Ta có: \(\left\{{}\begin{matrix}x+y=\dfrac{1}{2}\left(1\right)\\y+z=\dfrac{1}{3}\left(2\right)\\z+x=\dfrac{1}{4}\left(3\right)\end{matrix}\right.\)
Cộng (1); (2); (3) vế theo vế ta được:
\(2\left(x+y+z\right)=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\)
=> \(2\left(x+y+z\right)=\dfrac{13}{12}\)
=> \(x+y+z=\dfrac{13}{24}\)
+) Mà \(x+y=\dfrac{1}{2}\) => \(z=\dfrac{13}{24}-\dfrac{1}{2}\) = \(\dfrac{1}{24}\)
+) Mà y + z = \(\dfrac{1}{3}\) => \(\left\{{}\begin{matrix}y=\dfrac{1}{3}-\dfrac{1}{24}\\x=\dfrac{13}{24}-\dfrac{1}{3}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=\dfrac{7}{24}\\x=\dfrac{5}{24}\end{matrix}\right.\) (TM)
Vậy \(x=\dfrac{5}{24};y=\dfrac{7}{24};z=\dfrac{1}{24}\)
P/s: Bài này có nhiều cách giải lắm!