (x + y + 1) ta được

N = 10 2 (10 + 1) = 1100

Suy ra N > 1000 khi x = 10 – y

Đáp án cần chọn là: D

`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) \(\left(\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5\right):\dfrac{3}{5}ax^3\)

\(=\dfrac{\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5}{\dfrac{3}{5}ax^3}\)

\(=\dfrac{\dfrac{3}{5}ax^3\left(a^5+\dfrac{5}{7}a^2x-\dfrac{3}{2}x^2\right)}{\dfrac{3}{5}ax^3}\)

\(=a^5+\dfrac{5}{7}a^2x-\dfrac{3}{2}x^2\)

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

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

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

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

c) \(\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)

\(=\dfrac{6x^2-xy}{x}+\dfrac{2x^3y+3xy^2}{xy}-x\left(2x-1\right)\)

\(=\dfrac{x\left(6x-y\right)}{x}+\dfrac{xy\left(2x^2+3y\right)}{xy}-2x^2+x\)

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

\(=7x+2y\)

d) \(\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^2\right):\dfrac{3}{2}x^2y^3\)

\(=\dfrac{x^2-xy}{x}+\dfrac{6x^2y^5-9x^3y^4+15x^4y^2}{\dfrac{3}{2}x^2y^3}\)

\(=\dfrac{x\left(x-y\right)}{x}+\dfrac{\dfrac{3}{2}x^2y^2\left(4y^3-6xy^2+10x^2\right)}{\dfrac{3}{2}x^2y^3}\)

\(=x-y+\dfrac{4y^3-6xy^2+10x^2}{y}\)

Cái này y hệt cái đề mik thi:)

trời dài thế làm lâu phết đó nha hừm làm theo đúng công thức là được :)

b) 24x^2+6x^2y−2x−12yx−3y^2x+y 

tôi làm theo cách tìm tích số

nếu thấy đúng thì tick cho tôi nha