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\(a\text{)}\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y+z-x\right)\left[\left(x+y+z\right)^2+x\left(x+y+z\right)+x^2\right]-\left(y^3+z^3\right)\)
\(=\left(y+z\right)\left(3x^2+y^2+z^2+3xy+3xz+2yz\right)-\left(y+z\right)\left(y^2-yz+z^2\right)\)
\(=\left(y+z\right)\left(3x^2+y^2+z^2+3xy+3xz+2yz-y^2+yz-z^2\right)\)
\(=\left(y+z\right)\left(3x^2+3xy+3yz+3xz\right)\)
\(=3\left(y+z\right)\left(x^2+xy+yz+xz\right)\)
\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)
\(b\text{)}x^4+2012x^2+2011x+2012\)
\(=\left(x^4-x\right)+\left(2012x^2+2012x+2012\right)\)
\(=x\left(x^3-1\right)+2012\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2012\left(x^2+x+1\right)\)
\(=\left(x^2-x\right)\left(x^2+x+1\right)+2012\left(x^2+x+1\right)\)
\(=\left(x^2-x+2012\right)\left(x^2+x+1\right)\)
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)
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) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)^3+z^3+3.\left(x+y\right).z.\left(x+y+z\right)\right]-x^3-y^3-z^3\)
\(=\left[x^3+y^3+3xy.\left(x+y\right)+z^3+3\left(x+y\right).z.\left(x+y+z\right)\right]-x^3-y^3-z^3\)
\(=3xy\left(x+y\right)+3\left(x+y\right)z.\left(x+y+z\right)\)
\(=3.\left(x+y\right)\left(xy+zx+zy+z^2\right)\)
\(=3.\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
b) \(x^4+2012x^2+2011x+2012\)
\(=x^4-x+2012x^2+2012x+2012\)
\(=x.\left(x^3-1\right)+2012.\left(x^2+x+1\right)\)
\(=x.\left(x-1\right)\left(x^2+x+1\right)+2012.\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2012\right)\)