\(\left\{{}\begin{matrix}3xy=2\left(x+y\right)\\4yz=3\left(y+z\right)\\5zx=6\left(z+x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{3}{2}\\\dfrac{y+z}{yz}=\dfrac{4}{3}\\\dfrac{z+x}{zx}=\dfrac{5}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}+\dfrac{1}{x}=\dfrac{3}{2}\\\dfrac{1}{z}+\dfrac{1}{y}=\dfrac{4}{3}\\\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{5}{6}\end{matrix}\right.\)
Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b;\dfrac{1}{z}=c\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=\dfrac{3}{2}\\b+c=\dfrac{4}{3}\\a+c=\dfrac{5}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\\c=\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\\z=3\end{matrix}\right.\)
Vậy . . .
