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

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

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

Đây nè bạn.

a) Ta có: \(M=x^2-2xy+y^2-10x+10y\)

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

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

`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)]

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

Bài 2: 

a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)

b: \(5x^2+5xy-x-y\)

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

\(=\left(x+y\right)\left(5x-1\right)\)

c:\(-6x^2+7x-2\)

\(=-6x^2+3x+4x-2\)

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

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

1.

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

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

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

c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)

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

2.

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

b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)

c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)

3.

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

c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)

4.

a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

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

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

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

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

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

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

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

Kết quả: 27

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

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

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

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