\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2=xy+yz+zx\end{cases}}\)
Với \(x+y+z=0\)
\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\z+x=-y\end{cases}}\)
\(\Rightarrow\frac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{xyz}{\left(-z\right).\left(-x\right).\left(-y\right)}=-1\)
Với \(x^2+y^2+z^2=xy+yz+zx\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\)
\(\Rightarrow\frac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{x^3}{8x^3}=\frac{1}{8}\)
Trả lời :.................................
\(\frac{1}{8}................................................\)
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
