Đặt \(\dfrac{x+y}{3}=\dfrac{5-z}{1}=\dfrac{y+z}{2}=\dfrac{9+y}{5}=k\)
Suy ra : x = 3k - y ; z = 5 - k ; y = 5k - 9
=\(\dfrac{3k-5k+9+5k-9}{3}=\dfrac{5-\left(5-k\right)}{1}=\dfrac{5k-9+5-k}{2}=\dfrac{9+5k-9}{5}\)
= \(\dfrac{3k}{3}=\dfrac{k}{1}=\dfrac{4k-4}{2}=\dfrac{5k}{k}\)
= \(k=k=\dfrac{4\left(k-1\right)}{2}=k\)
=\(k=k=2\left(k-1\right)=k\)
= \(k=2k-1\)
Thiếu đề bài .