Ta có: 2x = 3y
\(\Rightarrow\dfrac{x}{y}=\dfrac{3}{2}\Rightarrow x=1,5y\) Thay vào \(\left|x-y\right|=5\)
=> \(\left|x-y\right|=\left|1,5y-y\right|=5\)
=> | 0,5y | = 5
TH1: 0,5y = -5 => y = -10 => \(\left\{{}\begin{matrix}x=-15\\z=-37,5\end{matrix}\right.\)
TH2: 0,5y = 5 => y = 10 => \(\left\{{}\begin{matrix}x=15\\z=37,5\end{matrix}\right.\)
Ta có : 2x=3y=5z
=> \(\dfrac{x}{3}=\dfrac{y}{2}\) =>\(\dfrac{x}{15}=\dfrac{y}{10}\)
\(\dfrac{y}{5}=\dfrac{z}{3}\) => \(\dfrac{y}{10}=\dfrac{z}{6}\)
=> \(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)
Ta có : |x-y|=5 => x-y = 5 ; -5
Áp dụng tính chất dãy tỉ số bằng nhau
*với x-y=5
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x-y}{15-10}=\dfrac{5}{5}=1\)
⇒\(\left\{{}\begin{matrix}x=15\\y=10\\z=16\end{matrix}\right.\)
*với x-y=-5
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x-y}{15-10}=\dfrac{-5}{5}=-1\)
⇒\(\left\{{}\begin{matrix}x=-15\\y=-10\\z=-6\end{matrix}\right.\)
Vậy x=10 ; y=15 ; z=6
x=-10 ; y=-15 ; z=-6
