Lời giải:
Từ đề bài ta dễ dàng suy ra \(x,y,z\neq 0\)
Đảo lại ta thu được hệ:
\(\left\{\begin{matrix} \frac{x+y}{xy}=\frac{1}{2}\\ \frac{y+z}{yz}=\frac{1}{4}\\ \frac{x+z}{xz}=\frac{1}{3}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{1}{x}+\frac{1}{y}=\frac{1}{2}(1)\\ \frac{1}{y}+\frac{1}{z}=\frac{1}{4}(2)\\ \frac{1}{x}+\frac{1}{z}=\frac{1}{3}(3)\end{matrix}\right.\)
Lấy \(\frac{(1)+(2)+(3)}{2}\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{\frac{1}{2}+\frac{1}{4}+\frac{1}{3}}{2}=\frac{13}{24}(4)\)
Lấy \((4)-(1)\Rightarrow \frac{1}{z}=\frac{13}{24}-\frac{1}{2}=\frac{1}{24}\Rightarrow z=24\)
Lấy \((4)-(2)\Rightarrow \frac{1}{x}=\frac{13}{24}-\frac{1}{4}=\frac{7}{24}\Rightarrow x=\frac{24}{7}\)
Lấy \((4)-(3)\Rightarrow \frac{1}{y}=\frac{13}{24}-\frac{1}{3}=\frac{5}{24}\Rightarrow y=\frac{24}{5}\)
Vậy \((x,y,z)=(\frac{24}{7}, \frac{24}{5}, 24)\)
