Dat  (x-y)²+(y-z)²+(x-z)²=A

=(x+y)³+z³-3x²y-3xy²-3xyz / A

=(x+y+z).(x²+2xy+y²-xy-yz+z²)-3xy(x+y+z) / A

=(x+y+z).(x²+y²+z²-xy-yz-xz) /A

=2(x+y+z).(x²+y²+z²-xy-yz-xz) /2A 

=(x+y+z)[ (x²-2xy+y²)+(y²-2yz+z²)+(x²-2xz+z²) / 2A

=(x+y+z).[ (x-y}²+(y-z)²+(x-z)² ] /2A

=(x+y+z). A /2A

=x+y+z /2

kimh thế

nếu mẫu toàn cộng thì còn làm đc

\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left(x+y+z\right)^3-3\left(x+y\right)z\left(x+y+z\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3xy-3yz-3zx\right]\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(=\frac{1}{2}\left(x+y+z\right)\left(x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\right)\)

\(=\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)

Suy ra \(\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{1}{2}\left(x+y+z\right)\)

\(\frac{x^3-y^3+z^3+3xzy}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

\(=\frac{\left(x-y\right)^3+z^3+3x^2y-3xy^2+3xyz}{2x^2+2y^2+2z^2+2xy+2yz-2xz}\)

\(=\frac{\left(x-y+z\right)\left[\left(x-y\right)^2+z^2-\left(x-y\right)z\right]+3xy\left(x-y+z\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\frac{\left(x-y+z\right)\left[x^2+y^2-2xy+z^2-xz+yz+3xy\right]}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\frac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-xz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\frac{x-y+z}{2}\)

x³−y³+z³+3xzy(x+y)²+(y+z)²+(z−x)² 

=(x−y)³+z³+3x²y−3xy²+3xyz2x²+2y²+2z²+2xy+2yz−2xz 

=(x−y+z)[(x−y)²+z²−(x−y)z]+3xy(x−y+z)2(x²+y²+z²+xy+yz−xz) 

=(x−y+z)[x²+y²−2xy+z²−xz+yz+3xy]2(x²+y²+z²+xy+yz−xz) 

=(x−y+z)(x²+y²+z²+xy+yz−xz)2(x²+y²+z²+xy+yz−xz) 

=x−y+z2