\(2x=3y\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{3}.\dfrac{1}{5}=\dfrac{y}{2}.\dfrac{1}{5}\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}\left(1\right)\)
\(3y=5z\)
\(\Leftrightarrow\dfrac{y}{5}=\dfrac{z}{3}\Rightarrow\dfrac{y}{5}.\dfrac{1}{2}=\dfrac{z}{3}.\dfrac{1}{2}\Rightarrow\dfrac{y}{10}=\dfrac{z}{6}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)
Vì \(\left|x-2y\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-2y=5\\x-2y=-5\end{matrix}\right.\)
* \(TH1:x-2y=5\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{2y}{20}=\dfrac{x-2y}{15-20}=\dfrac{5}{-5}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=-1\\\dfrac{y}{10}=-1\\\dfrac{z}{6}=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-1.15=-15\\y=-1.10=-10\\z=-1.6=-6\end{matrix}\right.\)
\(TH2:\left|x-2y\right|=-5\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{2y}{20}=\dfrac{x-2y}{15-20}=\dfrac{-5}{-5}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=1\\\dfrac{y}{10}=1\\\dfrac{z}{6}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1.15=15\\y=1.10=10\\z=1.6=6\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-15\\y=-10\\z=-6\end{matrix}\right.;\left\{{}\begin{matrix}x=15\\y=10\\z=6\end{matrix}\right.\)
