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a) \(=\left(x-5\right)\left(2+x+5-2x-1\right)=\left(x-5\right)\left(6-x\right)\)
e) \(=\left(ab^3c^2-a^2b^2c^2\right)+\left(ab^2c^3-a^2bc^3\right)=ab^2c^2\left(b-a\right)+abc^3\left(b-a\right)=abc^2\left(b-a\right)\left(b+c\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 ;)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)\)
\(3,7x-6x^2-2=-6x^2+7x-2=-6x^2+3x+4x-2=3x\left(-2x+1\right)+2\left(2x-1\right)\)
\(=3x\left(1-2x\right)-2\left(1-2x\right)=\left(1-2x\right)\left(3x-2\right)\)
\(2,5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2y+y^2x+x^2z+z^2x+y^2z+z^2y+2xyz\)..
\(=\left(x^2y+z^2y+2xyz\right)+\left(y^2x+y^2z\right)+\left(z^2x+x^2z\right)\).
\(=y\left(x+z\right)^2+y^2\left(x+z\right)+xz\left(x+z\right)\)
\(=\left(xy+yz\right)\left(x+z\right)+\left(x+z\right)\left(y^2+xz\right)\).
\(=\left(x+z\right)\left(xy+yz+y^2+xz\right)\).
\(=\left(x+z\right)\left[x\left(y+z\right)+y\left(y+z\right)\right]\).
\(=\left(x+z\right)\left(x+y\right)\left(y+z\right)\).
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)\(=x^2y+xy^2+y^2z+yz^2+x^2z+xz^2+2xyz\)
\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xz^2+xyz\right)+\left(y^2z+yz^2\right)\)
\(=xy\left(x+y+z\right)+xz\left(x+z+y\right)+yz\left(y+z\right)\)
\(=x\left(x+y+z\right)\left(y+z\right)+yz\left(y+z\right)\)
\(=\left(y+z\right)\left(x^2+xy+xz+yz\right)\)
\(=\left(y+z\right)\left[x\left(x+z\right)+y\left(x+z\right)\right]\)
\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
b, \(2x^2+2y^2-x^2z+z-y^2z-2\)
\(=\left(2x^2-x^2z\right)+\left(2y^2-y^2z\right)-\left(2-z\right)\)
\(=x^2\left(2-z\right)+y^2\left(2-z\right)-\left(2-z\right)\)
\(=\left(2-z\right)\left(x^2+y^2-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2y+xy^2+x^2z+y^2z+2xyz=z\left(x^2+2xy+y^2\right)+xy\left(x+y\right)=z\left(x+y\right)^2+xy\left(x+y\right)=\left(x+y\right)\left[z\left(x+y\right)+xy\right]=\left(x+y\right)\left(zx+zy+xy\right)\)
a, 5xy3-2xyz-15y2+6z
=5xy3-15y2+6z-2xyz
=5y2(xy-3)-2z(xy-3)
=(xy-3)(5y2-2z)
b, x2(y-z)+y2(z-x)+z2(x-y)
=x2y-x2z+y2z-y2x+z2(x-y)
=x2y-y2x-x2z+y2z+z2(x-y)
=xy(x-y)-z(x2-y2)+z2(x-y)
=xy(x-y)-z(x-y)(x+y)+z2(x-y)
=(x-y)[xy-z(x+y)+z2]
=(x-y)(xy-zx-zy+z2)
=(x-y)(xy-zx-zy+z2)
=(x-y)[x(y-z)-z(y-z)]
=(x-y)(y-z)(x-z)