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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 ;)
phân tích thành nhân tử: x(x-y)3-y(y-x)2-y2(x-y)
x(x-y)3-y(y-x)2-y2(x-y)
= (y-x)(y^2-2y-3x)
nah bạn chúc bạn học tốt nha
\(x^2+y^2-x^2y^2+xy-x-y.\)
\(=\left(x^2-x^2y^2\right)+\left(y^2-y\right)+\left(xy-x\right)\)
\(=x^2\left(1-y^2\right)+y\left(y-1\right)+x\left(y-1\right)\)
\(=x^2\left(1-y\right)\left(1+y\right)-y\left(1-y\right)-x\left(1-y\right)\)
\(=\left(1-y\right)\left(x^2\left(1+y\right)-y-x\right)\)
\(=\left(1-y\right)\left(x^2+x^2y-y-x\right)\)
\(=\left(1-y\right)\left[\left(x^2-x\right)+\left(x^2y-y\right)\right]\)
\(=\left(1-y\right)\left[x\left(x-1\right)+y\left(x^2-1\right)\right]\)
\(=\left(1-y\right)\left[x\left(x-1\right)+y\left(x-1\right)\left(x+1\right)\right]\)
\(=\left(1-y\right)\left[\left(x-1\right)\left(x+y\left(x-1\right)\right)\right]\)
\(=\left(1-y\right)\left(x-1\right)\left(x+yx+y\right)\)
1)
\(=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]\)
\(=\left(x+y-x+y\right)\left(x+y+x-y\right)\)
\(=2y\cdot2x=4xy\)
1)
\(=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]\)
\(=\left(x+y-x+y\right)\left(x+y+x-y\right)\)
\(=2y\cdot2x=4xy\)
2)
\(=\left[\left(3x+1\right)-\left(x+1\right)\right]\left[\left(3x+1\right)+\left(x+1\right)\right]\)
\(=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)
\(=2x\left(4x+2\right)\)
\(=8x^2+4x\)
a)\(4x^2-y^2+4x+1=\left(4x^2+4x+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2\)
\(=\left(2x+1-y\right)\left(2x+1+y\right)\)
b)\(x^3-x+y^3-y=\left(x^3+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
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)
\(x^2+y^2-x^2y^2+xy-x-y\)
\(=x^2-x^2y^2+y^2-y+xy-x\)
\(=x^2\left(1-y^2\right)+y\left(y-1\right)+x\left(y-1\right)\)
\(=x^2\left(1-y\right)\left(y+1\right)+y\left(y-1\right)+x\left(y-1\right)\)
\(=\left(y-1\right)\left[-x^2\left(y+1\right)+y-x\right]\)
\(=\left(y-1\right)\left[-x^2y-x^2+y-x\right]\)
\(x^2-x-y^2-y\\=(x^2-y^2)-(x+y)\\=(x-y)(x+y)-(x+y)\\=(x+y)(x-y-1)\)