1) \(\frac{x}{3}\)= \(\frac{y}{4}\); \(\frac{y}{5}\) =\(\frac{z}{7}\) và 2x + 3y -z
Ta có:\(\frac{x}{15}\) = \(\frac{y}{20}\); \(\frac{y}{20}\) = \(\frac{z}{28}\)
Theo tính chất dãy tỉ số bằng nhau :
\(\frac{x}{15}\) = \(\frac{y}{20}\) = \(\frac{z}{28}\) = \(\frac{2x}{30}\)= \(\frac{3y}{60}\) = \(\frac{2x+3y-z}{30+60-28}\) = \(\frac{124}{62}\) = 2
\(\Rightarrow\)\(\hept{\begin{cases}\frac{x}{15}=2\\\frac{y}{20}=2\\\frac{z}{28}=2\end{cases}}\) \(\Leftrightarrow\)\(\hept{\begin{cases}x=30\\y=40\\z=54\end{cases}}\)
Vậy ( x;y;z) = (30;40;54)