\(\left\{{}\begin{matrix}x+y+z=6\left(1\right)\\xy+yz-zx=7\\x^2+y^2+z^2=14\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}\left(x+y+z\right)^2=36\\xy+yz-xz=7\\x^2+y^2+z^2=14\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x^2+y^2+z^2+2\left(xy+yz+xz\right)=36\\xy+yz-xz=7\\x^2+y^2+z^2=14\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}14+2\left(xy+yz+xz\right)=36\\xy+yz-xz=7\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}xy+yz+xz=11\\xy+yz-xz=7\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}xy+yz=\frac{11+7}{2}=9\\xz=\frac{11-7}{2}=2\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y\left(x+z\right)=9\\x=\frac{2}{z}\end{matrix}\right.\)
=>\(y\left(\frac{2}{z}+z\right)=9\)
<=> \(y=\frac{9}{\frac{2}{z}+z}=\frac{9}{\frac{2+z^2}{z}}=\frac{9z}{2+z^2}\)
Thay \(x=\frac{2}{z},y=\frac{9z}{2+z^2}\) vào (1) có:
\(\frac{2}{z}+\frac{9z}{2+z^2}+z=6\)
<=> \(\frac{2\left(2+z^2\right)+9z^2+z^2\left(2+z^2\right)}{z\left(2+z^2\right)}=6\)
<=>\(4+2z^2+9z^2+2z^2+z^4=6z\left(2+z^2\right)\)
<=> \(z^4+13z^2+4-12z-6z^3=0\)
<=> \(z^4-3z^3+2z^2-3z^3+9z^2-6z+2z^2-6z+4=0\)
<=>\(z^2\left(z^2-3z+2\right)-3z\left(z^2-3z+2\right)+2\left(z^2-3z+2\right)=0\)
<=> \(\left(z^2-3z+2\right)^2=0\)
<=> \(z^2-3z+2=0\)<=> \(z\left(z-2\right)-\left(z-2\right)=0\)
<=> \(\left(z-1\right)\left(z-2\right)=0\)
=>\(\left[{}\begin{matrix}z=1\\z=2\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\frac{2}{z}=2,y=\frac{9z}{2+z^2}=3\\x=1,y=3\end{matrix}\right.\)
Vậy (x,y,z) \(\in\left\{\left(2,3,1\right),\left(1,3,2\right)\right\}\)
