\(a,\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\\ \Rightarrow\left\{{}\begin{matrix}x=45\\y=60\\z=84\end{matrix}\right.\\ b,\Rightarrow\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{z+y-3}{z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\\ \Rightarrow\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\z+y-3=2z\end{matrix}\right.\Rightarrow2x-1=y+z\)
Lại có \(\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\Rightarrow x+2x-1=\dfrac{1}{2}\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}+z+2=2y\\y+z+1=\dfrac{1}{2}\cdot2=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2y-z=\dfrac{5}{2}\\y+z=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=\dfrac{5}{6}\\z=-\dfrac{5}{6}\end{matrix}\right.\)
\(c,\Rightarrow\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{21}=\dfrac{5x+y-2z}{50+6-42}=\dfrac{28}{14}=2\\ \Rightarrow\left\{{}\begin{matrix}x=20\\y=12\\z=42\end{matrix}\right.\\ d,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3};\dfrac{x}{5}=\dfrac{z}{7}\\ \Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{14}=\dfrac{x-y+z}{10-15+14}=\dfrac{32}{9}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{320}{9}\\y=\dfrac{160}{3}\\z=\dfrac{448}{9}\end{matrix}\right.\)
