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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 ;)
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1) \(x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
\(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
2)\(5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(a+5\right)\)
\(a^3-a^2x-ay+xy=a^2\left(a-x\right)-y\left(a-x\right)=\left(a-x\right)\left(a^2-y\right)\)
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1)\(=x^2\left(x-y\right)-y\left(x-y\right)\\ =\left(x-y\right)\left(x^2-y\right)\)
2)\(=\left(x^2+x\right)-\left(2xy+2y\right)\\ =x\left(x+1\right)-2y\left(x+1\right)\\ =\left(x+1\right)\left(x-2y\right)\)
3)\(=\left(x^2+2.x.2y+4y^2\right)-y^2\\ =\left(x+2y\right)^2-y^2\\ =\left(x+2y+y\right)\left(x+2y-y\right)\)
1) x3 - x2y - xy + y2
= (x3 - x2y) - (xy - y2)
= x2.(x - y) - y.(x - y)
= (x - y).(x2 - y)
2) x2 - 2xy + x - 2y
= (x2 + x) - (2xy + 2y)
= x.(x + 1) - 2y.(x + 1)
= (x + 1).(x - 2y)
3) x2 + 4xy + 3y2
= x2 + 3xy + xy + 3y2
= (x2 + 3xy) + (xy + 3y2)
= x.(x + 3y) + y.(x + 3y)
= (x + 3y).(x + y)
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1 ) \(x^3-x^2y-xy+y^2\)
\(=\left(x^3-x^2y\right)-\left(xy-y^2\right)\)
\(=x^2.\left(x-y\right)-y.\left(x-y\right)\)
\(=\left(x-y\right).\left(x^2-y\right)\)
2 ) \(x^2-2xy+x-2y\)
\(=\left(x^2+x\right)-\left(2xy+2y\right)\)
\(=x.\left(x+1\right)-2y.\left(x+1\right)\)
\(=\left(x+1\right).\left(x-2y\right)\)
3 ) \(x^2+4xy+3y^2\)
\(=x^2+3xy+xy+3y^2\)
\(=\left(x^2+3xy\right)+\left(xy+3y^2\right)\)
\(=x.\left(x+3y\right)+y.\left(x+3y\right)\)
\(=\left(x+3y\right).\left(x+y\right)\)
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Bạn hội con bò gì đó ơi cho mk tham gia đc không vì là hội học hành nên .....
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1) Ta có: 2xy - x2 - y2 + 16
= -(x2 - 2xy + y2 - 16)
= -[(x - y)2 - 16]
= -(x - y - 4)(x - y + 4)
2) x3 + 2x2y + xy2 - 9x
= x(x2 + 2xy + y2 - 9)
= x[(x + y)2 - 9]
= x(x + y - 3)(x + y + 3)
3) x4 - 2x2 = x2(x2 - 2)
1. 2xy-x2-y2+16= -(x2-2xy+y2-16) = -(x2-2xy+y2)-16 = -(x-y)2-16= (x+y)2-42= (x+y-4).(x+y+4)
2. x3+2x2y+xy2-9x= (có sai đề không vậy?)