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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
a) \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
b) sửa đề nhé!
\(6x-9-x^2=-\left(x^2-6x+9\right)\)
\(=-\left(x-3\right)^2\)
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 )
a) \([(x-y)3 + (y-z)3]+ (z-x)3\)=\(\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^3\)
\(=\left(x-z\right)\left[\left(\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(x-z\right)^2\right)\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-y-y+z\right)+\left(y-z-x+z\right)\left(y-z+x-z\right)\right]=\left(x-z\right)\left[\left(x-2y+z\right)\left(x+z\right)-\left(x-y\right)\left(x+y-2z\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(x-2y+z-x-y+2z\right)=\left(x-z\right)\left(x-y\right)\left(z-y\right)3\)
b) \(=y^2\left(x^2y-x^3+z^3-z^2y\right)-z^2x^2\left(z-x\right)=y^2\left[-y\left(z^2-x^2\right)-\left(z^3-x^3\right)\right]-z^2x^2\left(z-x\right)\)
\(=y^2\left(z-x\right)\left(-yz-xy-z^2-zx-x^2\right)-z^2x^2\left(z-x\right)=\left(z-x\right)\left(-y^3z-xy^2-z^2y^2-xyz-x^2y^2-z^2x^2\right)\)
đến đây coi như là thành nhân tử rồi nha. em muốn gọn thì ráng ngồi nghĩ rồi tách nha. chỉ cần nhóm mấy cái có ngoặc giống nhau là đc. k khó đâu. chịu khó nghĩ để rèn luyện nha
c) \(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-x^4=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(\left(9a^3-6a^2\right)+\left(6a^2-4a\right)+\left(-9a+6\right)=3a^2\left(3a-2\right)+2a\left(3a-2\right)-3\left(3a-2\right)=\left(3a-2\right)\left(3a^2+2a-3\right)\)
d) em sửa đề đi. đề sai rồi. đồng nhất hệ số phải có dấu bằng nha.
có gì liên hệ chị. đúng nha ;)
a)\(9x^2y^3-3x^4y^2-6x^3y^2+18xy^4=3xy^2\left(3xy-x^3-2x^2+6y^2\right)\)
b)\(a^3x^2y^2-\frac{5}{2}a^3x^4+\frac{3}{2}a^4x^2y=a^3x^2\left(y^2-\frac{5}{2}x^2+\frac{3}{2}ay\right)\)
c)\(x^2+4xy-21y^2=\left(x^2+4xy+4y^2\right)-25y^2=\left(x+2y\right)^2-\left(5y\right)^2=\left(x+2y-5y\right)\left(x+2y+5y\right)=\left(x-3y\right)\left(x+7y\right)\)d)\(2x^4+4=2\left(x^4+4\right)=2\left(x^4+4x^2+4-4x^2\right)=2\left[\left(x^2+2\right)^2-\left(2x\right)^2\right]=2\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
a) 5x3y - 30x2y + 45xy
= 5xy.(x2 - 6x + 9)
= 5xy.(x2 - 2.3.x + 32)
= 5xy.(x-3)2
b) x3 - 2x2 - x+2
= x2.(x-2) - (x-2)
= (x-2).(x2-1)
= (x-2).(x-1).(x+1)
c) 3x.(x-y) - (y-x)2
= 3x.(x-y) - (x-y)2
= (x-y).( 3x - x + y)
= (x-y).( 2x+y)
`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)]