Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1/a ) = (x+y)3 -(x+y)
= (x+y)[(x+y)2+1]
c) = 5(x2-xy+y2)-20z2
=5(x-y)2-20z2
= 5 [ (x-y)2- 4z2 ]
=5(x-y-4z)(x-y+4z)
Bài 1:
a) x3-x+3x2y+3xy2+y3-y
=x3+2x2y-x2+xy2-xy+x2y+2xy2-xy+y3-y2+x2+2xy-x+y2-y
=x(x2+2xy-x+y2-y)+y(x2+2xy-x+y2-y)+(x2+2xy-x+y2-y)
=(x2+2xy-x+y2-y)(x+y+1)
=[x(x+y-1)+y(x+y-1)](x+y+1)
=(x+y-1)(x+y)(x+y+1)
c) 5x2-10xy+5y2-20z2
=-5(2xy-y2+4z2-2)
Bài 2:
5x(x-1)=x-1
=>5x2-6x+1=0
=>5x2-x-5x+1
=>x(5x-1)-(5x-1)
=>(x-1)(5x-1)=0
=>x=1 hoặc x=1/5
b) 2(x+5)-x2-5x=0
=>2(x+5)-x(x+5)=0
=>(2-x)(x+5)=0
=>x=2 hoặc x=-5
1)
a) (x+y)3-(x+y)= (x+y)(x+y-1)
b) xem lại đề câu B nha bạn
2)
a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc=0
(a+b)3+c3-3ab(a+b+c)=0
(a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c)=0
(a+b+c)(a2+b2+c2-xy-yz-xz)=0
Suy ra: a3+b3+c3=3abc
1. a) = (x+y)3 -(x+y) =(x+y)((x+y)2 -1)
= (x+y)(x+y+1)(x+y-1)
b) = 5(( x-y)2 - 4z2)
= 5( x-y +2z)(x-y-2z)
2. áp dụng ( a+b+c)3 = .....rồi biến đổi
a) x4 + 2x3 + x2 = x2.(x2 + 2x + 1) = x2(x + 1)2
b) x3 - x + 3x2y + 3xy2 + y3 - y = x3 + 3x2y + 3xy2 + y3 - x - y = (x + y)3 - (x + y) = (x + y)[(x + y)2 - 1] = (x + y - 1)(x + y)(x + y + 1)
c) 5x2 - 10xy + 5y2 - 20z2 = 5.(x2 - 2xy + y2 - 4z2) = 5[(x - y)2 - (2z)2] = 5(x - y - 2z)(x - y + 2z)
\(a,x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
\(b,x^3-x+3x^2y+3xy^2+y^3-y=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
\(c,5x^2-10xy+5y^2-20z^2=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left[\left(x-y+2z\right)\left(x-y-2z\right)\right]\)
1)\(x^4+2x^3+x^2\)
=\(\left(x^4+x^3\right)+\left(x^3+x^2\right)\)đật nhân tử chung ra
=\(x^2\left(x+1\right)^2\)
2) pt => \(\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
=\(\left(x+y\right)^3-\left(x+y\right)\)
=\(\left(x+y\right)\left(\left(x+y\right)^2+1\right)\)
3)chia tất cả cho 5 pt => \(x^2-2xy+y^2-4x^2\)
=\(\left(x+y\right)^2-4z^2\)
=\(\left(x+y+2z\right)\left(x+y-2z\right)\)
4)pt => \(2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
=\(2\left(x-y\right)-\left(x-y\right)^2\)
=\(\left(x-y\right)\left(2-x+y\right)\)
k chi nha
a, x4 + 2x3 + x2 = \(x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2=\left[x\left(x+1\right)\right]^2=\)\(\left(x^2+x\right)^2\)
b, x^3 - x + 3x^2y + 3xy^2+y^3-y
x^3 + 3x^2y + 3xy^2+y^3- x - y
(x+y)^3 - (x+y)
=(x+y)[ (x+y)^2 - 1]
=(x+y)(x+y+1)(x+y-1)
c, 5x^2 - 10xy + 5y^2 - 20(c hỗ này có dấu gì ko???) z^2
a, x4 + 2x3 +x2 = x4 +x3 +x3 +x2 =(x4+x3 )+(x3 +x2 ) =x3(x +1 ) + x2 (x+1 ) =(x+1)(x3+x2)
a) x4 + 2x3 + x2
= x2(x2 + 2x + 1)
= x2(x + 1)2
= [x(x + 1)]2
= (x2 + x)2
b) 5x3 - 10xy + 5y2 - 20z2
= 5(x3 - 2xy + y2 - 4z2)
c) x2y - xy2 + x3 - y3
= xy(x - y) + (x - y)(x2 + xy + y2)
= (x - y)(x2 + 2xy + y2)
= (x - y)(x + y)2
d) x2 - xy + 4x - 2y + 4
= (x2 + 4x + 4) - (xy + 2y)
= (x + 2)2 - y(x + 2)
= (x + 2)(x + 2 - y)
d) x2 - x - 6
= x2 - 3x + 2x - 6
= x(x - 3) + 2(x - 3)
= (x + 2)(x - 3)
f) 3x2 - 5x - 8
= 3x2 + 3x - 8x - 8
= 3x(x + 1) - 8(x + 1)
= (3x - 8)(x + 1)
g) x3 + 3x2 + 6x + 4
= (x3 + 3x2 + 3x + 1) + (3x + 3)
= (x + 1)3 + 3(x + 1)
= (x + 1)[(x + 1)2 + 3]
h) 3x3 - 5x2 - 6x + 8
= 3x3 - 3x2 - 2x2 - 6x + 8
= 3x3 - 3x2 - 2x2 + 2x - 8x + 8
= 3x2(x - 1) - 2x(x - 1) - 8(x - 1)
= (3x2 - 2x - 8)(x - 1)
a) \(x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(5x^2-10xy+5y^2-20z^2\) (đã sửa đề)
\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
c) \(x^2y-xy^2+x^3-y^3\)
\(=xy\left(x-y\right)+\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\left(x-y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)^2\)
d) \(x^2-xy+4x-2y+4\)
\(=\left(x^2+4x+4\right)-\left(xy+2y\right)\)
\(=\left(x+2\right)^2-y\left(x+2\right)\)
\(=\left(x+2\right)\left(x-y+2\right)\)
e) \(x^2-x-6=\left(x+2\right)\left(x-3\right)\)
f) \(3x^2-5x-8\)
\(=\left(3x^2+3x\right)-\left(8x+8\right)\)
\(=3x\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-8\right)\)
a) Sửa đề
\(x^4+2x^3+x^2\)
\(=\left(x^4+x^3\right)+\left(x^3+x^2\right)\)
\(=x^3\left(x+1\right)+x^2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x^2\right)\)
\(=\left(x+1\right).x^2\left(x+1\right)\)
\(=x^2\left(x+1\right)^2\)
b) \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
a.\(x^2\left(x^2+2x+1\right)\)
\(x^2\left(x+1\right)^2\)