ta có:\(\dfrac{x+y+1}{z}+1=\dfrac{x+z+2}{y}+1=\dfrac{y+z-3}{z}+1=\dfrac{1}{x+y+z}+1\Rightarrow\dfrac{x+y+z+1}{z}=\dfrac{x+y+z+2}{y}=\dfrac{x+y+z-3}{z}=\dfrac{1}{x+y+z}+1\)
\(\Rightarrow\dfrac{3\times\left(x+y+z\right)}{x+y+z}=\dfrac{1}{x+y+z}+1\Rightarrow3=\dfrac{1}{x+y+z}+1\Rightarrow\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\)mà: \(\left\{{}\begin{matrix}\dfrac{x+y+z+1}{x}=3\\\dfrac{x+y+z+2}{y}=3\\\dfrac{x+y+z-3}{z}=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=-\dfrac{5}{6}\end{matrix}\right.\)
vậy........
Bạn ơi ,bài này ở trong sách'' nâng cao và phát triển toán 7'' trang 20 (VD 10)
