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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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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)
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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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Ta có: \(A=\left(x^3+x^2\right)-\left(4x^2-4\right)\)
\(\Leftrightarrow A=x^2.\left(x+1\right)-4.\left(x-1\right).\left(x+1\right)\)
\(\Leftrightarrow A=\left(x+1\right).\left[x^2-4.\left(x-1\right)\right]\)
\(\Leftrightarrow A=\left(x+1\right).\left(x^2-4x+4\right)\)
\(\Leftrightarrow A=\left(x+1\right).\left(x-2\right)^2\)
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a, x4 - 5x2 + 4
= x4 - 4x2 - x2 + 4
= x2 . (x2 - 4) - (x2 - 4)
= (x2 - 4) . (x2 - 1)
= (x - 2) . (x + 2) . (x - 1) . (x + 1)
a)x4-5x2+4=x4-x2-4x2+4
=(x4-x2)-(4x2-4)
=x2(x2-1)-4(x2-1)
=(x2-1)(x2-4)
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\(x^2-y^2+4x+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(4x^2-y^2+8\left(y-2\right)\)
\(=4x^2-\left(y^2-8y+16\right)\)
\(=4x^2-\left(y-4\right)^2\)
\(=\left(2x+y-4\right)\left(2x-y+4\right)\)
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b: \(=x\left(x^4-y^4\right)+y\left(x^4-y^4\right)\)
\(=\left(x+y\right)\left(x^4-y^4\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x^2+y^2\right)\left(x+y\right)^2\cdot\left(x-y\right)\)
\(x^3y^3+x^2y^2+4=x^2y^2\left(xy+2\right)-xy\left(xy+2\right)+xy+2\)
\(=\left(xy+2\right)\left(x^2y^2-xy+1\right)\)
Đừng làm tắt bạn ơi