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x3+y(1-3x2)+x(3y2-1)-y3
= x3-3x2y+3xy2-y3+y-x
=(x-y)3 -(x-y)
=(x-y)(x2-2xy+y2-1)
cái chỗ kia giải thích dùm mìh đy : \(x^3-3x^2y+3xy^2-y^3+y-x\)
\(x^3+y\left(1-3x^2\right)+x\left(3y^2-1\right)-y^3\)
\(=x^3-3x^2y+3xy^2-y^3+y-x\)
\(=\left(x-y\right)^3-\left(x-y\right)\)
phân tích đa thức thành nhân tử cơ mà
=(x-y)3-(x-y)
=(x-y)[(x-y)2-1]
a , 3x2 + 3y2 - 6xy - 12
= 3 ( x2 + y2 - 2xy - 4 )
= 3 ( x - y )2 - 22
= 3 ( x - y + 2 ) ( x - y - 2 )
\(3y^3+6xy^2+3x^2y=3y\left(y^2+2xy+x^2\right)=3y\left(x+y\right)^2\)
\(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
\(x^3+3x^2-3x-1=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)=\left(x-1\right)\left(x^2+x+1+3x\right)\)
\(=\left(x-1\right)\left(x^2+4x+1\right)\)
Tham khảo nhé~
\(x^3-3x^2+3x-1-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)
\(x^3-3x^2+3x-1-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)
Rất vui vì giúp đc bạn <3
a) \(x^2yz+4zyx+4yz\)
\(=yz\left(x^2+4x+4\right)\)
\(=yz\left(x+2\right)^2\)
b) \(5x^4-3x^3y-45x^2y^2+27xy^3\)
\(=x\left(5x^3-3x^2y-45xy^2+27y^3\right)\)
\(x^3-3x^2+3x-1-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)
\(x^3-3x^2+3x-1-y^3\\ =\left(x-1\right)^3-y^3\\ =\left(x-1-y\right)\text{[ (x-1)^2+y(x-1)+y^2}\)
\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)
\(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3x^2-3y^2-2\left(x^2-2xy+y^2\right)\)
\(=3x^2-3y^2-2x^2+4xy-2y^2\)
\(=x^2+4xy-5y^2\)
\(=x^2+4xy+4y^2-9y^2\)
\(=\left(x+2y\right)^2-\left(3y\right)^2\)
\(=\left(x+2y-3y\right)\left(x+2y+3y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)