\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=\frac{2}{xy}-\frac{1}{z^2}\)
\(\Leftrightarrow\left(\frac{1}{x^2}+\frac{2}{zx}+\frac{1}{z^2}\right)+\left(\frac{1}{z^2}+\frac{2}{yz}+\frac{1}{y^2}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\)
Hai số hạng đều không âm nên ta được:
\(\left\{{}\begin{matrix}\frac{1}{x}=-\frac{1}{z}\\\frac{1}{y}=-\frac{1}{z}\end{matrix}\right.\Leftrightarrow x=y=-z\)
Thay vào phương trình đầu:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\Leftrightarrow\frac{1}{x}+\frac{1}{x}-\frac{1}{x}=2\Leftrightarrow\frac{1}{x}=2\Leftrightarrow x=\frac{1}{2}\)
Vậy \(x=y=\frac{1}{2};z=-\frac{1}{2}\)