K
Khách

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

a) Ta có: \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)\)

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

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

b) Ta có: \(\left(49x^2-81y^2\right):\left(7x+9y\right)\)

\(=\frac{\left(7x+9y\right)\left(7x-9y\right)}{7x+9y}\)

\(=7x-9y\)

c) Ta có: \(\left(x^3+3x^2y+3xy^2+y^3\right):\left(x+y\right)\)

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

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

d) Ta có: \(\left(x^3-3x^2y+3xy^2-y^3\right):\left(x^2-2xy+y^2\right)\)

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

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

e)Sửa đề: \(\left(8x^3+1\right):\left(2x+1\right)\)

Ta có: \(\left(8x^3+1\right):\left(2x+1\right)\)

\(=\frac{\left(2x+1\right)\left(4x^2-2x+1\right)}{2x+1}\)

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

f) Ta có: \(\left(8x^3-1\right):\left(4x^2+2x+1\right)\)

\(=\frac{\left(2x-1\right)\left(4x^2+2x+1\right)}{4x^2+2x+1}\)

\(=2x-1\)

2 tháng 9 2020

a, (x4 + 2x2y2 + y4) : (x2 + y2)

= (x2 + y2)2 : (x2 + y2)

= x2 + y2

b, (49x2 - 81y2) : (7x + 9y)

= (7x - 9y)(7x + 9y) : (7x + 9y)

= 7x - 9y

c, (x3 + 3x2y + 3xy2 + y3) : (x + y)

= (x + y)3 : (x + y)

= (x + y)2

d, (x3 - 3x2y + 3xy2 - y3) : (x2 - 2xy + y2)

= (x - y)3 : (x - y)2

= x - y

Phần e thiếu thì phải

f, (8x3 - 1) : (4x2 + 2x + 1)

= (2x - 1)(4x2 + 2x + 1) : (4x2 + 2x + 1)

= 2x - 1

Chúc bn học tốt!

28 tháng 10 2021

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

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

28 tháng 10 2021

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

11 tháng 8 2021

Đây nè bạn.

undefined

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)]

13 tháng 1 2018

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

15 tháng 10 2021

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

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

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

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

15 tháng 10 2021

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)

15 tháng 6 2023

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

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

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

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

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

 

2 tháng 1 2020

Kết quả: 27

21 tháng 10 2021

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

21 tháng 10 2021

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

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

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