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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) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a)
\(\left(y^3+8\right)+\left(y^2-4\right)\)
\(=\left(y+2\right)\left(y^2-2y+4\right)+\left(y-2\right)\left(y+2\right)\)
\(=\left(y+2\right)\left(y^2-2y+4+y+2\right)\)
\(=\left(y+2\right)\left(y^2+y+6\right)\)
b)
\(x^6-1=\)
\(=\left(x^3+1\right)\left(x^3-1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)\left(x-1\right)\left(x^2+x+1\right)\)
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 )
x6+3x4y2-8x3y3+3x2y4+y6= x6+3x4y2+3x2y4+y6-8x3y3=(x2+y2)3-(2xy)3
= (x2+y2-2xy)[(x2+y2)2+2xy(x2+y2)+(2xy)2]= (x-y)2(x4+6x2y2+y4+2x3y+2xy3)
(x2+y2-5)2-4x2y2-16xy-16=(x2+y2-5)2-(4x2y2+16xy+16)=(x2+y2-5)2-(2xy+4)2
=(x2+y2-5+2xy+4)(x2+y2-5-2xy-4)=(x2+2xy+y2-1)(x2-2xy+y2-9)=[(x+y)2-1][(x-y)2-32]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)
x4+324=x4+36x2+324-36x2=(x2+18)2-(6x)2=(x2+18-6x)(x2+18+6x)
\(A=xy+4\)
Bạn hội con bò gì đó ơi cho mk tham gia đc không vì là hội học hành nên .....