Question

Hệ phương trình có nghiệm là \(\left\{{}\begin{matrix}x+y+z=9\\x.y+y.z+z.x=27\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=81\\xy+yz+xz=27\\\dfrac{xy+xz+zy}{xyz}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2\left(xy+yz+xz\right)=81\\xy+yz+xz=27\\xyz=27\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+z^2=27\\xy+yz+xz=27\\xyz=27\end{matrix}\right.\Leftrightarrow x^2+y^2+z^2=xy+yz+xz=xyz\)

theo bđt ta có \(x^2+y^2+z^2\ge xy+xz+yz\)

để \(x^2+y^2+z^2=xy+xz+yz\) khi \(x=y=z=3\)