Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^3+3x^2y+3xy^2+y^3-x-y=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
Ta có : \(x^3+3x^2y+3xy^2+y^3-x-y.\)
\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
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
3x^2 +3y^2 -6xy -12
=3(x^2 - 2xy +y^2 - 2^2 )
=3 (x-y)^2 - 2^2
=3(x-y-2)(x-y+2)
3(x+y) -(x^2+2xy+y^2)
=3(x+y) -(x+y)^2
(x+y)(3-x-y)
a) \(\left(xy+1\right)^2-\left(x+y\right)\)
\(=\left(xy+1-x-y\right)\left(xy+1+x+y\right)\)
\(=\left[x\left(y-1\right)-\left(y-1\right)\right]\left[x\left(y+1\right)+\left(y+1\right)\right]\)
\(=\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)
b) \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x-y\right)\left(x+y\right)+\left(x-y\right)^2\right]\)
\(=2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\left(3x^2+y^2\right)\)
a) ( xy + 1 )2 - ( x + y )2
= [ ( xy + 1 ) - ( x + y ) ][ ( xy + 1 ) + ( x + y ) ]
= ( xy - x - y + 1 )( xy + x + y + 1 )
b) ( x + y )3 - ( x - y )3
C1. = x3 + 3x2y + 3xy2 + y3 - ( x3 - 3x2y + 3xy2 - y3 )
= x3 + 3x2y + 3xy2 + y3 - x3 + 3x2y - 3xy2 + y3
= 6x2y + 2y3
= 2y( 3x2 + y2 )
C2. = [ ( x + y ) - ( x - y ) ][ ( x + y )2 + ( x + y )( x - y ) + ( x - y )2 ]
= ( x + y - x + y )( x2 + 2xy + y2 + x2 - y2 + x2 - 2xy + y2 )
= 2y( 3x2 + y2 )
c) 3x4y2 + 3x3y2 + 3xy2 + 3y2
= 3( x4y2 + x3y2 + xy2 + y2 )
= 3[ ( x4y2 + x3y2 ) + ( xy2 + y2 ) ]
= 3[ x3y2( x + 1 ) + y2( x + 1 ) ]
= 3( x + 1 )( x3y2 + y2 )
= 3y2( x + 1 )( x3 + 1 )
= 3y2( x + 1 )( x + 1 )( x2 - x + 1 )
= 3y2( x + 1 )2( x2 - x + 1 )
d) 4( x2 - y2 ) - 8( x - ay ) - 4( a2 - 1 )
= 4[ ( x2 - y2 ) - 2( x - ay ) - ( a2 - 1 )
= 4( x2 - y2 - 2x + 2ay - a2 + 1 )
= 4[ ( x2 - 2x + 1 ) - ( y2 - 2ay + a2 ) ]
= 4[ ( x - 1 )2 - ( y - a )2 ]
= 4[ ( x - 1 ) - ( y - a ) ][ ( x - 1 ) + ( y + a ) ]
= 4( x - y + a - 1 )( x + y + a - 1 )
Bài 1 :
a) xy(x+y)+yz(y+z)+xz(x+z)+2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(y + x)
= (x + y)(xy + zx + zy + z²)
= (x + y)[x(y + z) + z(y + z)]
= (x + y)(y + z)(z + x)
b) \(x^3-x+3x^2y+3xy^2+y^3-x-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
Đã có kết quả
Bài 1,chữa phần a
xy(x+y)+yz(y+z)+xz(x+z)+2xyz
=[xy(x+y)+xyz]+[yz(y+z)+xyz]+xz(x+z)
=xy(x+y+z)+yz(x+y+z)+xz(x+z)
=y(x+y+z)(x+z)+xz(x+z)
=(x+z)(xy+y2+yz+xz)
=(x+z)(x+y)(y+z)
Chữa phần b
x3-x+3x2y+3xy2+y3-y
=(x+y)(x+y-1)(x+y+1)
Bài2
a3+b3+c3=(a+b)3-3ab(a+b)+c3=-c3-3ab(-c)+c3=3abc
Ai làm đúng như này ớ sẽ k
x^3 - x +3x^2y +3xy^2 + y^3 -y
=(x3+3x2y+3xy2+y3)+(-x-y)
=(x+y)3-(x+y)
=(x+y)[(x+y)2-1]
=(x+y)(x+y-1)(x+y+1)
\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(\Leftrightarrow\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^3-\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(\Leftrightarrow\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
\(x^3-x+3x^2y+3xy^2+y^3-y\\ =\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\\ =\left(x+y\right)^3-\left(x+y\right)\\ =\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left[\left(x^2+2xy+y^2\right)-1\right]\)