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\(x^3-3x^2+3x-1-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)
\(x^3-3x^2+3x-1-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)
Rất vui vì giúp đc bạn <3
\(x^3-3x^2+3x-1-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)
\(x^3-3x^2+3x-1-y^3\\ =\left(x-1\right)^3-y^3\\ =\left(x-1-y\right)\text{[ (x-1)^2+y(x-1)+y^2}\)
\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)
a) \(3^2\left(y-x\right)+6x^2\left(x-y\right)^2\)
\(=3\left(y-x\right)\left[3+2x^2\left(y-x\right)\right]\)
\(=3\left(y-x\right)\left(3+2x^2y-2x^3\right)\)
b) \(x^4-3x^3+3x-1\)
\(=\left(x^4+x^3\right)-\left(4x^3+4x^2\right)+\left(4x^2+4x\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3-4x^2+4x-1\right)\)
\(=\left(x+1\right)\left[\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(x-1\right)\right]\)
\(=\left(x+1\right)\left(x-1\right)\left(x^2-3x+1\right)\)
a)(x+y)2-(x-y)2
=(x+y-x+y)(x+y+x-y)
=2y.2x=4xy
b)(3x+1)2-(x+1)2
=(3x+1-x-1)(3x+1+x+1)
=2x.(4x+2)
=4x(2x+1)
c) x3+y3+z3-3xyz
= (x+y)3- 3xy(x+y) +z3-3xyz
=(x+y+z)( x2+2xy+y2-xz-yz+z2)-3xy(x+y+z)
=(x+y+z)(x2+y2+z2-xy-xz-yz)
Phân tích đa thức sau thành nhân tử :
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
b) \(x^3+y^3+z^3-3xyz\)
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,\(x^3-3x^2+3x-1-y^3=\left(x^3-1\right)-\left(3x^2-3x\right)-y^3\)
\(=\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)-y^3\)
\(=\left(x-1\right)\left(x^2-2x+1\right)-y^3\)
\(=\left(x-1\right)^3-y^3=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
....
=(x-1)3 - y3 = (x-y-1)( (x-1)2 +(x-1)y +y2))
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