NT
3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TH
4 tháng 11 2016
a , 3x2 + 3y2 - 6xy - 12
= 3 ( x2 + y2 - 2xy - 4 )
= 3 ( x - y )2 - 22
= 3 ( x - y + 2 ) ( x - y - 2 )
26 tháng 7 2018
\(e,x^2-y^2+2x+1=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
\(f,x^3+2x^2+2x+1=\left(x^3+1\right)+\left(2x^2+2x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+2x\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
30 tháng 9 2018
\(x^2-y^2+2x+1\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x-y+1\right)\left(x+y+1\right)\)
hk tốt
^^
NC
26 tháng 8 2020
Bài làm:
1) Ta có: \(2x^2+5xy+2y^2\)
\(=\left(2x^2+4xy\right)+\left(xy+2y^2\right)\)
\(=2x\left(x+2y\right)+y\left(x+2y\right)\)
\(=\left(2x+y\right)\left(x+2y\right)\)
2) Ta có: \(2x^2+2xy-4y^2\)
\(=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)\)
\(=2x\left(x-y\right)+4y\left(x-y\right)\)
\(=2\left(x+2y\right)\left(x-y\right)\)
G
26 tháng 8 2020
\(1)2x^2+5xy+2y^2=2x^2+4xy+xy+2y^2=\left(2x^2+4xy\right)+\left(xy+2y^2\right)=2x\left(x+2y\right)+y\left(x+2y\right)=\left(2x+y\right)\left(x+2y\right)\)\(2)2x^2+2xy-4y^2=2x^2+4xy-2xy-4y^2=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)=2x\left(x-y\right)+4y\left(x-y\right)=\left(2x+4y\right)\left(x-y\right)\)
15 tháng 10 2016
a) \(x^4-2x^3+2x-1\)
\(=x^4-x^3-x^3+2x-2+1\)
\(=\left(x^4-x^3\right)+\left(2x-2\right)-\left(x^3-1\right)\)
\(=x^3\left(x-1\right)+2\left(x-1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x^3+2-x^2-x-1\right)\)
\(=\left(x-1\right)\left(x^3-x^2-x+1\right)\)
\(=\left(x-1\right)\left[\left(x^3-x^2\right)-\left(x-1\right)\right]\)
\(=\left(x-1\right)\left[x^2\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(x^2-1\right)\left(x-1\right)\)
\(=\left(x-1\right)^2\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^3\left(x+1\right)\)
b) \(x^4+2x^3+2x^2+2x+1\)
\(=\left(x^4+2x^2+1\right)+\left(2x^3+2x\right)\)
\(=\left(x^2+1\right)^2+2x\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+1+2x\right)\)
\(=\left(x^2+1\right)\left(x+1\right)^2\)
KT
3
6 tháng 12 2019
x2 + 1 - y2 - 2x
= x2 - 2x + 1 - y2
=[x2 - 2x + 1] - y2
=[x-1]2 - y2
=[x-1-y][x-1+y]
CC
7 tháng 12 2019
a) \(x^2+1-y^2-2x=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)
b) \(64x^4+y^4=\left(8x^2\right)^2+\left(y^2\right)^2=\left(8x^2\right)^2+16x^2y^2+\left(y^2\right)^2-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)
LV
1
25 tháng 10 2017
Ta có : \(x^4+x^3+2x^2+x+1\)
\(=x^4+x^3+x^2+x^2+x+1\)
\(=x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2+1\right)\)
HN
25 tháng 8 2021
\(B=x^8+2x^5-2x^4+x^2-2x-100+10x\left(x^4+x\right)+\left(5x-1\right)^2\)
\(=x^8+2x^5-2x^4+x^2-2x-100+10x^5+25x^2-10x+1\)
\(=x^8+12x^5-2x^4+36x^2-12x-99\)
\(=x^8+6x^5+9x^4+6x^5+36x^2+54x-11x^4-66x-99\)
\(=x^4\left(x^4+6x+9\right)+6x\left(x^4+6x+9\right)-11\left(x^4+6x+9\right)\)
\(=\left(x^4+6x+9\right)\left(x^4+6x-11\right)\)
8 tháng 8 2018
\(x^3+2x^2+2x+1=\left(x^3+1\right)+\left(2x^2+2x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+2x\right)=\left(x+1\right)\left(x^2+x+1\right)\)
\(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
\(x^4+2x^3+2x^2+2x+1=x^4+x^2+2x^3+x^2+2x+1\)
\(=x^2\left(x^2+1\right)+2x\left(x^2+1\right)+\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+2x+1\right)\)
\(=\left(x^2+1\right)\left(x+1\right)^2\)
\(x^4-2x^3+2x-1=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1-2x\right)=\left(x^2-1\right)\left(x-1\right)^2\)
8 tháng 8 2018
\(x^3+2x^2+2x+1=\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\)
\(=x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2+x+1\right)\)
\(x^3-4x^2+12x-27\)
\(=\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(9x-27\right)\)
\(=x^2.\left(x-1\right)-3x.\left(x-1\right)+9.\left(x-3\right)\)
\(=\left(x-1\right).\left(x^2-3x\right)+9.\left(x-3\right)\)
\(=x.\left(x-1\right).\left(x-3\right)+9.\left(x-3\right)\)
\(=\left(x-3\right)\left[x.\left(x-1\right)+9\right]\)
=(x4-1)+(x4-x2)
=(x2-1)(x2+1)+x2(x2-1)
=(x2-1)(2x2+1)
=(x-1)(x+1)(2x2+1)
= (x4-1) + (x4-x2)
= (x2-1)(x2+1)+x2(x2-1)
= (x2-1)(2x2+1)
= (x-1)(x+1)(2x2+1)
Mình cững đồng ý với bạn
Như trên